├── LICENSE ├── README.md ├── homework ├── MIT Dynamic Programming II.pdf ├── homework_4_problem.pdf ├── homework_5_problem.pdf ├── hw4_solution.md ├── hw4_solution.pdf ├── hw5_solution.md └── hw5_solution.pdf ├── 实验1-棋盘覆盖 ├── Chessboard.cpp └── 实验1报告.pdf ├── 实验2-矩阵连乘问题 ├── MatrixMulti.cpp ├── 实验2、矩阵连乘问题.doc └── 实验2报告.pdf ├── 实验3-最长公共子序列 ├── LCS.cpp ├── 实验3、最长公共子序列.doc └── 实验3报告.pdf ├── 实验4-哈夫曼编码 ├── Huffman.cpp ├── 实验4、哈夫曼编码.doc └── 实验4报告.pdf ├── 实验5-Dijkstra算法 ├── Dijkstra.cpp ├── 实验5、Dijkstra算法.doc └── 实验5报告.pdf ├── 实验6-跳马问题 ├── Horse.cpp ├── 实验6、跳马问题.doc └── 实验6报告.pdf └── 实验7-装载问题 ├── Load.cpp ├── 实验7、装载问题.doc └── 实验7报告.pdf /LICENSE: -------------------------------------------------------------------------------- 1 | GNU GENERAL PUBLIC LICENSE 2 | Version 3, 29 June 2007 3 | 4 | Copyright (C) 2007 Free Software Foundation, Inc. 5 | Everyone is permitted to copy and distribute verbatim copies 6 | of this license document, but changing it is not allowed. 7 | 8 | Preamble 9 | 10 | The GNU General Public License is a free, copyleft license for 11 | software and other kinds of works. 12 | 13 | The licenses for most software and other practical works are designed 14 | to take away your freedom to share and change the works. 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It is safest 630 | to attach them to the start of each source file to most effectively 631 | state the exclusion of warranty; and each file should have at least 632 | the "copyright" line and a pointer to where the full notice is found. 633 | 634 | 635 | Copyright (C) 636 | 637 | This program is free software: you can redistribute it and/or modify 638 | it under the terms of the GNU General Public License as published by 639 | the Free Software Foundation, either version 3 of the License, or 640 | (at your option) any later version. 641 | 642 | This program is distributed in the hope that it will be useful, 643 | but WITHOUT ANY WARRANTY; without even the implied warranty of 644 | MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the 645 | GNU General Public License for more details. 646 | 647 | You should have received a copy of the GNU General Public License 648 | along with this program. If not, see . 649 | 650 | Also add information on how to contact you by electronic and paper mail. 651 | 652 | If the program does terminal interaction, make it output a short 653 | notice like this when it starts in an interactive mode: 654 | 655 | Copyright (C) 656 | This program comes with ABSOLUTELY NO WARRANTY; for details type `show w'. 657 | This is free software, and you are welcome to redistribute it 658 | under certain conditions; type `show c' for details. 659 | 660 | The hypothetical commands `show w' and `show c' should show the appropriate 661 | parts of the General Public License. Of course, your program's commands 662 | might be different; for a GUI interface, you would use an "about box". 663 | 664 | You should also get your employer (if you work as a programmer) or school, 665 | if any, to sign a "copyright disclaimer" for the program, if necessary. 