├── .gitignore ├── Assignment ├── 04-DMST.pdf ├── Assignment1 │ ├── CS216 Assignment1.md │ └── CS216 Assignment1.pdf ├── Assignment2 │ ├── 04-DMST.pdf │ ├── 05-leftist-heaps-markup.pdf │ ├── figure │ │ ├── figure1.png │ │ ├── figure2.png │ │ └── figure3.png │ ├── report.aux │ ├── report.log │ ├── report.out │ ├── report.pdf │ ├── report.synctex.gz │ ├── report.tex │ ├── report.toc │ └── 勘误.txt ├── Assignment3 │ ├── figure │ │ └── 1.png │ ├── report.aux │ ├── report.log │ ├── report.out │ ├── report.pdf │ ├── report.synctex.gz │ ├── report.tex │ └── report.toc ├── Assignment4 │ ├── figure │ │ ├── 1.png │ │ └── 2.png │ ├── report.aux │ ├── report.log │ ├── report.out │ ├── report.pdf │ ├── report.synctex.gz │ ├── report.tex │ └── report.toc ├── Assignment5 │ ├── figure │ │ ├── 1.png │ │ └── 2.png │ ├── report.aux │ ├── report.log │ ├── report.out │ ├── report.pdf │ ├── report.synctex.gz │ ├── report.tex │ └── report.toc ├── Assignments.txt └── temp_report.tex ├── Lab ├── Lab1 │ ├── Lab1-A.pdf │ ├── Lab1-B.pdf │ ├── a.cpp │ └── b.cpp ├── Lab10 │ ├── A.cpp │ ├── B.cpp │ ├── C.cpp │ ├── D.cpp │ ├── Lab10-A.pdf │ ├── Lab10-B.pdf │ ├── Lab10-C.pdf │ └── Lab10-D.pdf ├── Lab2 │ ├── Lab2-A.pdf │ ├── Lab2-B.pdf │ ├── a.cpp │ └── b.cpp ├── Lab3 │ ├── A.cpp │ ├── B.cpp │ ├── Lab3-A.png │ └── Lab3-B.png ├── Lab4 │ ├── A.cpp │ ├── B.cpp │ ├── Lab4-A.pdf │ └── Lab4-B.pdf ├── Lab5 │ ├── A.cpp │ ├── Lab5-A.pdf │ └── Lab5-B.pdf ├── Lab6 │ ├── A.cpp │ ├── B.cpp │ ├── Lab6-A.pdf │ └── Lab6-B.pdf ├── Lab7 │ ├── A.cpp │ ├── B.cpp │ ├── Lab7-A.pdf │ └── Lab7-B.pdf ├── Lab8 │ ├── A.cpp │ ├── B.cpp │ ├── Lab8-A.pdf │ └── Lab8-B.pdf ├── Lab9 │ ├── A.cpp │ ├── B.cpp │ ├── C.cpp │ ├── D.cpp │ ├── Lab9-A.pdf │ ├── Lab9-B.pdf │ ├── Lab9-C.pdf │ └── Lab9-D.pdf └── temp.cpp ├── README.md ├── assignmentscore.png └── labscore.png /.gitignore: -------------------------------------------------------------------------------- 1 | *.exe 2 | Textbook(Chinese).pdf 3 | Textbook(English).pdf 4 | Lecture* -------------------------------------------------------------------------------- /Assignment/04-DMST.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Assignment/04-DMST.pdf -------------------------------------------------------------------------------- /Assignment/Assignment1/CS216 Assignment1.md: -------------------------------------------------------------------------------- 1 | ## CS216 Assignment1 2 | 3 | ###### 12111012 匡亮 4 | 5 | #### 1. Problem 6 | 7 | [Link to sustech space](https://spaces.sustech.cloud/classes/1/assignment/lab1-b) 8 | 9 | #### 2. Algorithm Description 10 | 11 | Initially all $s\in S$ and $c\in C$ are free 12 | 13 | While there is a student $s$ who is free and hasn't applied for all colleges 14 | 15 |         Choose such a student $s$ 16 | 17 |         Let $c$ be the highest-ranked college in $s$'s list which he hasn't applied for 18 | 19 |         If $s_{s,c}>0\land c_{c,s}>0$ then 20 | 21 |                 If $c$ is not full then 22 | 23 |                         Add $s$ to $c$'s student list $student_c$ 24 | 25 |                 Else if $c_{c,s}>\min_{s'\in student_c}\{c_{c,s'}\}$ then 26 | 27 |                         Remove $s'$ from $student_c$ 28 | 29 |                         $s'$ becomes free 30 | 31 |                         Add $s$ to $student_c$ 32 | 33 |                 Else then 34 | 35 |                         $s$ remains free 36 | 37 |                 Endif 38 | 39 |         Endif 40 | 41 | Endwhile 42 | 43 | Return the set of list $student$ 44 | 45 | #### 3. Time Complexity Analysis 46 | 47 | Assume there are $n$ students and $m$ colleges. 48 | 49 | In the worst case, all the students will try to apply for all colleges, and each try cost $O(\log n)$ time as we need to use a priority queue to maintain the worst student of each college. Therefore, the total time complexity is $O(mn\log n)$. 50 | 51 | #### 4. Stability Definition 52 | 53 | The definition below referred to [Lecture01](https://sakai.sustech.edu.cn/access/content/group/7b887991-0c42-450f-8506-9a379cd75e6b/Lecture01-introduction-stable-matching.pdf) P30. 54 | 55 | A matching $S$ in this problem is **unstable** if there exists college $c$ and student $s$ such 56 | that all the following holds: 57 | 58 | + $c_{c,s}>0$ and $s_{s,c}>0$; 59 | 60 | + Either $c$ is not full, or $c$ prefers $s$ to at least one of its enrolled students; 61 | 62 | + Either $s$ is unmatched, or $s$ prefers $c$ to the college he has applied for. 63 | 64 | Also a matching $S$ in this problem is **unstable** if for some $(s,c)\in S$ that $c_{c,s}<0$ or $s_{s,c}<0$. 65 | 66 | #### 5. Correctness Proof 67 | 68 | Obviously we will never match a pair $(s,c)$ that $c_{c,s}<0$ or $s_{s,c}<0$. 69 | 70 | If there exists college $c$ and student $s$ makes $S$ unstable, then there are two possibilities. 71 | 72 | First, $s$ has never applied for $c$. In this case, $s$ is not free at the end, and has been accepted by some other college $c'$, which means $s_{s,c'}>s_{s,c}$ as $s$ applied for $c'$ earlier than $c$. 73 | 74 | Second, $s$ used to be accepted by $c$ but was refused later. In this case, as the student was once kicked out of the list, all the students in the list, including the students joined later, have a higher rank than him. 75 | 76 | The two possibilities above are both contradict with $(s,c)$ makes $S$ unstable. Therefore, $S$ is stable. 77 | 78 | #### 6. Optimality Anlysis 79 | 80 | The anlysis below referred to [Lecture01](https://sakai.sustech.edu.cn/access/content/group/7b887991-0c42-450f-8506-9a379cd75e6b/Lecture01-introduction-stable-matching.pdf) P26-28. 81 | 82 | **Claim.** The matching algorithm in this problem is **student-optimal**. 83 | 84 | **Proof.** Let's prove by contradiction. Suppose $y$ is **the first** student refused by a valid college, and $a$ is **the first** college to do so. 85 | 86 | When $a$ rejected $y$, $a$ has accepted a set of students $student_a$. Since they are never rejected by any valid college, they prefer $a$ to any other valid college. 87 | 88 | Let $S$ be a stable matching in which $a$ accepted $y$, call its student list now $student_{a}'$, then for any $z\in{student_a-student_a'}$, $z$ prefers $y$ to his college $b$, and $y$ prefers $z$ to $a$, therefore $S$ is not stable. 89 | 90 | Therefore, student will never be rejected by a valid college. As students apply for colleges in order of preference, students will always be accepted by their favorite valid college. So the matching algorithm gives us a **student-optimal** matching. 91 | -------------------------------------------------------------------------------- /Assignment/Assignment1/CS216 Assignment1.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Assignment/Assignment1/CS216 Assignment1.pdf -------------------------------------------------------------------------------- /Assignment/Assignment2/04-DMST.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Assignment/Assignment2/04-DMST.pdf -------------------------------------------------------------------------------- /Assignment/Assignment2/05-leftist-heaps-markup.pdf: -------------------------------------------------------------------------------- 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Time\end{large}}} 20 | 21 | \author{匡亮(12111012)} 22 | 23 | \date{March 24, 2023} 24 | 25 | \begin{document} 26 | 27 | \maketitle 28 | 29 | \renewcommand\abstractname{Abstract} 30 | \begin{abstract} 31 | 32 | This is my report of CS216 Assignment2, including: 33 | 34 | \begin{itemize} 35 | \item[1.] What the Directed Minimum Spanning Tree problem is. 36 | \item[2.] How Edmond's Algorithm solves this problem and how Tarjan's implementation improves it to $O(m\log n)$. 37 | \item[3.] A sample code in C++ and detailed time complexity analysis. 38 | \item[4.] A little expansion. 39 | \end{itemize} 40 | 41 | \end{abstract} 42 | 43 | \newpage 44 | 45 | \renewcommand\contentsname{Contents} 46 | \tableofcontents 47 | 48 | \newpage 49 | 50 | \section{Problem Definition} 51 | 52 | Let $G=(V,E,w)$ be a weighted directed graph, where $w:E\to\mathbb{R}$ is the cost function. Let $r\in V$. A directed spanning tree (DST) of G rooted at $r$, is a subgraph $T$ of $G$ such that the undirected version of $T$ is a tree and $T$ contains a directed path from $r$ to any other vertex in $V$. The one with the minimum total cost is called the minimum directed spanning tree (MDST). The problem is to find a MDST on a given graph $G$ and a given root $r$. 53 | 54 | For convenience, in the whole report we assume $|V|=n,|E|=m$ $O(\log m)=O(\log n)$. 55 | 56 | \section{Algorithm} 57 | 58 | \subsection{Important lemma} 59 | 60 | The whole algorithm is based on a subtle lemma. 61 | 62 | \begin{lemma} 63 | For each vertex $v\in V/\{r\}$, let $e_v$ be the entering edge of $v$ with minimum cost, and let $E'$ be the edge set of all $n-1$ different $e_v$, then: 64 | \begin{itemize} 65 | \item [1.] If there is no rings in $E'$, then $E'$ is the MDST. 66 | \item [2.] Otherwise, for each circle $C_i\in E'$, there is a MDST containing $|C_i|-1$ edges in $C_i$. 67 | \end{itemize} 68 | \end{lemma} 69 | 70 | \begin{proof} 71 | The first one is trival. We need an entering edge for each vertex except $r$, and fortunately we independently minimized each of them, then we literally get the MDST. 72 | 73 | For the second one, it is trival for the case $|C|=1$. Otherwise, let the circle $C=v_1\to v_2\to...\to v_k\to v_1$ and W.L.O.G let $v_1$ be one of the closest vertices to $r$ on the circle, i.e. there is not an ancestor of $v_1$ in $C$ (note that the edge $(v_k,v_1)\not\in T$). Assume that $e_{v_{i+1}}=(v_i,v_{i+1})\not\in T, 1\le i 220 | #define For(_, L, R) for(int _ = L; _ <= R; ++_) 221 | using namespace std; 222 | const int MAXN = 400000 + 10; 223 | const long long INF = 1ll << 40; 224 | 225 | typedef pair, pair> Edge; // An edge in the graph 226 | #define W first.first 227 | #define W0 first.second 228 | #define U second.first 229 | #define V second.second 230 | #define EDGE(u, v, w) make_pair(make_pair(w, w), make_pair(u, v)) 231 | const Edge EDGENULL = EDGE(0, 0, 0); 232 | 233 | template 234 | struct Node { // Leftist Tree Node 235 | Node *ls, *rs; 236 | int npl; // Null-Path Length 237 | T val; 238 | long long tag; 239 | Node() { ls = rs = NULL; npl = -1; } 240 | Node(T v): Node() { npl = 0; val = v; tag = 0; } 241 | }; 242 | template 243 | inline void gettag(Node *a, long long gtag) { 244 | if(a != NULL) { 245 | a->tag += gtag; 246 | a->val.W += gtag; 247 | } 248 | } 249 | template 250 | inline void pushdown(Node *a) { 251 | if(a->tag) { 252 | gettag(a->ls, a->tag); 253 | gettag(a->rs, a->tag); 254 | a->tag = 0; 255 | } 256 | } 257 | template 258 | Node* merge(Node *a, Node *b) { 259 | if(a == NULL) return b; 260 | if(b == NULL) return a; 261 | if(a->val > b->val) swap(a, b); 262 | pushdown(a); 263 | a->rs = merge(a->rs, b); 264 | if(a->ls == NULL || a->rs->npl > a->ls->npl) swap(a->ls, a->rs); 265 | a->npl = a->rs == NULL ? 