├── LICENSE ├── README.md ├── answer ├── 1706 │ ├── Assignment01_Solution.pdf │ ├── Assignment02_Solution.pdf │ ├── Assignment03_Solution.pdf │ └── Assignment04_Solution.pdf ├── 1806 │ ├── solution1.pdf │ ├── solution2.pdf │ ├── solution3.pdf │ └── solution4.pdf ├── 1823 │ ├── solution1.pdf │ ├── solution2.pdf │ ├── solution3.pdf │ └── solution4.pdf ├── 1906 │ ├── 计算机学院《算法设计与分析》第一次作业参考答案.pdf │ ├── 计算机学院《算法设计与分析》第三次作业参考答案.pdf │ ├── 计算机学院《算法设计与分析》第二次作业参考答案.pdf │ └── 计算机学院《算法设计与分析》第四次作业参考答案.pdf ├── 1923 │ ├── Solution1.pdf │ ├── Solution2.pdf │ ├── Solution3.pdf │ └── Solution4.pdf └── 2006 │ ├── 6系-sol3-release.pdf │ ├── 6系_hw4_sol_release.pdf │ ├── 第一次算法作业-参考答案.pdf │ └── 第二次算法作业-参考答案.pdf ├── exam └── 2022-算法期末试卷.pdf ├── homework ├── 算法设计与分析week11第三次作业_使用模板_100.pdf ├── 算法设计与分析week11第四次作业_使用模板_100.pdf ├── 算法设计与分析week3第一次作业_使用模板_100.pdf └── 算法设计与分析week7第二次作业_使用模板_100.pdf └── homework_template_latex ├── Algorithm_Homework.aux ├── Algorithm_Homework.log ├── Algorithm_Homework.out ├── Algorithm_Homework.pdf ├── Algorithm_Homework.synctex.gz └── Algorithm_Homework.tex /LICENSE: -------------------------------------------------------------------------------- 1 | MIT License 2 | 3 | Copyright (c) 2022 YJK 4 | 5 | Permission is hereby granted, free of charge, to any person obtaining a copy 6 | of this software and associated documentation files (the "Software"), to deal 7 | in the Software without restriction, including without limitation the rights 8 | to use, copy, modify, merge, publish, distribute, sublicense, and/or sell 9 | copies of the Software, and to permit persons to whom the Software is 10 | furnished to do so, subject to the following conditions: 11 | 12 | The above copyright notice and this permission notice shall be included in all 13 | copies or substantial portions of the Software. 14 | 15 | THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR 16 | IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, 17 | FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE 18 | AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER 19 | LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, 20 | OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE 21 | SOFTWARE. 22 | -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # BUAA-Algorithm-2022 作业 2 | 3 | 4 | 5 | ![license: MIT](https://img.shields.io/badge/license-MIT-green) 6 | 7 | ## 课程简介 8 | 9 | 《算法设计与分析》是北航计算机学院大三上学期核心专业课之一。本课程不要求具体代码的实现（如C/Java等），重在算法设计、复杂度分析、证明、伪代码设计等。 10 | 11 | ## 课程作业简介 12 | 13 | 实验部分分为四次作业，以下简介每次作业的要求： 14 | 15 | - 第一次作业 ：时间复杂度分析+分治算法 16 | - 第二次作业 ：动态规划 17 | - 第三次作业 ：贪心算法 18 | - 第四次作业 ：P与NP理论+图论 19 | 20 | ## 仓库简介 21 | 22 | ### 作业完成情况 23 | 24 | 四次作业均100/100。 25 | 26 | ### 仓库结构介绍 27 | 28 | - answer：部分年份的标准答案 29 | - exam：2022年试题 30 | - homework：笔者2022年的作业批改发回版，有助教的分数和批注 31 | - 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for UK/EU standard size 10 | \usepackage[a4paper,top=1.5cm,bottom=1.5cm,left=2cm,right=2cm,marginparwidth=1.75cm]{geometry} 11 | 12 | % Useful packages 13 | \usepackage{amsmath} 14 | \usepackage{amssymb} 15 | % 引入{amssymb}包，避免输入\nexists(存在否定)这个符号时 报未定义符号的错误 16 | \usepackage{graphicx} 17 | \usepackage{listings} 18 | \usepackage[ruled]{algorithm2e} 19 | \usepackage[colorlinks=true, allcolors=blue]{hyperref} 20 | 21 | \SetKwProg{Function}{function}{:}{end} 22 | 23 | % TODO 个人信息参数全局变量 24 | % -------------- define personal variables begin -------------- 25 | \newcommand\myCollege{[college name]} 26 | \newcommand\myId{[author id]} 27 | \newcommand\myName{[author name]} 28 | \newcommand\myHomeworkId{[homework id]} 29 | % -------------- define personal variables end -------------- 30 | 31 | \title{\heiti \myCollege《算法设计与分析》第\myHomeworkId 次作业} % 文章标题 32 | \author{\myCollege \quad \myId \quad \myName} % 作者信息 33 | 34 | \begin{document} 35 | \maketitle 36 | 37 | \section{[作业题目1]} % TODO 作业题目1名字 38 | 39 | \subsection{[作业题目1.1]} % TODO 作业题目1.1名字 40 | 41 | % TODO 大括号样例 42 | 43 | $$ 44 | T(n)=\left\{ 45 | \begin{array}{l} 46 | 1\ ,\ n=1\\ 47 | T(n-2)+3n\ ,\ n>1 48 | \end{array} 49 | \right. 50 | $$ 51 | 52 | $T(n)=T(n-2)+3n=T(n-4)+3n+3(n-2)=T(n-6)+3n+3(n-2)+3(n-4)=...$ 53 | 54 | \qquad\ $=3n+3(n-2)+3(n-4)+...+9+1=\frac{1}{2}(3n+9)(\frac{3n-9}{6}+1)+1=O(n^2)$ 55 | 56 | 由此可知，原式渐进上界为$O(n^2)$。 57 | 58 | 59 | 60 | \qquad$n=1$时，对$c\ge1$结论显然成立； 61 | 62 | \qquad$n>1$时，有：$T(n)=T(n/2)+2^n$ 63 | 64 | \qquad\qquad\qquad\qquad\qquad\quad$\le{c}2^\frac{n}{2}+2^n$ 65 | 66 | \qquad\qquad\qquad\qquad\qquad\quad$=c2^n-(c-1)2^n+c2^\frac{n}{2}$ 67 | 68 | \qquad\qquad\qquad\qquad\qquad\quad$=c2^n-((c-1)2^\frac{n}{2}-c)2^\frac{n}{2}$ 69 | 70 | \qquad\qquad\qquad\qquad\qquad\quad$\le{c}2^n-(2(c-1)-c)2^\frac{n}{2}$ 71 | 72 | \qquad\qquad\qquad\qquad\qquad\quad$=c2^n-(c-2)2^\frac{n}{2}$ 73 | 74 | 显然，只要$c\ge2$，就有$T(n)\le{c2^n}$成立，从而原式渐进上界为$O(2^n)$。 75 | 76 | \subsection{[作业题目1.2]} % TODO 作业题目1.2名字 77 | 78 | $$ 79 | T(n)=\left\{ 80 | \begin{array}{l} 81 | 1\ ,\ n=1\\ 82 | 8T(n/4)+2n\ ,\ n>1 83 | \end{array} 84 | \right. 85 | $$ 86 | 87 | 使用主方法计算渐进上界，注意到$2n=O(n)$，由上式可知$a=8,b=4,d=1$，由$d 240 | #include 241 | using namespace std; 242 | 243 | struct Node 244 | { 245 | int l,r,len; 246 | 247 | inline Node(int l,int r) 248 | { 249 | this->l=l,this->r=r; 250 | len=r-l+1; return; 251 | } 252 | 253 | inline bool operator<(const Node& rhs) const 254 | { 255 | if (this->lenlen>rhs.len) return false; 257 | return this->l>rhs.l; 258 | } 259 | 260 | inline bool operator>(const Node& rhs) const 261 | { 262 | if (this->len>rhs.len) return true; 263 | if (this->lenl pq; 276 | Node node(0,n-1); pq.push(node); 277 | for (int i=1,l,r;i<=n;i++) 278 | { 279 | Node node=pq.top(); pq.pop(); 280 | l=node.l,r=node.r; 281 | // printf("%d %d\n",l,r); 282 | int mid=(l+r)>>1; 283 | a[mid]=i; 284 | if (mid-1>=l) {Node newNode(l,mid-1); pq.push(newNode);} 285 | if (r>=mid+1) {Node newNode(mid+1,r); pq.push(newNode);} 286 | } 287 | for (int i=0;i 297 | #include 298 | using namespace std; 299 | 300 | int a[10005]; 301 | int tot,fst[10005],nxt[10005],val[10005]; 302 | 303 | void build(int l,int r) 304 | { 305 | nxt[++tot]=fst[r-l+1],fst[r-l+1]=tot; 306 | int mid=(l+r)>>1; val[tot]=mid; 307 | if (r>=mid+1) build(mid+1,r); 308 | if (mid-1>=l) build(l,mid-1); 309 | return; 310 | } 311 | 312 | int main() 313 | { 314 | int n,cnt=0; 315 | freopen("testcase.txt","r",stdin); 316 | scanf("%d",&n); 317 | build(0,n-1); 318 | for (int i=n;i>0;i--) 319 | for (int j=fst[i];j;j=nxt[j]) 320 | a[val[j]]=++cnt; 321 | for (int i=0;i