666 | For more information on this, and how to apply and follow the GNU GPL, see 667 | . 668 | 669 | The GNU General Public License does not permit incorporating your program 670 | into proprietary programs. If your program is a subroutine library, you 671 | may consider it more useful to permit linking proprietary applications with 672 | the library. If this is what you want to do, use the GNU Lesser General 673 | Public License instead of this License. But first, please read 674 | . 675 | -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # algorithm 2 | ### SHU算法设计与分析课程实验
3 | ### 含代码实现与报告
4 | ### 题目描述见word文档
5 | ### 报告里对解法作了简单陈述 6 | -------------------------------------------------------------------------------- /homework/MIT Dynamic Programming II.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/RuoShui66/algorithm/934cc8fd142cefa640c35ad0018486a1004f5eca/homework/MIT Dynamic Programming II.pdf -------------------------------------------------------------------------------- /homework/homework_4_problem.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/RuoShui66/algorithm/934cc8fd142cefa640c35ad0018486a1004f5eca/homework/homework_4_problem.pdf -------------------------------------------------------------------------------- /homework/homework_5_problem.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/RuoShui66/algorithm/934cc8fd142cefa640c35ad0018486a1004f5eca/homework/homework_5_problem.pdf -------------------------------------------------------------------------------- /homework/hw4_solution.md: -------------------------------------------------------------------------------- 1 | ## 算法作业4 2 | 3 | ### 第1题 最大m子段和问题 4 | 5 | 定义`dp[i][j]`为已经分了 j 段，且最后一个数字下标为 i 的最大结果 6 | 7 | 考虑两种情况的转移 8 | 9 | - 从该数字开始，开启一个新子段 10 | - 该数字加入前面的值最大的最后一个子段 11 | 12 | **时间复杂度分析**：程序主体复杂度上限为状态转移的for循环，第一层执行次数为m，第二层执行次数为n，所以复杂度为O(m*n) 13 | 14 | ```c++ 15 | const int N=5e3+10,inf=-1e9; 16 | int n,m; 17 | int dp[N][N],a[N]; 18 | void solve(){ 19 | for (int i=1; i<=m; i++) { 20 | for (int j=1; j<=n; j++) { 21 | dp[j][i]=inf; 22 | } 23 | } 24 | int ans=inf; 25 | for (int i=1; i<=m; i++) { 26 | int mx=inf; 27 | for (int j=1; j<=n; j++) { 28 | dp[j][i]=dp[j-1][i]+a[j];//情况1 29 | if (j>=i) { 30 | mx=max(dp[j-1][i-1],mx); 31 | dp[j][i]=max(dp[j][i],mx+a[j]);//情况2 32 | } 33 | } 34 | } 35 | for (int i=1; i<=n; i++) { 36 | ans=max(ans,dp[i][m]);//所有a[i]结尾的子序列且分了m段中的最大值 37 | } 38 | } 39 | ``` 40 | 41 | ### 第 2 题 交替硬币游戏 42 | 43 | 采用记忆化搜索的形式进行动态规划，定义`dp[i][j]`为对于区间i～j的最大价值，那么最后`dp[1][n]`即为先手所能取得的最大值，若该最大值比后手值大，则决定先手，否则后手 44 | 45 | 考虑状态转移，对于当前i～j的区间，我方有两种选择，即拿走最左边的或拿走最右边的，对方也可以选择拿走最左边的或拿走最右边的，总共有四种转移的情况，对四种值取最大值即位当前状态的最大值。 46 | 47 | **时间复杂度分析**：因每个状态计算时，从它的四个子状态取最大值转移过来，对于相同的状态，采用了记忆化避免重复计算，所以时间复杂度就是状态总数，即O(n^2^) 48 | 49 | ```c++ 50 | int n; 51 | int dp[N][N],v[N]; 52 | int dfs(int l,int r){ 53 | int &val=dp[l][r]; 54 | if (val!