0 : a->rs->npl + 1; 266 | return a; 267 | } 268 | 269 | int n, m, r; 270 | 271 | Node* node[MAXN]; // Leftist Tree root of each super-vertex 272 | Edge in[MAXN]; // In-edge of each node during the contracting process 273 | int parent[MAXN]; // Each (super-)vertex in which super-vertex 274 | vector children[MAXN]; // The children of each super-vertex (empty for original vertex) 275 | inline void addEdge(int u, int v, long long w) { 276 | node[v] = merge(node[v], new Node(EDGE(u, v, w))); 277 | } 278 | 279 | // Use a Union Set to maintain which super-vertex each vertex is in 280 | int f[MAXN]; 281 | int F(int n) { 282 | return n == f[n] ? n : f[n] = F(f[n]); 283 | } 284 | 285 | /* 286 | * To make the code more general, instead of calculate the total value, 287 | * we want to actually construct the DMST, which will not change our 288 | * time complexity but cost a little more time. 289 | */ 290 | Edge tree[MAXN]; // The finally in-edge of each node on DMST 291 | void expand_ring(int nod); 292 | 293 | /* 294 | * expand_node: We found an original vertex, who is the in-node of 295 | * several layers of super-vertices, and the highest layer of which 296 | * is root. Now we want to calculate the total value of these layers, 297 | * i.e. the total value inside the super-vertex root. 298 | */ 299 | void expand_node(int nod, int root) { 300 | while(nod != root) { 301 | expand_ring(nod); 302 | nod = parent[nod]; 303 | } 304 | } 305 | 306 | /* 307 | * expand_ring: We found a (super-)vertex, who is the in-node of 308 | * his parent super-vertex. Now we want to calculate the total 309 | * value of this layer. 310 | */ 311 | void expand_ring(int nod) { 312 | for(auto peer : children[parent[nod]]) 313 | if(peer != nod) { 314 | tree[in[peer].V] = in[peer]; // report an edge 315 | expand_node(in[peer].V, peer); 316 | } 317 | } 318 | 319 | int main() { 320 | /* 321 | * input 322 | * The first input n, m, r representing the number of vertices and edges, 323 | * and the id of the root. The m lines below input u, v, w each line 324 | * represent an edge E(u, v) = w in the graph. 325 | */ 326 | cin >> n >> m >> r; 327 | For(i, 1, m) { 328 | int u, v, w; 329 | cin >> u >> v >> w; 330 | addEdge(u, v, w); 331 | } 332 | For(i, 1, n) addEdge(i, i % n + 1, INF); // Make the graph strongly connected 333 | // initialize 334 | For(i, 1, n) f[i] = i; 335 | // contract 336 | int a = 1; // The (super-)vertex we are considering 337 | int cnt = n; 338 | while(node[a] != NULL) { // While LeftistTree[a] is not empty 339 | // get and delete min 340 | Edge edge = node[a]->val; 341 | pushdown(node[a]); 342 | node[a] = merge(node[a]->ls, node[a]->rs); 343 | int b = F(edge.U); // b is the super-vertex which u is in 344 | if(b != a) { 345 | in[a] = edge; 346 | if(in[b] == EDGENULL) a = b; // append the link 347 | else { 348 | int c = ++cnt; // c is the new super-vertex 349 | f[c] = c; 350 | while(a != c) { // When a == c, we have returned the start point 351 | parent[a] = c; 352 | f[a] = c; 353 | children[c].push_back(a); 354 | gettag(node[a], -in[a].W); 355 | node[c] = merge(node[c], node[a]); 356 | a = F(in[a].U); 357 | } 358 | a = c; 359 | } 360 | } 361 | } 362 | // expand 363 | expand_node(r, cnt); 364 | // now, tree[t] for 1 <= t <= n (except tree[r]) contains all edges in DMST 365 | long long ans = 0; 366 | For(i, 1, n) 367 | if(i != r) ans += tree[i].W0; 368 | if(ans >= INF) cout << -1 << endl; 369 | else cout << ans << endl; 370 | return 0; 371 | } 372 | \end{lstlisting} 373 | 374 | \section{Extensions} 375 | 376 | What if we are not given a root $r$ but required to find an optimal $r$ so that the total cost of our MDST is minimized? We can add a virture super source vertex $s$, add edges $(s,1),(s,2),...,(s,n)$ with weight $INF$, and again make the whole graph strongly connected. Then, use the algorithm above to solve the problem with the new graph and root $s$. If the total cost is greater than $2\times INF$, then there is no legal solution. Otherwise, we ignore the only edge beginning from $s$, and the other edges form the optimal MDST. 377 | 378 | \newpage 379 | 380 | \renewcommand\refname{References} 381 | \begin{thebibliography}{99} 382 | \bibitem[Uri Zwick(2013)]{MDST} Directed Minimum Spanning Trees 383 | 384 | https://www.cs.princeton.edu/courses/archive/spring13/cos528/directed-mst-1.pdf 385 | 386 | \bibitem[Brian Curless(2008)]{LH} Lefist Heaps 387 | 388 | https://courses.cs.washington.edu/courses/cse326/08sp/lectures/markup/05-leftist-heaps-markup.pdf 389 | \end{thebibliography} 390 | 391 | \end{document} 392 | -------------------------------------------------------------------------------- /Assignment/Assignment2/report.toc: -------------------------------------------------------------------------------- 1 | \contentsline {section}{\numberline {1}Problem Definition}{3}{section.1}% 2 | \contentsline {section}{\numberline {2}Algorithm}{3}{section.2}% 3 | \contentsline {subsection}{\numberline {2.1}Important lemma}{3}{subsection.2.1}% 4 | \contentsline {subsection}{\numberline {2.2}Algorithm - Contract part}{4}{subsection.2.2}% 5 | \contentsline {subsection}{\numberline {2.3}Algorithm - Expand part}{6}{subsection.2.3}% 6 | \contentsline {subsubsection}{\numberline {2.3.1}Expand a vertex}{6}{subsubsection.2.3.1}% 7 | \contentsline {subsubsection}{\numberline {2.3.2}Expand a ring}{6}{subsubsection.2.3.2}% 8 | \contentsline {subsection}{\numberline {2.4}Time Complexity Analysis}{7}{subsection.2.4}% 9 | \contentsline {section}{\numberline {3}Data Structures}{8}{section.3}% 10 | \contentsline {subsection}{\numberline {3.1}Lefist Heaps}{8}{subsection.3.1}% 11 | \contentsline {subsection}{\numberline {3.2}Disjoint Set Unions}{9}{subsection.3.2}% 12 | \contentsline {section}{\numberline {4}C++ Code}{10}{section.4}% 13 | \contentsline {section}{\numberline {5}Extensions}{13}{section.5}% 14 | -------------------------------------------------------------------------------- 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A time-domain graph is often more intuitive but complex and hard to modify, so we want a quick method to change a signal between time-domain and frequency-domain. 61 | 62 | By \textbf{fourier expansion}, we know that a period funtion $f(x)=f(x+T)$ can be expand to $f(t)=\sum_{n=-\infty}^{+\infty}c_ne^{in\omega t},c_n=\frac{1}{T}\int_{0}^{T}f(t)e^{-in\omega t}dt$. So, if our signal is the sum of some trigonometric functions with the same base frequency (i.e. their frequency are in the set $\{\omega,2\omega,3\omega,...,n\omega\}$), then, to change between time-domain and frequency-domain equals to transform between the function $f(x)$ and its corresponding sequence $c$. 63 | 64 | Look at the two equations above, it is not hard to find that transform between $f(x)$ and sequence $c$ is DFT and IDFT, therefore we can use FFT to speed up this procedure. So far, we have a quick method to split a signal into several basic waves. In the third section we will introduce more applications of this method. 65 | 66 | \section{Arbitrary-precision multiplication} 67 | 68 | We have learned in the lecture that we can use FFT to convert a polynomial between its coefficient representation and point-value representation, which can speed up multiplication of two polynomials. Then if we let $x=10$ and use a polynomial to represent a large number, we can use FFT to compute the product of two large numbers. However, FFT uses floating point numbers to represent complex numbers in its calculation, which is slow, redundant and imprecise sometimes. So in this section, I will introduce a variant of FFT, called \textbf{NTT (Number-Theoretic Transform)}, which only use integer in its calcualtion. 69 | 70 | Let's first recall why we need complex numbers in FFT. When we divide $A(x)=a_0+a_1x+a_2x^2+a_3x^3...$ into $A_{even}(x)=a_0+a_2x+...$ and $A_{odd}(x)=a_1+a_3x+...$, then calculate $A(x)$ by $A(x)=A_{even}(x^2)+xA_{odd}(x^2),A(-x)=A_{even}(x^2)-xA_{odd}(x^2)$, we want the points we chose are still pairwise negative after we squared them, which is impossible in integer domain, but possible in complex domain. However, if all coefficients in the polynomial are integers, such property is also possible in modulo domain. 71 | 72 | By Fermat's little theorem, we konw that $a^{p-1}\equiv 1\bmod p$ when $p$ is a prime number and $0>} 19 | \providecommand \oddpage@label [2]{} 20 | \@writefile{toc}{\contentsline {section}{\numberline {1}Core idea and critical procedures of Tarjan's trick}{3}{section.1}\protected@file@percent } 21 | \@writefile{toc}{\contentsline {section}{\numberline {2}Running time and space complexity analysis}{3}{section.2}\protected@file@percent } 22 | \@writefile{toc}{\contentsline {section}{\numberline {3}Pseudocode}{4}{section.3}\protected@file@percent } 23 | \@writefile{loa}{\contentsline {algocf}{\numberline {1}{\ignorespaces SPFA Improved by Tarjan's Trick}}{4}{algocf.1}\protected@file@percent } 24 | \@writefile{toc}{\contentsline {section}{\numberline {4}How it speeds up SPFA}{4}{section.4}\protected@file@percent } 25 | \@writefile{lof}{\contentsline {figure}{\numberline {1}{\ignorespaces An example forcing SPFA to run for $\Omega (nm)$ time.}}{5}{figure.1}\protected@file@percent } 26 | \newlabel{1}{{1}{5}{An example forcing SPFA to run for $\Omega (nm)$ time}{figure.1}{}} 27 | \@writefile{lof}{\contentsline {figure}{\numberline {2}{\ignorespaces An explanation for why SPFA runs slowly on this example.