=-1) { 55 | return val; 56 | } 57 | if (l==r) { 58 | return val=v[l]; 59 | } 60 | if (l>r) { 61 | return 0; 62 | } 63 | int mx=0; 64 | mx=max({v[l]+dfs(l+1, r-1),v[l]+dfs(l+2, r),v[r]+dfs(l+1, r-1),v[r]+dfs(l, r-2)}); 65 | return val=mx; 66 | } 67 | void solve(){ 68 | memset(dp, -1, sizeof dp); 69 | int sum=0; 70 | for (int i=1; i<=n; i++) { 71 | sum+=v[i]; 72 | } 73 | int first=dfs(1,n); 74 | int second=sum-first; 75 | if (first>second) { 76 | cout<<"先手"; 77 | } 78 | else cout<<"后手"; 79 | } 80 | ``` 81 | 82 | ### 第3题 编辑距离 83 | 84 | 定义`dp[i][j]`为将a中1-i的子串变成b中1-j的子串的最小操作次数 85 | 86 | 初始化： 87 | 88 | `dp[0][i]`如果a初始长度就是0，那么只能用插入操作让它变成b 89 | `dp[i][0]`同样地，如果b的长度是0，那么a只能用删除操作让它变成b 90 | 91 | 状态转移： 92 | 93 | - a[i]删掉之后a[1-i]和b[1-j]匹配,所以之前要先做到a[1-(i-1)]和b[1-j]匹配， 94 | 95 | 所以`dp[i][j]=dp[i-1][j]+1` 96 | 97 | - 插入之后a[i]与b[j]完全匹配，所以插入的就是b[j],那填之前a[1-i]和b[1-(j-1)]匹配 98 | 99 | `dp[i][j]=dp[i][j-1]+1` 100 | 101 | - 把a[i]改成b[j]之后想要a[1-i]与b[1-j]匹配,那么修改这一位之前，a[1-(i-1)]应该与b[1-(j-1)]匹配`dp[i][j]=dp[i-1][j-1] + 1` 102 | 103 | 但是如果本来a[i]与b[j]这一位上就相等，那么不用改，即`dp[i][j]=dp[i-1][j-1]` 104 | 105 | **时间复杂度分析**：分析程序主体代码可知，复杂度瓶颈在于计算所有的状态，而对于每一个状态的计算，它可以用三种之前的状态转移过来，该转移是近似O(1)的，所以总时间复杂度是O(nm) 106 | 107 | ```c++ 108 | string a,b; 109 | void solve(){ 110 | n=a.size(); 111 | m=b.size(); 112 | for(int i=0,j=0;j<=m;j++) dp[i][j]=j;//a为空，只用插入操作 113 | for(int i=0,j=0;i<=n;i++) dp[i][j]=i;//b为空，只用删除操作 114 | for (int i=1; i<=n; i++) { 115 | for (int j=1; j<=m; j++) { 116 | dp[i][j]=min(dp[i-1][j]+1,dp[i][j-1]+1); 117 | if (a[i-1]==b[j-1]) { 118 | dp[i][j]=min(dp[i-1][j-1],dp[i][j]); 119 | } 120 | else dp[i][j]=min(dp[i-1][j-1]+1,dp[i][j]); 121 | } 122 | } 123 | cout<width) { 24 | return inf; 25 | } 26 | return pow((width-len), 3); 27 | } 28 | void solve(){ 29 | memset(dp, 0x3f, sizeof dp); 30 | dp[n+1]=0; 31 | for (int i=n; i>=1; i--) { 32 | for (int j=i+1; j<=n+1; j++) { 33 | if (dp[i]>n; 64 | for (int i=1; i<=n; i++) { 65 | cin>>s[i]>>f[i]; 66 | } 67 | vector> v; 68 | for (int i=1; i<=n; i++) { 69 | v.push_back({s[i],1}); 70 | v.push_back({f[i],0}); 71 | } 72 | sort(v.begin(), v.end()); 73 | int mx=0,cnt=0; 74 | for (auto [x,id]: v) { 75 | if (id==0) { 76 | cnt--; 77 | } 78 | else cnt++; 79 | mx=max(mx,cnt); 80 | } 81 | cout<`来维护，用`upper_bound()`找到大于当前数的第一个，在它之前的则为小于等于它最大的（若它前面已经没有元素，则代表分配失败）。时间复杂度O(nlogn)，所以总时间复杂度是O(nlogn) 91 | 92 | ```c++ 93 | const int N=2e5+10; 94 | int n,pos[N]; 95 | set s; 96 | struct node{ 97 | int ddl; 98 | int penalty,id; 99 | bool operator < (const node &b)const{ 100 | if (penalty==b.penalty) { 101 | return ddlb.penalty; 104 | } 105 | }task[N]; 106 | void align(int x){ 107 | int dead=task[x].ddl; 108 | auto it=s.upper_bound(dead); 109 | if (it==s.begin()) { 110 | return; 111 | } 112 | it=prev(it); 113 | pos[x]=*it; 114 | s.erase(it); 115 | } 116 | 117 | void solve(){ 118 | cin>>n; 119 | for (int i=1; i<=n; i++) { 120 | s.insert(i); 121 | } 122 | for (int i=1; i<=n; i++) { 123 | cin>>task[i].ddl; 124 | task[i].id=i; 125 | } 126 | for (int i=1; i<=n; i++) { 127 | cin>>task[i].penalty; 128 | } 129 | sort(task+1, task+1+n); 130 | memset(pos, -1, sizeof pos); 131 | for (int i=1; i<=n; i++) { 