}}{5}{figure.2}\protected@file@percent } 28 | \newlabel{2}{{2}{5}{An explanation for why SPFA runs slowly on this example}{figure.2}{}} 29 | \bibcite{textbook}{Jon Kleinberg / Éva Tardos(2005)} 30 | \bibcite{article}{Stefan Lewandowski(2010)} 31 | \gdef \@abspage@last{6} 32 | -------------------------------------------------------------------------------- /Assignment/Assignment4/report.out: -------------------------------------------------------------------------------- 1 | \BOOKMARK 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-------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Assignment/Assignment4/report.pdf -------------------------------------------------------------------------------- /Assignment/Assignment4/report.synctex.gz: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Assignment/Assignment4/report.synctex.gz -------------------------------------------------------------------------------- /Assignment/Assignment4/report.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | 3 | \usepackage{ctex} 4 | \usepackage{graphicx} 5 | \usepackage{float} 6 | \usepackage{datetime} 7 | \usepackage{hyperref} 8 | \usepackage{amssymb} 9 | \usepackage{amsthm} 10 | \newtheorem{theorem}{Theorem} 11 | \newtheorem{corollary}{Colollary} 12 | \newtheorem{lemma}{Lemma} 13 | \newtheorem{definition}{Definition} 14 | \renewcommand\proofname{Proof} 15 | \renewcommand\figurename{Figure} 16 | \usepackage{listings} 17 | \usepackage{xcolor} 18 | \usepackage[linesnumbered, ruled]{algorithm2e} 19 | \usepackage[legalpaper, margin=1.4in]{geometry} 20 | 21 | \title{CS216 Assignment4 \\ {\begin{large} Description of the subtree disassembly trick proposed by Tarjan \end{large}}} %title 22 | 23 | \author{匡亮(12111012)} 24 | 25 | \date{May 1, 2023} % date 26 | 27 | \begin{document} 28 | 29 | \maketitle 30 | 31 | \renewcommand\abstractname{Abstract} 32 | \begin{abstract} 33 | 34 | This is my report of CS216 Assignment4, including a description of the subtree disassembly trick proposed by Tarjan: 35 | 36 | \begin{itemize} 37 | \item[1.] Its core idea and critical procedures. 38 | \item[2.] Running time and space complexity analysis. 39 | \item[3.] Its pseudocode. 40 | \item[4.] How it speeds up SPFA. 41 | \end{itemize} 42 | 43 | \end{abstract} 44 | 45 | \newpage % contents page 46 | 47 | \renewcommand\contentsname{Contents} 48 | \tableofcontents 49 | 50 | \newpage % 1st page here 51 | 52 | \section{Core idea and critical procedures of Tarjan's trick} 53 | 54 | For convenience, in the whole report we assume the terminus vertex $t$ is reachable from every vertex in $G$, and our algorithm will immediately terminate if we find a negative ring (though some vertices cannot reach it, which means they still have a valid non-infinite minimum distance to the terminus vertex). Under these conditions, Tarjan's trick can significantly speed up SPFA without change the algorithm a lot. 55 | 56 | Here I don't repeat what we have learned in our courses (like how SPFA works) and begin from SPFA. 57 | 58 | In SPFA, we maintain an array $first$ for each vertex, and check whether there is a ring after $n$ iterations to determine whether there is a negative ring. However, if we have found a ring during the iterations, we can immediately terminate the algorithm and report there is a negative ring. Therefore, the basic idea of this trick is try to detect whether there is a ring when modifying $first$ with an acceptable extra cost. 59 | 60 | Obviously, before we find a ring, $(v, first[v]),\forall v\not=t$ is a tree with root $t$. When we set $first[v]\leftarrow w$, we can do DFS in $v$'s subtree and check whether $w$ is in $v$'s subtree, if so, we can terminate our algorithm. However, this DFS will cost an extra $O(n)$ time, so we should expect it to do more for us. We can further observe that before the modification, the minimum distance of all vertices in $v$'s subtree (call them $x$) is evaluated based on the minimum distance of $v$, which has been just modified. Therefore, the minimum distance of all of them are out of date. We can set them \textbf{dormant}, then they will not be used to update others' minimum distance before their own minimum distance are updated. At the same time we can remove them from the tree (set $first[x]\leftarrow null$), so that they will not be set dormant by other ancestors again. They will add themselves back to the tree automatically when their minimum distance are updated (because they will update $first[x]$ at that time). 61 | 62 | \section{Running time and space complexity analysis} 63 | 64 | Based on SPFA, we just store one more tree with $O(1)$ space per vertex, so the space complexity is still $O(n)$. 65 | 66 | Add/Remove a vertex to/from the tree can be easily done in $O(1)$ time with linked-list. Adding only happens when $first[v]$ are updated for some $v$, and removing happens no more than adding because vertices will not be removed twice, so the time complexity is still $O(nm)$. 67 | 68 | \newpage 69 | 70 | \section{Pseudocode} 71 | 72 | \begin{algorithm} 73 | \KwData{G, t} 74 | $n=$ number of nodes in $G$\; 75 | $foundNegativeRing=$ False\; 76 | Array $M[V]$\; 77 | Initialize $M[t]=0$ and $M[v]=\infty$ for all other $v\in V$\; 78 | Initialize $first[t]=t$ and $first[v]=null$ for all other $v\in V$\; 79 | Initialize a tree with only root vertex $t$\; 80 | Set all vertices except $t$ dormant in the first iteration\; 81 | \For {$i=1,...,n-1$} { 82 | Set all vertices dormant in next iteration\; 83 | \For {$w\in V$ where $w$ is not dormant in this iteration} { 84 | \For {all edges $(v,w)$} { 85 | $M[v]=\min(M[v],c_{vw}+W[w])$\; 86 | \If {this change the value of $M[v]$} { 87 | $first[v]=w$\; 88 | \For {all vertex $x$ in $v$'s subtree} { 89 | \If {$x$ is equal to $w$} { 90 | $foundNegativeRing=$ True\; 91 | End the algorithm\; 92 | } 93 | Set $x$ dormant in next iteration\; 94 | Remove $x$ from the tree\; 95 | } 96 | Set $v$ not dormant in next iteration\; 97 | Add $v$ to the tree as $w$'s child\; 98 | } 99 | } 100 | } 101 | \If {no value changed in this iteration} { 102 | End the algorithm\; 103 | } 104 | \tcc { If $foundNegativeRing$ is true, then users should ignore the returned $M,first$. } 105 | } 106 | \KwResult{$foundNegativeRing,M,first$} 107 | \caption{SPFA Improved by Tarjan's Trick} 108 | \end{algorithm} 109 | 110 | \section{How it speeds up SPFA} 111 | 112 | If there is a negative ring, the algorithm will terminate in advance, so it is speeded up without a doubt. 113 | 114 | Even if there is no negative rings, this trick can still speed SPFA up. \textbf{Figure 1.} below shows a classic example. 115 | 116 | \begin{figure}[h] % 'h' means insert here 117 | \centering 118 | \includegraphics[height=2cm,width=8cm]{figure/1.png} 119 | \caption{An example forcing SPFA to run for $\Omega(nm)$ time.} 120 | \label{1} 121 | \end{figure} 122 | 123 | In this example, the terminus vertex is on the leftmost, and the orange numbers are weights. If we update all vertices from left to right, we will always first update the vertex on the lower left corner of a triangle, and then the vertex on the upper angle, which will update the vertex on the lower left corner in the next iteration. In short, the vertices on the lower layer have been always misleading in the whole procedure. Actually, in the $a$th round for $a<\frac{n}{2}$, there are $2a-1$ vertices being updated in the last iteration. However, if we use Tarjan's trick, there will always be $O(1)$ vertices need to be operated in each iteration. Thouth the misleading still exists, it will not bring extra cost to us because we can always immediately detect it and set the misled vertices dormant. 124 | 125 | We can think in another way. Recall that SPFA calculates the minimum distance from $v$ to $t$ with no more than $i$ edges for all vertices $v$ in the $i$th iteration. In this example, the $i$th lower layer vertex from right has $i$ different minimum distances, each of them use different numbers of edges, so its value will be changed for $i$ times during the procedure. \textbf{Figure 2.} below shows an example for $i=3$. But if we use Tarjan's trick, we will keep focusing on minimizing the final answer and ignore those useless local answer. 126 | 127 | \begin{figure}[h] % 'h' means insert here 128 | \centering 129 | \includegraphics[height=2cm,width=8cm]{figure/2.png} 130 | \caption{An explanation for why SPFA runs slowly on this example.} 131 | \label{2} 132 | \end{figure} 133 | 134 | \newpage % reference page 135 | 136 | \renewcommand\refname{References} % change '&' in the link to '\&' 137 | \begin{thebibliography}{99} 138 | \bibitem[Jon Kleinberg / Éva Tardos(2005)]{textbook} Algorithm Design (P.304 - 307) 139 | 140 | \bibitem[Stefan Lewandowski(2010)]{article} Shortest Paths and Negative Cycle Detection in Graphs with Negative Weights - I: The Bellman-Ford-Moore Algorithm Revisited 141 | https://d-nb.info/1014960916/34 142 | 143 | \end{thebibliography} 144 | 145 | \end{document} 146 | -------------------------------------------------------------------------------- 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Suppose we have a set $X$ of $n$ boolean variables $X=\{x_1,x_2,...,x_n\}$, each element of which can be $0$ or $1$. Use $\bar{x}$ to identify the negation of $x$. A \textbf{clause} $C$ of length $l$ is a disjunction of $l$ distinct terms $t_1\lor t_2\lor ...\lor t_n$, where $t_i\in\{x_1,x_2,...,x_n,\bar{x_1},\bar{x_2},...,\bar{x_n}\}$. A \textbf{truth assignment} is a function $v:X\to\{0,1\}$. For a collection of clause $C_1,C_2,...,C_k$, we say \textit{$v$ is the satisfying truth assignment} of them, if and only if $v$ causes all of the $C_i$ to evaluate to $1$, which means $C_1\land C_2\land ...\land C_k=1$. The problem that \textit{whether there exists a satisfying truth assignment of the given clauses} is called the \textit{SAT problem}. Specially, if all the clauses have length $l=3$, we call this problem \textit{3-SAT problem}. 50 | 51 | SAT problem is obviously an NP problem. A certificate for a yes-input is a satisfying truth assignment, with which we can calculate the value of $C_1\land ...\land C_k$ in $O(nk)$ time and check whether it equals to $1$. In fact, SAT problem is also an NPC problem but we will not prove here. 52 | 53 | \subsection{3-Color Problem} 54 | 55 | 3-Color problem, similarly, is a special case of Graph Color problem. Suppose we have a graph $G$. A \textit{k-coloring} of $G$ is a function $f:V\to\{1,2,...,k\}$ so that for every edge $(u,v)$ in $G$, we have $f(u)\not=f(v)$. If $G$ has a \textit{k-coloring}, then we will say that it is a \textit{k-colorable graph}. The problem that \textit{whether the given graph is a k-colorable graph} is called the \textit{Graph Color problem}. Specially, we call the problem with $k=3$ \textit{3-Color problem}. 56 | 57 | Graph Color problem is also an NP problem. A certificate for a yes-input is a k-coloring function $f$, with which we can check whether all edges have different colors on its two vertices in $O(|E|)$ time. In this assignment, we will problem that 3-SAT problem $\leq_p$ 3-Color problem, so if we later prove 3-SAT problem is in NPC, then we immediately know 3-Color problem is in NPC. 58 | 59 | \section{Proof} 60 | 61 | To prove this proposition, let's use a \textbf{3-Color problem solver black box} for polynomial number of times and other polynomial number of standard computational steps to solve a 3-SAT problem instance. 62 | 63 | Note that if a graph is 3-colorable, then every triangle (ring of three vertices) in it has three different colors on its three vertices. Therefore, we can build a basic graph model. We first build a triangle with three auxiliary vertices, true node, false node and base node. The base node will automatically take a color different from the true node and false node, so similarly, we then build a triangle with $v_i$, $\bar{v_i}$ and base node for $1\le i\le n$. Therefore, one of them will take the same color with the true node, and the other will take the same color with the false node. 64 | 65 | \begin{figure}[h] 66 | \centering 67 | \includegraphics[height=7cm,width=8cm]{figure/1.png} 68 | \caption{The figure 8.11 from the textbook.} 69 | \label{1} 70 | \end{figure} 71 | 72 | Then, let's build a subgraph for each clause, to ensure that the subgraph is 3-colorable if and only if the clause can be true. The subgraph is showed below for an example $C_i=v_1\lor\bar{v_2}\lor v_3$, and we can easily enumerate the eight cases and find that the subgraph is 3-colorable if and only if at least one of them is true. 73 | 74 | \begin{figure}[h] 75 | \centering 76 | \includegraphics[height=6cm,width=8cm]{figure/2.png} 77 | \caption{The figure 8.12 from the textbook.} 78 | \label{2} 79 | \end{figure} 80 | 81 | Now, we can build the basic graph and the subgraph for all the clauses, and use the black box to give it a 3-color or tell us that there is no solution. If the black box gives us a solution, then based on our former analysis, each vertex will take the same color with either the true node or the false node, and at least one of each clause will take the same color with the true node, so we can define a satisfying truth assignment based on the coloring. On the contrary, if the black box tells us that there is no solution, then the 3-SAT problem is unsatisfiable, because if it is, then we can also based on the satisfying truth assignment to give the graph a 3-coloring. 82 | 83 | Finally, we managed to use a 3-Color problem solver black box and polynomial steps to either solve the 3-SAT problem or prove there is no possible solution, and the black box uses $O(n)$ sized input graph, and only $O(1)$ other steps needed. Therefore, we proved that 3-SAT problem $\leq_p$ 3-Color problem. 84 | 85 | \newpage % reference page 86 | 87 | \renewcommand\refname{References} % change '&' in the link to '\&' 88 | \begin{thebibliography}{99} 89 | \bibitem[Jon Kleinberg / Éva Tardos(2005)]{textbook} Algorithm Design (Chapter 8) 90 | \end{thebibliography} 91 | 92 | \end{document} 93 | -------------------------------------------------------------------------------- /Assignment/Assignment5/report.toc: -------------------------------------------------------------------------------- 1 | \contentsline {section}{\numberline {1}Introduction}{3}{section.1}% 2 | \contentsline {subsection}{\numberline {1.1}3-SAT Problem}{3}{subsection.1.1}% 3 | \contentsline {subsection}{\numberline {1.2}3-Color Problem}{3}{subsection.1.2}% 4 | \contentsline {section}{\numberline {2}Proof}{3}{section.2}% 5 | -------------------------------------------------------------------------------- /Assignment/Assignments.txt: -------------------------------------------------------------------------------- 1 | Assignment 1 2 | 3 | In this assignment, you are required to apply the learned knowledge to analyze the extended Gale-Shapley algorithm used in Lab01B (see Lab01B-sol.pdf in the Resources for the algorithm outline). In particular, try to answer the following questions: 4 | 5 | 1. What is the time complexity? 6 | 7 | 2. How do we define stability for the college/student setting? Note that this setting allows one-to-many matching (e.g., one college can accept many students) and incomplete preference list (e.g., some colleges may be unacceptable by a student). Hint: Please refer to page 30 of Lecture01. 8 | 9 | 3. Prove that the extended GS algorithm indeed outputs a stable matching. 10 | 11 | 4. Does the student-proposing GS algorithm produce a student-optimal matching? 12 | 13 | Please submit your analysis here as a single PDF file. 14 | 15 | Assignment 2 16 | 17 | In this assignment, you need to understand and then describe a complicated algorithm. The goal of this assignment is to practice your writing/presentation skills. (This is crucial for your future success!) 18 | 19 | We learned in class that Chu-Liu's algorithm finds a min-cost arborescence in O(mn) time, but this is not very efficient. The attached lecture notes describe a more efficient algorithm/implementation that runs in O(m log n) time, which is much faster! 20 | 21 | Please read section 5 of the attached file and do the following: 22 | 23 | 1. Before writing, think how the lecture notes present the algorithm, e.g., why 5.1 is needed before 5.2 and 5.3. 24 | 25 | 2. Describe the algorithm implementation in your own words. Please do not copy-paste those from the lecture notes, because then you will learn nothing. Here the goal is to beat the lecture notes and explain the algorithm in a more clear way. 26 | 27 | 3. Please fill out the missing details about the underlying data structures, e.g., how to meld heaps? 28 | 29 | 4. Write your pseudocode or real program for the O(m log n) algorithm and then analyze its time complexity. Please be precise, e.g., specify the time complexity for each crucial step in your code. 30 | 31 | Note: The algorithm may involve several data structures and you can omit the description of those you already learned from your data structure course, e.g., union find set, binary heap, etc. However, more advanced data structures like leftist heaps have to be explained in sufficient details, i.e., imagine the readers only know the basic data structures and your goal is to let them understand. 32 | 33 | PS: Please feel free to describe an even faster algorithm instead, e.g., the O(m + n log n) algorithm proposed by Gabow, Galil, Spencer, and Tarjan in 1986. If you choose to do so, you don't need to describe the attached O(m log n) algorithm anymore, but please follow the same requirements as shown above. 34 | 35 | Please write in English and submit your writing as a single PDF file. Your writing will be graded based on quality, e.g., correctness, completeness, clarity, conciseness, etc. 36 | 37 | Note: There could be mistakes in the attached lecture notes, which could also appear in other online resources you may access when completing a task. You can treat this as part of your challenge. 38 | 39 | Attached file: 04-DMST.pdf 40 | 41 | Assignment 3 42 | 43 | We learned in class that fast Fourier transform (FFT) is designed to efficiently compute the discrete Fourier transform (and its inverse) of a sequence. FFT is a very popular algorithm with a lot of applications. 44 | 45 | In this assignment, you are asked to write a report to describe your favorite FFT applications. The goal of this assignment is to learn practical algorithms by yourself, systemize the learned knowledge, and explain it in a clear way. 46 | 47 | Your report will be graded based on quality, e.g., clarity, scope, succinctness, technicality, etc. We will also consider the importance of your selected FFT applications. Typically, we expect your report to have at least 5 applications and 2~3 of them should be described in sufficient technical details. Note that you could also introduce some variants of the FFT algorithm and discuss their applications. Please write your report in English. 48 | 49 | Hope you enjoy this assignment! 50 | 51 | Assignment 4 52 | 53 | In class we learned Bellman-Ford-Moore algorithm and its efficient implementation: the push-based implementation with O(n) extra space (also known as SPFA). We also learned how one can extend SPFA to detect negative cycles. However, in practice this implementation is still not fast enough. 54 | 55 | In this assignment, you are asked to understand and describe an efficient algorithm/implementation to find shortest paths and detect negative cycles. This algorithm combines Bellman-Ford-Moore with a so-called subtree disassembly trick proposed by Tarjan. Please read pages 304-307 in your textbook and do the following. 56 | 57 | 1. Write the pseudocode of this algorithm and explain its core idea and critical procedures in a clear way. 58 | 59 | 2. Analyze its running time and space complexity. 60 | 61 | 3. Explain in your words why in practice Tarjan's trick gains a considerable speedup compared to SPFA, even for graphs with no negative cycles. Note that the worst-case running time of this improved algorithm is nonetheless unchanged, i.e., O(mn). 62 | 63 | This should not be a complicated assignment. Please try to make sure your answers are clear and to the pint. Please write your answers in English and submit a single PDF file. 64 | 65 | Assignment 5 66 | 67 | In class we showed how to do polynomial-time reductions between problems, with simple examples. Actually, in your textbook, there are several more involved reduction examples. 