132 | align(i); 133 | } 134 | int ans=0; 135 | for (int i=1; i<=n; i++) { 136 | if (pos[i]==-1) { 137 | ans+=task[i].penalty; 138 | cout<<"第"< 2 | using namespace std; 3 | int board[1100][1100]; 4 | int id = 1; 5 | 6 | void fillboard(int x,int y,int tx,int ty,int size)//(x,y)为当前棋盘的左上角,(tx,ty)为特殊方格所在的坐标 7 | //size为当前棋盘的宽度 8 | { 9 | if(size==1) return ;//递归边界 10 | int t=id++;//骨牌号 11 | int len=size/2;//偏离距离 12 | //覆盖左上角子棋盘 13 | if(tx=y+len) 23 | fillboard(x,y+len,tx,ty,len);//特殊方格在此棋盘中 24 | else //此棋盘中无特殊方格，用t号L型骨牌覆盖左下角 25 | { 26 | board[x+len-1][y+len]=t; 27 | //覆盖其余方格 28 | fillboard(x,y+len,x+len-1,y+len,len); 29 | } 30 | //覆盖左下角子棋盘 31 | if(tx>=x+len && ty=x+len && ty>=y+len)//特殊方格在此棋盘中 41 | fillboard(x+len,y+len,tx,ty,len); 42 | else //此棋盘中无特殊方格，用t号L型骨牌覆盖左上角 43 | { 44 | board[x+len][y+len]=t; 45 | //覆盖其余方格 46 | fillboard(x+len,y+len,x+len,y+len,len); 47 | } 48 | } 49 | 50 | int main() 51 | { 52 | int i,j; 53 | int k,cas=0; 54 | while(cin>>k) 55 | { 56 | cas++; 57 | id=1; 58 | int size = 1<>x>>y; 61 | board[x][y]=0; 62 | fillboard(1, 1, x, y, size); 63 | cout<<"Case "< 2 | using namespace std; 3 | const int N=105; 4 | int n,dp[N][N],pos[N][N],p[N]; 5 | const int inf=(1<<30); 6 | int chain(int l,int r) 7 | { 8 | int &v=dp[l][r]; 9 | if(v>0)return v; 10 | if(l==r)return 0; 11 | int mn=inf; 12 | pos[l][r]=l;//pos[l][r]表示矩阵l到r在pos[l][r]后分开 13 | for(int k=l;k>n) { 44 | memset(dp, 0, sizeof dp); 45 | cas++; 46 | for(int i=0;i<=n;i++) 47 | cin>>p[i]; 48 | cout<<"Case "< 2 | #ifdef DEBUG 3 | #define debug(x) cerr<<__LINE__<<"行 [" << #x << "] = "< pii; 9 | int n,m; 10 | const int N=210,M=70,NPOS=-1;//NPOS为数组默认值，M为字母表长度，N为序列长度 11 | int func(char ch){ 12 | if (ch>='a'&&ch<='z') { 13 | return ch-'a'; 14 | } 15 | if (ch>='A'&&ch<='Z') { 16 | return ch-'A'+26; 17 | } 18 | if (ch>='0'&&ch<='9') { 19 | return ch-'0'+52; 20 | } 21 | return 0; 22 | } 23 | char func2(int x){ 24 | if (x>=0&&x<=25) { 25 | return 'a'+x; 26 | } 27 | if (x>25&&x<=51) { 28 | return 'A'+x-26; 29 | } 30 | if (x>=52) { 31 | return '0'+x-52; 32 | } 33 | return '0'; 34 | } 35 | class SequenceAM {// 序列自动机 Sequence Auto Machine 36 | public: 37 | int nxt[N][M]; // 核心数组,代表从第i个位置开始，字符j出现的第一个位置 38 | string s; // 当前串 39 | void init() { // 初始化序列自动机 40 | fill(nxt[s.size()], nxt[s.size()]+M, NPOS); 41 | for (int i=(int)s.size()-1; i>=0; i--) { // 倒序初始化 42 | memcpy(nxt[i], nxt[i+1], sizeof nxt[i]); 43 | nxt[i][func(s[i])]=i+1; // 更新数组 44 | } 45 | } 46 | }; 47 | SequenceAM a,b; 48 | int dp[N][N],len,id,cnt,path[N][N]; 49 | string ans,s1,s2; 50 | void dfs(int x,int y){ 51 | if (ans.size()==len) { 52 | cnt++; 53 | for (auto x: ans) { 54 | cout<=dp[i][j-1]){ 74 | dp[i][j]=dp[i-1][j]; 75 | path[i][j]=1; 76 | } 77 | else{ 78 | dp[i][j]=dp[i][j-1]; 79 | path[i][j]=2; 80 | } 81 | } 82 | } 83 | return dp[n][m]; 84 | } 85 | void solve(){ 86 | memset(dp, 0, sizeof dp); 87 | memset(path, 0, sizeof path); 88 | cnt=0; 89 | id++; 90 | cin>>n>>m; 91 | s1.resize(n); 92 | s2.resize(m); 93 | for (auto &x: s1) { 94 | cin>>x; 95 | } 96 | for (auto &x: s2) { 97 | cin>>x; 98 | } 99 | len=cal_len(); 100 | cout<<"Case "<>t; 129 | while (t--) { 130 | solve(); 131 | } 132 | return 0; 133 | } 134 | 135 | -------------------------------------------------------------------------------- /实验3-最长公共子序列/实验3、最长公共子序列.doc: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/RuoShui66/algorithm/934cc8fd142cefa640c35ad0018486a1004f5eca/实验3-最长公共子序列/实验3、最长公共子序列.doc -------------------------------------------------------------------------------- /实验3-最长公共子序列/实验3报告.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/RuoShui66/algorithm/934cc8fd142cefa640c35ad0018486a1004f5eca/实验3-最长公共子序列/实验3报告.pdf -------------------------------------------------------------------------------- /实验4-哈夫曼编码/Huffman.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | const int N = 110; 4 | int cas=1; 5 | struct node{ 6 | int val,id; 7 | vector leaves; 8 | bool operator < (const node & b) const{ //双关键字排序 9 | if(val==b.val) return id < b.id; 10 | return val > b.val; 11 | } 12 | }tr[N]; 13 | void solve(){ 14 | int n; 15 | cin>>n; 16 | vector res(N); 17 | priority_queue heap; 18 | for(int i=1,w;i<=n;++i){ 19 | node tmp; 20 | cin>>w; 21 | tr[i].val = w; 22 | tmp.val = w; 23 | tmp.id = i; 24 | tmp.leaves.push_back(i); 25 | heap.push(tmp); //初始将每个结点视作子树 26 | } 27 | int idx = n+1; 28 | while(!heap.empty()){ 29 | node r = heap.top(); 30 | heap.pop(); 31 | if(heap.empty()) 32 | break; 33 | node l = heap.top(); 34 | heap.pop(); //取出权值最小的两个子树 35 | node t; 36 | t.val = r.val + l.val; //合并权值 37 | t.id = idx++; 38 | for(int i=0;i<(int)r.leaves.size();++i){//右子树 39 | int id = r.leaves[i]; 40 | res[id].push_back('1'); 41 | t.leaves.push_back(id);//把叶子结点加到新合成的节点中 42 | } 43 | for(int i=0;i < (int)l.leaves.size(); ++i){//左子树 44 | int id = l.leaves[i]; 45 | res[id].push_back('0'); 46 | t.leaves.push_back(id); 47 | } 48 | heap.push(t); 49 | } 50 | cout<<"Case "<>T; 60 | while(T--){ 61 | solve(); 62 | } 63 | return 0; 64 | } 65 | 66 | -------------------------------------------------------------------------------- /实验4-哈夫曼编码/实验4、哈夫曼编码.doc: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/RuoShui66/algorithm/934cc8fd142cefa640c35ad0018486a1004f5eca/实验4-哈夫曼编码/实验4、哈夫曼编码.doc -------------------------------------------------------------------------------- /实验4-哈夫曼编码/实验4报告.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/RuoShui66/algorithm/934cc8fd142cefa640c35ad0018486a1004f5eca/实验4-哈夫曼编码/实验4报告.pdf -------------------------------------------------------------------------------- /实验5-Dijkstra算法/Dijkstra.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | typedef pair pii; 4 | const int N=2e5+10; 5 | int n,m,dist[N],pre[N],cnt[N];//dist数组记录距离 pre数组记录路径前一个点 cnt数组记录路径上边的数量 6 | vector h[N]; 7 | 8 | bool cmp(int x,int y){//比较到达x和y的路径字典序 9 | vector a,b; 10 | a.push_back(x); 11 | b.push_back(y); 12 | while (pre[a.back()]) {//把前面的节点都存入vector中比较 13 | a.push_back(pre[a.back()]); 14 | b.push_back(pre[b.back()]); 15 | } 16 | //路径是从后往前得到的,所以反转之后比较 17 | reverse(a.begin(), a.end()); 