68 | 69 | In this assignment, you are required to pick your favorite reduction from the following examples and write the proof in details. The proofs are explained in your textbook, but you must write the proof in your own words. Of course, you are encouraged to prove all of them (but one is enough for this assignment). The goal of this assignment is to understand how reduction works. Please describe the problem that you pick in sufficient details. 70 | 71 | 1. 3-SAT ≤ p Directed-Hamiltonian-Cycle 72 | 73 | 2. 3-SAT ≤ p 3D-Matching 74 | 75 | 3. 3-SAT ≤ p 3-Color 76 | 77 | 4. 3-SAT ≤ p Subset-Sum 78 | 79 | Note: It is not hard to show that all of the above problems are in NP. Later you will learn that 3-SAT is in NP-complete and hence those problems are actually in NP-complete. 80 | 81 | Please write in English and submit a single PDF file. -------------------------------------------------------------------------------- /Assignment/temp_report.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | 3 | \usepackage{ctex} 4 | \usepackage{graphicx} 5 | \usepackage{float} 6 | \usepackage{datetime} 7 | \usepackage{hyperref} 8 | \usepackage{amssymb} 9 | \usepackage{amsthm} 10 | \newtheorem{theorem}{Theorem} 11 | \newtheorem{corollary}{Colollary} 12 | \newtheorem{lemma}{Lemma} 13 | \newtheorem{definition}{Definition} 14 | \renewcommand\proofname{Proof} 15 | \renewcommand\figurename{Figure} 16 | \usepackage{listings} 17 | \usepackage{xcolor} 18 | \usepackage[linesnumbered, ruled]{algorithm2e} 19 | \usepackage[legalpaper, margin=1.4in]{geometry} 20 | 21 | \title{CS216 Assignment0 \\ {\begin{large} What is the title \end{large}}} %title 22 | 23 | \author{匡亮(12111012)} 24 | 25 | \date{April 21, 2023} % date 26 | 27 | \begin{document} 28 | 29 | \maketitle 30 | 31 | \renewcommand\abstractname{Abstract} 32 | \begin{abstract} 33 | 34 | This is my report of CS216 Assignment0, including: 35 | 36 | \begin{itemize} 37 | \item[1.] whatwhat 38 | \item[2.] 39 | \end{itemize} 40 | 41 | \end{abstract} 42 | 43 | \newpage % contents page 44 | 45 | \renewcommand\contentsname{Contents} 46 | \tableofcontents 47 | 48 | \newpage % 1st page here 49 | 50 | 51 | 52 | \newpage % reference page 53 | 54 | \renewcommand\refname{References} % change '&' in the link to '\&' 55 | \begin{thebibliography}{99} 56 | \bibitem[who(year)]{short} title 57 | 58 | http://linklink.xxx 59 | \end{thebibliography} 60 | 61 | \end{document} 62 | -------------------------------------------------------------------------------- /Lab/Lab1/Lab1-A.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab1/Lab1-A.pdf -------------------------------------------------------------------------------- /Lab/Lab1/Lab1-B.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab1/Lab1-B.pdf -------------------------------------------------------------------------------- /Lab/Lab1/a.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | #define For(_, L, R) for(int _ = L; _ <= R; ++_) 3 | using namespace std; 4 | const int MaxN = 1000 + 10; 5 | int N; 6 | int Map[MaxN][MaxN], Point[MaxN][MaxN], Lastpro[MaxN], Ans1[MaxN]; 7 | int Map2[MaxN][MaxN], Point2[MaxN][MaxN], Nowpro[MaxN], Ans2[MaxN]; 8 | char Name[MaxN][13], Name2[MaxN][13], str[13]; 9 | queue Q; 10 | const int MaxM = 20000 + 10; 11 | int Id[MaxM], To[MaxM][52], Cnt; 12 | inline int toInt(char c) { 13 | if(c <= 'Z') return c - 'A'; 14 | return c - 'a' + 26; 15 | } 16 | void Insert(int id, char *strs, int len) { 17 | int Nod = 0; 18 | For(i, 0, len - 1) { 19 | int toint = toInt(strs[i]); 20 | if(To[Nod][toint] == 0) To[Nod][toint] = ++Cnt; 21 | Nod = To[Nod][toint]; 22 | } 23 | Id[Nod] = id; 24 | } 25 | int getId(char *strs, int len) { 26 | int Nod = 0; 27 | For(i, 0, len - 1) Nod = To[Nod][toInt(strs[i])]; 28 | return Id[Nod]; 29 | } 30 | int main() { 31 | scanf("%d", &N); 32 | For(i, 1, N) { 33 | scanf("%s", Name[i]); 34 | Insert(i, Name[i], strlen(Name[i])); 35 | } 36 | For(i, 1, N) { 37 | scanf("%s", Name2[i]); 38 | Insert(i, Name2[i], strlen(Name2[i])); 39 | } 40 | For(i, 1, N) 41 | Q.push(i); 42 | For(i, 1, N) 43 | For(j, 1, N) { 44 | memset(str, 0, sizeof str); 45 | scanf("%s", str); 46 | Map[i][j] = getId(str, strlen(str)); 47 | Point[i][Map[i][j]] = N - j; 48 | } 49 | For(i, 1, N) 50 | For(j, 1, N) { 51 | memset(str, 0, sizeof str); 52 | scanf("%s", str); 53 | Map2[i][j] = getId(str, strlen(str)); 54 | Point2[i][Map2[i][j]] = N - j; 55 | } 56 | while(!Q.empty()) { 57 | int Man = Q.front(); Q.pop(); 58 | int Woman = Map[Man][++Lastpro[Man]]; 59 | if(Nowpro[Woman] == 0) Nowpro[Woman] = Man; 60 | else if(Point2[Woman][Man] > Point2[Woman][Nowpro[Woman]]) { 61 | Q.push(Nowpro[Woman]); 62 | Nowpro[Woman] = Man; 63 | } else Q.push(Man); 64 | } 65 | For(i, 1, N) { 66 | Ans1[Nowpro[i]] = i; 67 | Ans2[i] = Nowpro[i]; 68 | } 69 | For(i, 1, N) 70 | For(j, 1, N) 71 | if(Point[i][j] > Point[i][Ans1[i]] && Point2[j][i] > Point2[j][Ans2[j]]) 72 | return !printf("Wrong Answer\n"); 73 | For(i, 1, N) { 74 | printf("%s %s", Name[i], Name2[Ans1[i]]); 75 | if(i != N) printf("\n"); 76 | } 77 | return 0; 78 | } 79 | /* 80 | 3 81 | aaa bbbb c 82 | a bb cc 83 | cc a bb 84 | a bb cc 85 | bb cc a 86 | aaa bbbb cc 87 | aaa bbbb cc 88 | aaa bbbb cc 89 | 90 | */ 91 | -------------------------------------------------------------------------------- /Lab/Lab1/b.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | #define For(_, L, R) for(int _ = L; _ <= R; ++_) 3 | using namespace std; 4 | const int MaxN = 1000 + 10; 5 | int N, M; 6 | int Cap[MaxN]; 7 | pair Point1[MaxN][MaxN]; 8 | int lastApply[MaxN]; 9 | int Point2[MaxN][MaxN]; 10 | queue Q; 11 | priority_queue> enrolled[MaxN]; 12 | 13 | bool Enroll[MaxN][MaxN]; 14 | int Point[MaxN][MaxN]; 15 | int enrollSchool[MaxN], worstStudent[MaxN], enrollNumber[MaxN]; 16 | 17 | inline void Check() { 18 | For(i, 1, N) 19 | For(j, 1, M) 20 | if(Enroll[i][j]) { 21 | if(Point[i][j] < 0) printf("Student %d would prefer not to enroll school %d.\n", i, j); 22 | if(Point2[j][i] < 0) printf("School %d would prefer not accept student %d.\n", j, i); 23 | } else { 24 | if(Point[i][enrollSchool[i]] < Point[i][j] && Point2[j][i] > 0 && (enrollNumber[j] < Cap[j] || worstStudent[j] < Point2[j][i])) printf("Student %d and school %d formed an unstable pair.\n", i, j); 25 | } 26 | } 27 | 28 | int main() { 29 | 30 | // freopen("b.in", "r", stdin); 31 | // freopen("b.out", "w", stdout); 32 | 33 | scanf("%d%d", &N, &M); 34 | For(i, 1, M) scanf("%d", &Cap[i]); 35 | For(i, 1, N) Q.push(i); 36 | For(i, 1, N) { 37 | For(j, 1, M) { 38 | scanf("%d", &Point1[i][j].first); 39 | Point[i][j] = Point1[i][j].first; 40 | Point1[i][j].second = j; 41 | } 42 | sort(Point1[i] + 1, Point1[i] + 1 + M); 43 | reverse(Point1[i] + 1, Point1[i] + 1 + M); 44 | } 45 | For(i, 1, M) 46 | For(j, 1, N) 47 | scanf("%d", &Point2[i][j]); 48 | while(!Q.empty()) { 49 | int Student = Q.front(); Q.pop(); 50 | if(lastApply[Student] == M) continue; 51 | int School = Point1[Student][++lastApply[Student]].second; 52 | if(Point1[Student][lastApply[Student]].first < 0 || Point2[School][Student] < 0) Q.push(Student); 53 | else if((int)enrolled[School].size() < Cap[School]) enrolled[School].push(make_pair(-Point2[School][Student], Student)); 54 | else if(-enrolled[School].top().first < Point2[School][Student]) { 55 | Q.push(enrolled[School].top().second); 56 | enrolled[School].pop(); 57 | enrolled[School].push(make_pair(-Point2[School][Student], Student)); 58 | } 59 | else Q.push(Student); 60 | } 61 | For(i, 1, M) { 62 | printf("%d ", (int)enrolled[i].size()); 63 | // enrollNumber[i] = (int)enrolled[i].size(); 64 | while(!enrolled[i].empty()) { 65 | printf("%d ", enrolled[i].top().second); 66 | // Enroll[enrolled[i].top().second][i] = true; 67 | // worstStudent[i] = min(worstStudent[i], -enrolled[i].top().first); 68 | // if(enrollSchool[enrolled[i].top().second]) printf("Student %d enrolled two schools %d and %d.\n", enrolled[i].top().second, i, enrollSchool[enrolled[i].top().second]); 69 | // enrollSchool[enrolled[i].top().second] = i; 70 | enrolled[i].pop(); 71 | } 72 | // if(i != M) printf("\n"); 73 | printf("\n"); 74 | } 75 | 76 | // Check(); 77 | 78 | return 0; 79 | } 80 | /* 81 | 1 1 82 | 1 83 | -1 84 | -1 85 | 86 | */ 87 | -------------------------------------------------------------------------------- /Lab/Lab10/A.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | const int MaxN = 1000000 + 10; 12 | const long long Inf = 1ll << 32; 13 | int N, M, S, T; 14 | int Cnt = 1, Head[MaxN], To[MaxN], Next[MaxN]; 15 | long long W[MaxN], Ans; 16 | inline void Add(int u, int v, long long w) { 17 | To[++Cnt] = v; 18 | W[Cnt] = w; 19 | Next[Cnt] = Head[u]; 20 | Head[u] = Cnt; 21 | } 22 | int Now[MaxN]; // Present-arc optimization 23 | int Dis[MaxN]; // Layer graph 24 | bool Bfs() { 25 | For(i, 1, N) Dis[i] = N, Now[i] = Head[i]; 26 | Dis[S] = 0; 27 | queue Q; Q.push(S); 28 | while(!Q.empty()) { 29 | int Nod = Q.front(); Q.pop(); 30 | for(int i = Head[Nod]; i ; i = Next[i]) 31 | if(W[i] > 0 && Dis[To[i]] == N) { 32 | Q.push(To[i]); 33 | Dis[To[i]] = Dis[Nod] + 1; 34 | } 35 | } 36 | return Dis[T] != N; 37 | } 38 | long long Dfs(int Nod, long long Limit) { 39 | if(Nod == T) return Limit; 40 | long long Tot = 0; 41 | for(int i = Now[Nod]; i && Limit ; i = Next[i]) { 42 | Now[Nod] = i; 43 | if(W[i] > 0 && Dis[To[i]] == Dis[Nod] + 1) { 44 | long long Flow = Dfs(To[i], min(Limit, W[i])); 45 | if(Flow == 0) Dis[To[i]] = N; 46 | else { 47 | W[i] -= Flow; 48 | W[i^1] += Flow; 49 | Tot += Flow; 50 | Limit -= Flow; 51 | } 52 | } 53 | } 54 | return Tot; 55 | } 56 | int main() { 57 | ios_base::sync_with_stdio(false); 58 | cin.tie(0); 59 | cout.tie(0); 60 | cin >> N >> M >> S >> T; 61 | For(i, 1, M) { 62 | int u, v; 63 | long long w; 64 | cin >> u >> v >> w; 65 | Add(u, v, w); 66 | Add(v, u, 0); 67 | } 68 | while(Bfs()) Ans += Dfs(S, Inf); 69 | cout << Ans << endl; 70 | return 0; 71 | } -------------------------------------------------------------------------------- /Lab/Lab10/B.