18 | reverse(b.begin(), b.end()); 19 | for (int i=0; i<(int)a.size(); i++) { 20 | if (a[i]==b[i]) {//节点相等则继续比较 21 | continue; 22 | } 23 | if (a[i],greater> heap; 36 | heap.push({0,st}); 37 | dist[st]=0; 38 | cnt[st]=0; 39 | while (heap.size()) { 40 | auto [distance,cur]=heap.top(); 41 | heap.pop(); 42 | for (auto [x,w]: h[cur]) { 43 | if (distance+w>dist[x]) { 44 | continue;//路径长度更大则不用更新 45 | } 46 | if (distance+wcnt[cur]+1) {//如果路段数更少,则更新 58 | heap.push({dist[x],x}); 59 | pre[x]=cur; 60 | cnt[x]=cnt[cur]+1; 61 | continue; 62 | } 63 | //路段数相等,比较路径节点字典序 64 | if (cmp(cur,pre[x])) { 65 | heap.push({dist[x],x}); 66 | pre[x]=cur; 67 | } 68 | } 69 | } 70 | if (dist[ed]==0x3f3f3f3f) { 71 | return -1; 72 | } 73 | return dist[ed]; 74 | } 75 | 76 | void printsfp(int ed){ 77 | stack st; 78 | st.push(ed); 79 | while (pre[st.top()]!=0) { 80 | st.push(pre[st.top()]); 81 | } 82 | while (st.size()) { 83 | cout<"; 87 | } 88 | } 89 | } 90 | 91 | int main(){ 92 | int cas=1; 93 | while (cin>>n) { 94 | for (int i=1; i<=n; i++) { 95 | for (int j=1; j<=n; j++) { 96 | int w; 97 | cin>>w; 98 | if (w!=-1) { 99 | h[i].push_back({j,w}); 100 | } 101 | } 102 | } 103 | int st,ed; 104 | cin>>st>>ed; 105 | cout<<"Case "<"< 2 | using namespace std; 3 | const int N=1e3+10; 4 | int n=8; 5 | int dx[8]={1,1,2,2,-1,-1,-2,-2},dy[8]={2,-2,1,-1,2,-2,1,-1}; 6 | int dist[N][N]; 7 | bool state[N][N]; 8 | int solve(int x,int y,int ex,int ey){ 9 | queue> q; 10 | q.push({x,y}); 11 | memset(dist, -1, sizeof dist); 12 | memset(state, 0, sizeof state); 13 | dist[x][y]=0; 14 | state[x][y]=true; 15 | while (q.size()) { 16 | auto [x2,y2]=q.front(); 17 | q.pop(); 18 | for (int i=0; i<8; i++) { 19 | int tx=x2+dx[i],ty=y2+dy[i]; 20 | if (tx<1||tx>n||ty<1||ty>n||state[tx][ty]) { 21 | continue; 22 | } 23 | state[tx][ty]=true; 24 | q.push({tx,ty}); 25 | dist[tx][ty]=dist[x2][y2]+1; 26 | } 27 | } 28 | return dist[ex][ey]; 29 | } 30 | int main() 31 | { 32 | ios::sync_with_stdio(false); 33 | cin.tie(nullptr); 34 | string st,ed; 35 | while (cin>>st) { 36 | cin>>ed; 37 | int sx,sy,ex,ey; 38 | sx=st[1]-'0'; 39 | sy=st[0]-'a'+1; 40 | ex=ed[1]-'0'; 41 | ey=ed[0]-'a'+1; 42 | int dist=solve(sx,sy,ex,ey); 43 | cout<"< 2 | using namespace std; 3 | const int N=30; 4 | int n,w[N],c1,c2,sum; 5 | int bestw;//第一艘船能装的最大装载重量 6 | string best,cur;//最佳分配方案 当前分配方案 7 | void dfs(int u,int curw){ 8 | if (u==n+1) {//已经分配完 9 | if (sum-curw>c2) { 10 | return;//剩余货物第二艘船装不下 不合法的装配方案 11 | } 12 | if (curw>bestw) { 13 | bestw=curw; 14 | best=cur; 15 | } 16 | if (curw==bestw&&cur>best) {//如果存在多种最大的方案 选择字典序更大的 17 | best=cur; 18 | } 19 | return; 20 | } 21 | if (curw+w[u]<=c1) { 22 | cur[u]='1';//能选则选上该货物 23 | dfs(u+1, curw+w[u]); 24 | cur[u]='0';//回溯时恢复现场 25 | } 26 | dfs(u+1, curw);//不选该货物 27 | } 28 | 29 | int main(){ 30 | int cas=1; 31 | while (cin>>n) { 32 | best.resize(n+1, '0'); 33 | cur.resize(n+1, '0'); 34 | sum=0; 35 | bestw=0; 36 | for (int i=1; i<=n; i++) { 37 | cin>>w[i]; 38 | sum+=w[i]; 39 | } 40 | cin>>c1>>c2; 41 | dfs(1,0); 42 | cout<<"Case "<