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | const int MaxN = 1000000 + 10; 12 | const long long Inf = 1ll << 32; 13 | int N, M, S, T; 14 | int Cnt = 1, Head[MaxN], To[MaxN], Next[MaxN]; 15 | long long W[MaxN], Ans; 16 | inline void Add(int u, int v, long long w) { 17 | To[++Cnt] = v; 18 | W[Cnt] = w; 19 | Next[Cnt] = Head[u]; 20 | Head[u] = Cnt; 21 | } 22 | int Now[MaxN]; // Present-arc optimization 23 | int Dis[MaxN]; // Layer graph 24 | bool Bfs() { 25 | For(i, 1, N) Dis[i] = N, Now[i] = Head[i]; 26 | Dis[S] = 0; 27 | queue Q; Q.push(S); 28 | while(!Q.empty()) { 29 | int Nod = Q.front(); Q.pop(); 30 | for(int i = Head[Nod]; i ; i = Next[i]) 31 | if(W[i] > 0 && Dis[To[i]] == N) { 32 | Q.push(To[i]); 33 | Dis[To[i]] = Dis[Nod] + 1; 34 | } 35 | } 36 | return Dis[T] != N; 37 | } 38 | long long Dfs(int Nod, long long Limit) { 39 | if(Nod == T) return Limit; 40 | long long Tot = 0; 41 | for(int i = Now[Nod]; i && Limit ; i = Next[i]) { 42 | Now[Nod] = i; 43 | if(W[i] > 0 && Dis[To[i]] == Dis[Nod] + 1) { 44 | long long Flow = Dfs(To[i], min(Limit, W[i])); 45 | if(Flow == 0) Dis[To[i]] = N; 46 | else { 47 | W[i] -= Flow; 48 | W[i^1] += Flow; 49 | Tot += Flow; 50 | Limit -= Flow; 51 | } 52 | } 53 | } 54 | return Tot; 55 | } 56 | int K; 57 | int main() { 58 | ios_base::sync_with_stdio(false); 59 | cin.tie(0); 60 | cout.tie(0); 61 | cin >> N >> M >> K; 62 | S = 1; T = ++N; 63 | For(i, 1, K) { 64 | int g; cin >> g; ++g; 65 | Add(g, T, 1); Add(T, g, 0); 66 | } 67 | For(i, 1, M) { 68 | int u, v; cin >> u >> v; ++u; ++v; 69 | Add(u, v, 1); Add(v, u, 1); 70 | } 71 | while(Bfs()) Ans += Dfs(S, Inf); 72 | cout << Ans << endl; 73 | return 0; 74 | } -------------------------------------------------------------------------------- /Lab/Lab10/C.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | const int MaxN = 1000000 + 10; 12 | const long long Inf = 1ll << 32; 13 | int N, M, S, T; 14 | int Cnt = 1, Head[MaxN], To[MaxN], Next[MaxN]; 15 | long long W[MaxN], Ans; 16 | inline void _Add(int u, int v, long long w) { 17 | To[++Cnt] = v; 18 | W[Cnt] = w; 19 | Next[Cnt] = Head[u]; 20 | Head[u] = Cnt; 21 | } 22 | inline void Add(int u, int v, long long w) { 23 | _Add(u, v, w); 24 | _Add(v, u, 0); 25 | } 26 | int Now[MaxN]; // Present-arc optimization 27 | int Dis[MaxN]; // Layer graph 28 | bool Bfs() { 29 | For(i, 1, N) Dis[i] = N, Now[i] = Head[i]; 30 | Dis[S] = 0; 31 | queue Q; Q.push(S); 32 | while(!Q.empty()) { 33 | int Nod = Q.front(); Q.pop(); 34 | for(int i = Head[Nod]; i ; i = Next[i]) 35 | if(W[i] > 0 && Dis[To[i]] == N) { 36 | Q.push(To[i]); 37 | Dis[To[i]] = Dis[Nod] + 1; 38 | } 39 | } 40 | return Dis[T] != N; 41 | } 42 | long long Dfs(int Nod, long long Limit) { 43 | if(Nod == T) return Limit; 44 | long long Tot = 0; 45 | for(int i = Now[Nod]; i && Limit ; i = Next[i]) { 46 | Now[Nod] = i; 47 | if(W[i] > 0 && Dis[To[i]] == Dis[Nod] + 1) { 48 | long long Flow = Dfs(To[i], min(Limit, W[i])); 49 | if(Flow == 0) Dis[To[i]] = N; 50 | else { 51 | W[i] -= Flow; 52 | W[i^1] += Flow; 53 | Tot += Flow; 54 | Limit -= Flow; 55 | } 56 | } 57 | } 58 | return Tot; 59 | } 60 | int Map[1005][1005]; 61 | long long Val[1005][1005]; 62 | int main() { 63 | ios_base::sync_with_stdio(false); 64 | cin.tie(0); 65 | cout.tie(0); 66 | cin >> N >> M; 67 | int cnt = 0; 68 | long long Sum = 0; 69 | For(i, 1, N) 70 | For(j, 1, M) { 71 | cin >> Val[i][j]; 72 | Sum += Val[i][j]; 73 | Map[i][j] = ++cnt; 74 | } 75 | S = N * M + 1; 76 | T = S + 1; 77 | For(i, 1, N) 78 | For(j, 1, M) 79 | if((i ^ j) & 1) { 80 | Add(S, Map[i][j], Val[i][j]); 81 | if(i != 1) Add(Map[i][j], Map[i-1][j], Inf); 82 | if(i != N) Add(Map[i][j], Map[i+1][j], Inf); 83 | if(j != 1) Add(Map[i][j], Map[i][j-1], Inf); 84 | if(j != M) Add(Map[i][j], Map[i][j+1], Inf); 85 | } else 86 | Add(Map[i][j], T, Val[i][j]); 87 | N = T; 88 | while(Bfs()) Ans += Dfs(S, Inf); 89 | cout << Sum - Ans << endl; 90 | return 0; 91 | } -------------------------------------------------------------------------------- /Lab/Lab10/D.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | const int MaxN = 1000000 + 10; 12 | const long long Inf = 1ll << 32; 13 | int N, M, S, T; 14 | int Cnt = 1, Head[MaxN], To[MaxN], Next[MaxN]; 15 | long long W[MaxN], Ans; 16 | inline void _Add(int u, int v, long long w) { 17 | To[++Cnt] = v; 18 | W[Cnt] = w; 19 | Next[Cnt] = Head[u]; 20 | Head[u] = Cnt; 21 | } 22 | inline void Add(int u, int v, long long w) { 23 | _Add(u, v, w); 24 | _Add(v, u, 0); 25 | } 26 | int Now[MaxN]; // Present-arc optimization 27 | int Dis[MaxN]; // Layer graph 28 | bool Bfs() { 29 | For(i, 1, N) Dis[i] = N, Now[i] = Head[i]; 30 | Dis[S] = 0; 31 | queue Q; Q.push(S); 32 | while(!Q.empty()) { 33 | int Nod = Q.front(); Q.pop(); 34 | for(int i = Head[Nod]; i ; i = Next[i]) 35 | if(W[i] > 0 && Dis[To[i]] == N) { 36 | Q.push(To[i]); 37 | Dis[To[i]] = Dis[Nod] + 1; 38 | } 39 | } 40 | return Dis[T] != N; 41 | } 42 | long long Dfs(int Nod, long long Limit) { 43 | if(Nod == T) return Limit; 44 | long long Tot = 0; 45 | for(int i = Now[Nod]; i && Limit ; i = Next[i]) { 46 | Now[Nod] = i; 47 | if(W[i] > 0 && Dis[To[i]] == Dis[Nod] + 1) { 48 | long long Flow = Dfs(To[i], min(Limit, W[i])); 49 | if(Flow == 0) Dis[To[i]] = N; 50 | else { 51 | W[i] -= Flow; 52 | W[i^1] += Flow; 53 | Tot += Flow; 54 | Limit -= Flow; 55 | } 56 | } 57 | } 58 | return Tot; 59 | } 60 | int n, x[MaxN], a[MaxN], dp[MaxN]; 61 | int main() { 62 | ios_base::sync_with_stdio(false); 63 | cin.tie(0); 64 | cout.tie(0); 65 | cin >> n; 66 | For(i, 1, n) cin >> a[i]; 67 | For(i, 1, n) cin >> x[i]; 68 | // DP part 69 | For(i, 1, n) 70 | For(j, 0, i - 1) 71 | if(a[j] <= a[i]) 72 | Checkmax(dp[i], dp[j] + 1); 73 | int k = 0; 74 | For(i, 1, n) 75 | Checkmax(k, dp[i]); 76 | cout << k << endl; 77 | // Flow part 78 | N = (n + 1) << 1; 79 | S = N - 1; 80 | T = N; 81 | For(i, 1, n) Add((i << 1) - 1, i << 1, x[i]); 82 | For(i, 1, n) { 83 | if(dp[i] == 1) Add(S, (i << 1) - 1, Inf); 84 | if(dp[i] == k) Add(i << 1, T, Inf); 85 | For(j, i + 1, n) 86 | if(a[i] <= a[j] && dp[i] + 1 == dp[j]) 87 | Add(i << 1, (j << 1) - 1, Inf); 88 | } 89 | while(Bfs()) Ans += Dfs(S, Inf); 90 | cout << Ans << endl; 91 | return 0; 92 | } -------------------------------------------------------------------------------- /Lab/Lab10/Lab10-A.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab10/Lab10-A.pdf -------------------------------------------------------------------------------- /Lab/Lab10/Lab10-B.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab10/Lab10-B.pdf -------------------------------------------------------------------------------- /Lab/Lab10/Lab10-C.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab10/Lab10-C.pdf -------------------------------------------------------------------------------- /Lab/Lab10/Lab10-D.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab10/Lab10-D.pdf -------------------------------------------------------------------------------- /Lab/Lab2/Lab2-A.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab2/Lab2-A.pdf -------------------------------------------------------------------------------- /Lab/Lab2/Lab2-B.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab2/Lab2-B.pdf -------------------------------------------------------------------------------- /Lab/Lab2/a.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | #define For(_, L, R) for(int _ = L; _ <= R; ++_) 3 | using namespace std; 4 | int main() { 5 | int t; 6 | cin >> t; 7 | while(t--) { 8 | int n; 9 | cin >> n; 10 | if(n % 6) printf("Alice\n"); 11 | else printf("Bob\n"); 12 | } 13 | return 0; 14 | } 15 | 16 | -------------------------------------------------------------------------------- /Lab/Lab2/b.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | #define For(_, L, R) for(int _ = L; _ <= R; ++_) 3 | using namespace std; 4 | int main() { 5 | int N, K; 6 | cin >> N >> K; 7 | if (K > N / 2) K = N - K + 1; 8 | printf("%d\n", N == 1 ? 1 : 2 * K); 9 | return 0; 10 | } 11 | 12 | -------------------------------------------------------------------------------- /Lab/Lab3/A.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | #define For(_, L, R) for(int _ = L; _ <= R; ++_) 3 | using namespace std; 4 | const int MaxN = 1000 + 10; 5 | int N, Sum, T[MaxN], W[MaxN]; 6 | vector Time[MaxN]; 7 | priority_queue Q; 8 | int main() { 9 | ios_base::sync_with_stdio(false); 10 | cin.tie(0); cout.tie(0); 11 | cin >> N; 12 | For(i, 1, N) cin >> T[i]; 13 | For(i, 1, N) cin >> W[i]; 14 | For(i, 1, N) Time[T[i]].push_back(W[i]); 15 | For(i, 1, N) { 16 | int sz = (int)Time[i].size(); 17 | For(j, 1, sz) { 18 | int Game = Time[i][j - 1]; 19 | Q.push(-Game); 20 | if((int)Q.size() > i) { 21 | Sum -= Q.top(); 22 | Q.pop(); 23 | } 24 | } 25 | } 26 | cout << Sum << endl; 27 | return 0; 28 | // Time Complexity: O(NlogN) 29 | } 30 | 31 | -------------------------------------------------------------------------------- /Lab/Lab3/B.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | #define For(_, L, R) for(int _ = L; _ <= R; ++_) 3 | using namespace std; 4 | const int MaxN = 1000000 + 10; 5 | int N; 6 | long long A[MaxN], C[MaxN], Avg, Mid, Ans; 7 | int main() { 8 | ios_base::sync_with_stdio(false); 9 | cin.tie(0); cout.tie(0); 10 | cin >> N; 11 | For(i, 1, N) { 12 | cin >> A[i]; 13 | Avg += A[i]; 14 | } 15 | Avg /= N; 16 | For(i, 1, N - 1) C[i] = C[i - 1] + A[i] - Avg; 17 | nth_element(C + 1, C + (N + 1) / 2, C + N + 1); 18 | Mid = C[(N + 1) / 2]; 19 | For(i, 1, N) Ans += abs(C[i] - Mid); 20 | cout << Ans << endl; 21 | return 0; 22 | } 23 | 24 | -------------------------------------------------------------------------------- /Lab/Lab3/Lab3-A.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab3/Lab3-A.png -------------------------------------------------------------------------------- /Lab/Lab3/Lab3-B.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab3/Lab3-B.png -------------------------------------------------------------------------------- /Lab/Lab4/A.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | #define For(_, L, R) for(int _ = L; _ <= R; ++_) 3 | #define SQU(x) ((x)*(x)) 4 | using namespace std; 5 | const int MaxN = 1000 + 10; 6 | int N, K, M; 7 | long long X[MaxN], Y[MaxN]; 8 | int Find[MaxN]; 9 | int F(int N) { 10 | return N == Find[N] ? N : Find[N] = F(Find[N]); 11 | } 12 | vector>> Edge; 13 | int main() { 14 | ios_base::sync_with_stdio(false); 15 | cin.tie(0); cout.tie(0); 16 | cin >> N >> K; 17 | For(i, 1, N) cin >> X[i] >> Y[i]; 18 | M = N; 19 | For(i, 1, N) Find[i] = i; 20 | For(i, 1, N) For(j, i + 1, N) 21 | Edge.push_back(make_pair(SQU(X[i] - X[j]) + SQU(Y[i] - Y[j]), make_pair(i, j))); 22 | sort(Edge.begin(), Edge.end()); 23 | for(auto e : Edge) { 24 | long long w = e.first; 25 | int u = e.second.first, v = e.second.second; 26 | if(F(u) == F(v)) continue; 27 | if(M == K) return not printf("%.2lf\n", sqrt(w)); 28 | --M; 29 | Find[Find[u]] = Find[v]; 30 | } 31 | } 32 | -------------------------------------------------------------------------------- /Lab/Lab4/B.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | #define For(_, L, R) for(int _ = L; _ <= R; ++_) 3 | using namespace std; 4 | const int MaxN = 100000 + 10; 5 | int N, Fa[MaxN], R, Find[MaxN]; 6 | vector To[MaxN]; 7 | long long Cost[MaxN], Sum[MaxN], C[MaxN], Siz[MaxN]; 8 | int F(int N) { 9 | return N == Find[N] ? N : Find[N] = F(Find[N]); 10 | } 11 | priority_queue> Q; 12 | bool Vis[MaxN]; 13 | int main() { 14 | ios_base::sync_with_stdio(false); 15 | cin.tie(0); cout.tie(0); 16 | cin >> N >> R; 17 | For(i, 1, N) cin >> C[i]; 18 | For(i, 1, N - 1) { 19 | int U, V; 20 | cin >> U >> V; 21 | To[U].push_back(V); 22 | Fa[V] = U; 23 | } 24 | For(i, 1, N) { 25 | Q.push(make_pair((double)C[i], i)); 26 | Siz[i] = 1; 27 | Sum[i] = Cost[i] = C[i]; 28 | Find[i] = i; 29 | } 30 | Vis[R] = true; 31 | while(!Q.empty()) { 32 | int Nod = Q.top().second; Q.pop(); 33 | if(Vis[Nod]) continue; 34 | Vis[Nod] = true; 35 | int fa = F(Fa[Nod]); 36 | Find[Nod] = fa; 37 | Cost[fa] += Cost[Nod] + Siz[fa] * Sum[Nod]; 38 | Siz[fa] += Siz[Nod]; 39 | Sum[fa] += Sum[Nod]; 40 | Q.push(make_pair((double)Sum[fa] / Siz[fa], fa)); 41 | } 42 | cout << Cost[R] << endl; 43 | return 0; 44 | } 45 | /* 46 | 5 1 47 | 1 2 1 2 4 48 | 1 2 49 | 1 3 50 | 2 4 51 | 3 5 52 | 53 | 5 1 54 | 1 1 1 2 1 55 | 1 2 56 | 2 3 57 | 3 5 58 | 1 4 59 | 60 | */ 61 | -------------------------------------------------------------------------------- /Lab/Lab4/Lab4-A.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab4/Lab4-A.pdf -------------------------------------------------------------------------------- /Lab/Lab4/Lab4-B.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab4/Lab4-B.pdf -------------------------------------------------------------------------------- /Lab/Lab5/A.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | #define For(_, L, R) for(int _ = L; _ <= R; ++_) 3 | using namespace std; 4 | const int Ans[5][5] = { 5 | {1, 1, 1, 1, 1}, 6 | {0, 1, 1, 1, 1}, 7 | {0, 0, 2, 1, 1}, 8 | {0, 0, 0, 0, 1}, 9 | {0, 0, 0, 0, 0}}, 10 | dx[8] = {-2, -2, -1, -1, 1, 1, 2, 2}, 11 | dy[8] = {1, -1, 2, -2, 2, -2, 1, -1}; 12 | int Map[5][5], x, y; 13 | int v() { 14 | int V = 0; 15 | For(i, 0, 4) 16 | For(j, 0, 4) 17 | V += (Map[i][j] != Ans[i][j]); 18 | return V == 2 ? 1 : V; 19 | } 20 | bool IDAx(int dep, int lim) { 21 | if(dep == lim) return !(bool)v(); 22 | For(i, 0, 7) { 23 | int xx = x, yy = y; 24 | x += dx[i], y += dy[i]; 25 | if(x >= 0 && y >= 0 && x <= 4 && y <= 4) { 26 | swap(Map[x][y], Map[xx][yy]); 27 | if(v() + dep <= lim) 28 | if(IDAx(dep + 1, lim)) 29 | return true; 30 | swap(Map[x][y], Map[xx][yy]); 31 | } 32 | x = xx, y = yy; 33 | } 34 | return false; 35 | } 36 | int main() { 37 | int T; cin >> T; while(T--) { 38 | For(i, 0, 4) 39 | For(j, 0, 4) { 40 | char C = getchar(); 41 | while(C != '0' && C != '1' && C != '*') C = getchar(); 42 | if((Map[i][j] = (C == '1' ? 1 : (C == '0' ? 0 : 2))) == 2) x = i, y = j; 43 | } 44 | bool notfound = true; 45 | For(i, 0, 15) 46 | if(IDAx(0, i)) { 47 | cout << i << endl; 48 | notfound = false; 49 | break; 50 | } 51 | if(notfound) cout << -1 << endl; 52 | } 53 | return 0; 54 | } 55 | /* 56 | 1 57 | 11111 58 | 01111 59 | 00111 60 | 0000* 61 | 00000 62 | 63 | */ 64 | -------------------------------------------------------------------------------- /Lab/Lab5/Lab5-A.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab5/Lab5-A.pdf -------------------------------------------------------------------------------- /Lab/Lab5/Lab5-B.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab5/Lab5-B.pdf -------------------------------------------------------------------------------- /Lab/Lab6/A.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | const int MaxN = 1000000 + 10; 12 | int N, M, T, C; 13 | int Del[MaxN], Time[MaxN], Val[MaxN]; 14 | priority_queue Q, D; 15 | int Top() { 16 | while(!Q.empty() && !D.empty() && Q.top() == D.top()) 17 | Q.pop(), D.pop(); 18 | return Q.empty() ? 0 : Q.top(); 19 | } 20 | int main() { 21 | ios_base::sync_with_stdio(false); 22 | cin.tie(0); 23 | cout.tie(0); 24 | cin >> N >> M >> T; 25 | For(i, 1, M) { 26 | if(Del[i]) { 27 | D.push(-i); 28 | Val[Del[i]] = 0; 29 | Time[Del[i]] = 0; 30 | Del[i] = 0; 31 | --C; 32 | } 33 | int op, k, v; 34 | cin >> op; 35 | if(op == 1) { 36 | cin >> k; 37 | if(Val[k]) { 38 | printf("%d\n", Val[k]); 39 | Del[Time[k]] = 0; 40 | D.push(-Time[k]); 41 | Del[Time[k] = i + T + 1] = k; 42 | Q.push(-Time[k]); 43 | } else printf("-1\n"); 44 | } else { 45 | cin >> k >> v; 46 | if(Val[k]) { 47 | Val[k] = v; 48 | Del[Time[k]] = 0; 49 | D.push(-Time[k]); 50 | Del[Time[k] = i + T + 1] = k; 51 | Q.push(-Time[k]); 52 | } else { 53 | if(C == N) { 54 | int t = Del[-Top()]; 55 | Q.pop(); 56 | Del[Time[t]] = 0; 57 | Time[t] = 0; 58 | Val[t] = 0; 59 | } else ++C; 60 | Del[Time[k] = i + T + 1] = k; 61 | Val[k] = v; 62 | Q.push(-Time[k]); 63 | } 64 | } 65 | } 66 | int t; 67 | while(t = -Top()) 68 | printf("%d ", Val[Del[t]]), Q.pop(); 69 | printf("\n"); 70 | return 0; 71 | } 72 | -------------------------------------------------------------------------------- /Lab/Lab6/B.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | struct Point { long long X, Y; }; 12 | bool Cmp(Point &A, Point &B) { return A.X == B.X ? A.Y < B.Y : A.X < B.X; } 13 | const int MaxN = 3000000 + 10; 14 | long long Mindist = (long long)9e16; 15 | double mindist = 3e8; 16 | int N, Q[MaxN]; 17 | Point P[MaxN], Reg[MaxN]; 18 | inline long long Squ(long long X) { return X * X; } 19 | inline void Update(int i, int j) { 20 | long long Newdist = Squ(P[i].X - P[j].X) + Squ(P[i].Y - P[j].Y); 21 | if(Mindist > Newdist) mindist = sqrt(Mindist = Newdist); 22 | } 23 | void Merge(int L, int Mid, int R) { 24 | int l = L, r = Mid + 1, reg = L; 25 | while(l <= Mid && r <= R) 26 | if(P[l].Y < P[r].Y) Reg[reg++] = P[l++]; 27 | else Reg[reg++] = P[r++]; 28 | while(l <= Mid) Reg[reg++] = P[l++]; 29 | while(r <= R) Reg[reg++] = P[r++]; 30 | For(i, L, R) P[i] = Reg[i]; 31 | } 32 | void Solve(int L, int R) { 33 | if(L == R) return; 34 | int Mid = (L + R) >> 1, Midx = P[Mid].X, Qtop = 0; 35 | Solve(L, Mid); Solve(Mid + 1, R); Merge(L, Mid, R); 36 | For(i, L, R) 37 | if(Midx - mindist < P[i].X && P[i].X < Midx + mindist) { 38 | for(int j = Qtop; j && P[i].Y - P[Q[j]].Y < mindist; --j) Update(i, Q[j]); 39 | Q[++Qtop] = i; 40 | } 41 | } 42 | int main() { 43 | ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); 44 | cin >> N; For(i, 1, N) cin >> P[i].X >> P[i].Y; 45 | sort(P + 1, P + 1 + N, Cmp); 46 | Solve(1, N); 47 | printf("%lld\n", Mindist); 48 | return 0; 49 | } 50 | -------------------------------------------------------------------------------- /Lab/Lab6/Lab6-A.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab6/Lab6-A.pdf -------------------------------------------------------------------------------- /Lab/Lab6/Lab6-B.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab6/Lab6-B.pdf -------------------------------------------------------------------------------- /Lab/Lab7/A.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | const int MaxN = 2100000 + 10; 12 | struct cp { 13 | double x, y; 14 | cp (double X = 0, double Y = 0) { x = X, y = Y; } 15 | }; 16 | cp operator + (cp a,cp b) { return cp(a.x + b.x, a.y + b.y); } 17 | cp operator - (cp a,cp b) { return cp(a.x - b.x, a.y - b.y); } 18 | cp operator * (cp a,cp b) { return cp(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x); } 19 | cp A[MaxN], B[MaxN], C[MaxN]; 20 | const double pi = acos(-1.0); 21 | int N, M, Lim = 1, L; 22 | int R[MaxN]; 23 | void FFT(cp *A, int type) { 24 | For(i, 0, Lim - 1) 25 | if(i < R[i]) swap(A[i], A[R[i]]); 26 | for(int i = 1; i < Lim; i <<= 1) { 27 | cp wn = cp(cos(pi / i), type * sin(pi / i)); 28 | for(int R = i << 1, j = 0; j < Lim; j += R) { 29 | cp w = cp(1, 0); 30 | for(int k = 0; k < i; ++k, w = w * wn) { 31 | cp x = A[j + k], y = w * A[j + k + i]; 32 | A[j + k] = x + y; 33 | A[j + k + i] = x - y; 34 | } 35 | } 36 | } 37 | } 38 | int main() { 39 | ios_base::sync_with_stdio(false); 40 | cin.tie(0); 41 | cout.tie(0); 42 | cin >> N; 43 | For(i, 1, N) { 44 | cin >> A[i].x; 45 | B[i].x = 1.0 / i / i; 46 | C[N - i].x = A[i].x; 47 | } 48 | while(Lim <= (N << 1)) Lim <<= 1, ++L; 49 | For(i, 0, Lim - 1) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1)); 50 | FFT(A, 1); FFT(B, 1); FFT(C, 1); 51 | For(i, 0, Lim - 1) A[i] = A[i] * B[i], C[i] = C[i] * B[i]; 52 | FFT(A, -1); FFT(C, -1); 53 | For(i, 1, N) printf("%.5lf\n", (A[i].x - C[N - i].x) / Lim); 54 | return 0; 55 | } 56 | 57 | -------------------------------------------------------------------------------- /Lab/Lab7/B.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | const int MaxN = 1100000 + 10; 12 | const int Mod = 998244353; 13 | inline int Pow(int A, int B) { 14 | int Ans = 1; 15 | while(B) { 16 | if(B & 1) Ans = 1ll * Ans * A % Mod; 17 | A = 1ll * A * A % Mod; 18 | B >>= 1; 19 | } 20 | return Ans; 21 | } 22 | const int G = 3; 23 | const int iG = Pow(G, Mod - 2); 24 | int N, M, Lim = 1, L, R[MaxN]; 25 | void NTT(int *A, int type) { 26 | For(i, 0, Lim - 1) 27 | if(i < R[i]) swap(A[i], A[R[i]]); 28 | for(int i = 1; i < Lim; i <<= 1) { 29 | int wn = Pow(type == 1 ? G : iG, (Mod - 1) / (i << 1)); 30 | for(int R = i << 1, j = 0; j < Lim; j += R) { 31 | int w = 1; 32 | for(int k = 0; k < i; ++k, w = 1ll * w * wn % Mod) { 33 | int x = A[j + k], y = 1ll * w * A[j + k + i] % Mod; 34 | A[j + k] = (x + y) % Mod; 35 | A[j + k + i] = (x - y + Mod) % Mod; 36 | } 37 | } 38 | } 39 | if(type == -1) { 40 | int inv = Pow(Lim, Mod - 2); 41 | For(i, 0, Lim - 1) 42 | A[i] = 1ll * A[i] * inv % Mod; 43 | } 44 | } 45 | int A[MaxN], B[MaxN], A2[MaxN], B2[MaxN], A3[MaxN], B3[MaxN], C[MaxN]; 46 | char S[MaxN]; 47 | int Charmap[300]; 48 | int main() { 49 | Charmap['?'] = 0, Charmap['0'] = 1, Charmap['1'] = 2; 50 | cin >> S; 51 | N = strlen(S); 52 | For(i, 0, N - 1) { 53 | A[i] = Charmap[S[i]]; 54 | A2[i] = A[i] * A[i]; 55 | A3[i] = A2[i] * A[i]; 56 | B[i] = Charmap[S[N - 1 - i]]; 57 | B2[i] = B[i] * B[i]; 58 | B3[i] = B2[i] * B[i]; 59 | } 60 | while(Lim <= N + N + 2) Lim <<= 1, ++L; 61 | For(i, 0, Lim - 1) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1)); 62 | NTT(A, 1); NTT(B, 1); NTT(A2, 1); NTT(B2, 1); NTT(A3, 1); NTT(B3, 1); 63 | For(i, 0, Lim - 1) 64 | C[i] = ((1ll * A[i] * B3[i] + 1ll * B[i] * A3[i] - 2ll * A2[i] * B2[i]) % Mod + Mod) % Mod; 65 | NTT(C, -1); 66 | long long Ans = 0; 67 | For(i, 1, N / 2) 68 | if(C[i - 1] == 0) 69 | Ans ^= 1ll * i * i; 70 | printf("%lld\n", Ans); 71 | return 0; 72 | } 73 | -------------------------------------------------------------------------------- /Lab/Lab7/Lab7-A.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab7/Lab7-A.pdf -------------------------------------------------------------------------------- /Lab/Lab7/Lab7-B.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab7/Lab7-B.pdf -------------------------------------------------------------------------------- /Lab/Lab8/A.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | const int MaxN = 100 + 10; 12 | const int MaxM = 40000 + 10; 13 | int N, C, V[MaxN], W[MaxN], S[MaxN]; 14 | struct Queue { 15 | int Lim, Bias, T[MaxM], W[MaxM], Head, Tail; 16 | void Push(int t, int w) { 17 | w -= Bias; 18 | while(Head < Tail && W[Tail] <= w) --Tail; 19 | W[++Tail] = w; 20 | T[Tail] = t; 21 | } 22 | void Init(int lim, int w) { 23 | Bias = Head = T[1] = 0; 24 | Tail = 1; 25 | W[1] = w; 26 | Lim = lim; 27 | } 28 | int Pop(int t) { 29 | while(Head < Tail && T[Head + 1] + Lim < t) ++Head; 30 | return Head == Tail ? -(1 << 30) : W[Head + 1] + Bias; 31 | } 32 | } q[MaxN]; 33 | int Dp[MaxM], Reg[MaxM]; 34 | int main() { 35 | ios_base::sync_with_stdio(false); 36 | cin.tie(0); 37 | cout.tie(0); 38 | cin >> N >> C; 39 | For(i, 1, N) cin >> W[i] >> V[i] >> S[i]; 40 | For(i, 1, N) { 41 | For(j, 0, C) 42 | if(j < V[i]) q[j].Init(S[i], Reg[j] = Dp[j]); 43 | else { 44 | int id = j % V[i], t = j / V[i]; 45 | q[id].Bias += W[i]; 46 | Reg[j] = max(Dp[j], q[id].Pop(t)); 47 | q[id].Push(t, Dp[j]); 48 | } 49 | memcpy(Dp, Reg, sizeof Dp); 50 | } 51 | int Ans = 0; 52 | For(i, 0, C) 53 | Checkmax(Ans, Dp[i]); 54 | cout << Ans << endl; 55 | return 0; 56 | } 57 | 58 | -------------------------------------------------------------------------------- /Lab/Lab8/B.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | const int Mod = 12345678; 12 | inline void Qmod(int &A) { 13 | if(A >= Mod) A -= Mod; 14 | } 15 | int N, M, K; 16 | int Dp[155][155][25][25]; 17 | int main() { 18 | ios_base::sync_with_stdio(false); 19 | cin.tie(0); 20 | cout.tie(0); 21 | cin >> N >> M >> K; 22 | Dp[0][0][0][0] = 1; 23 | For(i, 0, N) For(j, 0, M) For(m, 0, K) For(n, 0, K) 24 | Qmod(Dp[i + 1][j][m + 1][max(0, n - 1)] += Dp[i][j][m][n]), 25 | Qmod(Dp[i][j + 1][max(0, m - 1)][n + 1] += Dp[i][j][m][n]); 26 | int Ans = 0; 27 | For(i, 0, K) For(j, 0, K) 28 | Qmod(Ans += Dp[N][M][i][j]); 29 | cout << Ans << endl; 30 | return 0; 31 | } 32 | 33 | -------------------------------------------------------------------------------- /Lab/Lab8/Lab8-A.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab8/Lab8-A.pdf -------------------------------------------------------------------------------- /Lab/Lab8/Lab8-B.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab8/Lab8-B.pdf -------------------------------------------------------------------------------- /Lab/Lab9/A.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | const int MaxN = 200000 + 10; 12 | int N; 13 | int Head[MaxN], To[MaxN], Next[MaxN], Cnt; 14 | inline void Add(int U, int V) { 15 | To[++Cnt] = V; 16 | Next[Cnt] = Head[U]; 17 | Head[U] = Cnt; 18 | } 19 | long long V[MaxN], Plus[MaxN], Minus[MaxN]; 20 | void Dfs(int Nod, int Fa) { 21 | for(int i = Head[Nod]; i ; i = Next[i]) 22 | if(To[i] != Fa) { 23 | Dfs(To[i], Nod); 24 | Checkmax(Plus[Nod], Plus[To[i]]); 25 | Checkmax(Minus[Nod], Minus[To[i]]); 26 | } 27 | long long Now = V[Nod] + Plus[Nod] - Minus[Nod]; 28 | if(Now < 0) Plus[Nod] -= Now; 29 | else Minus[Nod] += Now; 30 | } 31 | int main() { 32 | ios_base::sync_with_stdio(false); 33 | cin.tie(0); 34 | cout.tie(0); 35 | cin >> N; 36 | For(i, 1, N - 1) { 37 | int U, V; 38 | cin >> U >> V; 39 | Add(U, V); 40 | Add(V, U); 41 | } 42 | For(i, 1, N) cin >> V[i]; 43 | Dfs(1, 0); 44 | cout << Plus[1] + Minus[1] << endl; 45 | return 0; 46 | } -------------------------------------------------------------------------------- /Lab/Lab9/B.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | const int MaxN = 27000 + 10; 12 | int N, M; 13 | int Head[MaxN], To[MaxN], Next[MaxN], V[MaxN], P[MaxN], Cnt; 14 | double W[MaxN], Dis[MaxN]; 15 | inline void Add(int S, int T) { 16 | To[++Cnt] = T; 17 | Next[Cnt] = Head[S]; 18 | Head[S] = Cnt; 19 | } 20 | bool Vis[MaxN]; 21 | bool Spfa(int Nod) { 22 | Vis[Nod] = true; 23 | for(int i = Head[Nod]; i ; i = Next[i]) { 24 | if(Dis[To[i]] > Dis[Nod] + W[i]) { 25 | if(Vis[To[i]]) return false; 26 | else { 27 | Dis[To[i]] = Dis[Nod] + W[i]; 28 | if(!Spfa(To[i])) return false; 29 | } 30 | } 31 | } 32 | Vis[Nod] = false; 33 | return true; 34 | } 35 | int main() { 36 | ios_base::sync_with_stdio(false); 37 | cin.tie(0); 38 | cout.tie(0); 39 | cin >> N >> M; 40 | For(i, 1, M) { 41 | int S, T; 42 | cin >> S >> T >> V[i]; 43 | Add(S, T); 44 | } 45 | For(i, 1, M) cin >> P[i]; 46 | For(i, 1, N) Add(0, i); 47 | double L = 0.0, R = 256.0, Mid; 48 | For(_, 1, 15) { 49 | Mid = (L + R) / 2; 50 | For(i, 1, M) W[i] = Mid * P[i] - V[i]; 51 | memset(Dis, 127, sizeof Dis); 52 | memset(Vis, 0, sizeof Vis); 53 | Dis[0] = 0.0; 54 | if(Spfa(0)) R = Mid; 55 | else L = Mid; 56 | } 57 | if(fabs(L) < 1e-3) printf("-1\n"); 58 | else printf("%.1f\n", L); 59 | return 0; 60 | } -------------------------------------------------------------------------------- /Lab/Lab9/C.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | int N, Mod; 12 | inline void Qmod(int &Ans) { 13 | if(Ans >= Mod) Ans -= Mod; 14 | } 15 | int Dp[3][256][256]; 16 | const int sp[8] = {2, 3, 5, 7, 11, 13, 17, 19}; // small primes 17 | vector Num[505]; 18 | int main() { 19 | ios_base::sync_with_stdio(false); 20 | cin.tie(0); 21 | cout.tie(0); 22 | cin >> N >> Mod; 23 | For(i, 2, N) { 24 | int lp = -1, j = i; // largest prime factor 25 | for(int k = 2; k * k <= j; ++k) { 26 | if(j % k == 0) Checkmax(lp, k); 27 | while(j % k == 0) j /= k; 28 | } 29 | Checkmax(lp, j); 30 | Num[lp].push_back(i); 31 | } 32 | Dp[0][0][0] = 1; 33 | For(i, 2, 500) if(Num[i].size()) { 34 | memcpy(Dp[1], Dp[0], sizeof Dp[1]); 35 | memcpy(Dp[2], Dp[0], sizeof Dp[2]); 36 | for(auto p : Num[i]) { 37 | int s = 0; 38 | For(i, 0, 7) if(p % sp[i] == 0) s |= 1 << i; 39 | _For(i, 255, 0) _For(j, 255, 0) if(!(i & j)) { 40 | if(!(j & s)) Qmod(Dp[1][i | s][j] += Dp[1][i][j]); 41 | if(!(i & s)) Qmod(Dp[2][i][j | s] += Dp[2][i][j]); 42 | } 43 | } 44 | _For(i, 255, 0) _For(j, 255, 0) 45 | Dp[0][i][j] = (1ll * Mod + Dp[1][i][j] + Dp[2][i][j] - Dp[0][i][j]) % Mod; 46 | } 47 | int Ans = 0; 48 | _For(i, 255, 0) _For(j, 255, 0) Qmod(Ans += Dp[0][i][j]); 49 | cout << Ans << endl; 50 | return 0; 51 | } -------------------------------------------------------------------------------- /Lab/Lab9/D.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | const int MaxN = 500 + 10; 12 | int N, M; 13 | char S[MaxN]; 14 | int Dp[MaxN][MaxN]; 15 | int main() { 16 | ios_base::sync_with_stdio(false); 17 | cin.tie(0); 18 | cout.tie(0); 19 | cin >> M; 20 | cin >> (S + 1); 21 | For(i, 1, M) 22 | if(S[i] != S[i - 1]) 23 | S[++N] = S[i]; 24 | For(i, 1, N) { 25 | Dp[i][i] = 1; 26 | For(j, i + 1, N) 27 | Dp[i][j] = 1000; 28 | } 29 | For(l, 2, N) 30 | For(i, 1, N - l + 1) { 31 | int j = i + l - 1; 32 | For(k, i, j - 1) 33 | Checkmin(Dp[i][j], Dp[i][k] + Dp[k + 1][j]); 34 | if(S[i] == S[j]) { 35 | For(k, i + 1, j - 1) 36 | Checkmin(Dp[i][j], Dp[i + 1][k] + Dp[k + 1][j]); 37 | For(k, i, j - 2) 38 | Checkmin(Dp[i][j], Dp[i][k] + Dp[k + 1][j - 1]); 39 | } 40 | } 41 | cout << Dp[1][N] << endl; 42 | return 0; 43 | } -------------------------------------------------------------------------------- /Lab/Lab9/Lab9-A.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab9/Lab9-A.pdf -------------------------------------------------------------------------------- /Lab/Lab9/Lab9-B.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab9/Lab9-B.pdf -------------------------------------------------------------------------------- /Lab/Lab9/Lab9-C.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab9/Lab9-C.pdf -------------------------------------------------------------------------------- /Lab/Lab9/Lab9-D.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/Lab/Lab9/Lab9-D.pdf -------------------------------------------------------------------------------- /Lab/temp.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | #define For(_,l,r) for(int _ = l; _ <= r; ++_) 4 | #define _For(_,r,l) for(int _ = r; _ >= l; --_) 5 | template inline bool Checkmax(R &A, R B) { 6 | return A < B ? A = B, 1 : 0; 7 | } 8 | template inline bool Checkmin(R &A, R B) { 9 | return A > B ? A = B, 1 : 0; 10 | } 11 | int main() { 12 | ios_base::sync_with_stdio(false); 13 | cin.tie(0); 14 | cout.tie(0); 15 | } -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # Algorithm-Lab 2 | SUSTech 2023 Spring CS216 Algorithm Design and Analysis(H) 3 | 4 | ~~本来是英文的，但我英文太差还是不要恶心自己和大家了~~ 5 | 6 | 南方科技大学 2023 春季 CS216 算法设计与分析(H) 7 | 8 | 备份了所有 Lab problems 和 Assignments 以及对应的我完成的内容 9 | 10 | **一定不要**将其中的任何内容作为你自己的作业来提交，否则自负查重风险 11 | 12 | 完成情况： 13 | 14 | ![lab](labscore.png) 15 | 16 | ![assignment](assignmentscore.png) 17 | 18 | 期末考试 96 ，总评 98 19 | -------------------------------------------------------------------------------- /assignmentscore.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/assignmentscore.png -------------------------------------------------------------------------------- /labscore.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/DeerInForestovo/Algorithm-Lab/ebf27db571ec45a3af3de11e0295c33e0593301b/labscore.png --------------------------------------------------------------------------------