题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3718
题意：有一堆答题情况和正确答案，问每一个答题情况的正确率最大是多少。
给每一对答案和答题情况的字母做映射，每次映射权值+1，这样会构造出一个二分图。在这个二分图上做最大匹配结果除以题目数量就是正确率。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int maxm = 10100; 5 const int maxn = 80; 6 const int inf = 0x3f3f3f3f; 7 int n, m, k; 8 int nx,ny; 9 int G[maxn][maxn]; 10 int linker[maxn],lx[maxn],ly[maxn]; 11 int slack[maxn]; 12 bool visx[maxn],visy[maxn]; 13 char ans[maxm]; 14 char stu[maxm]; 15 16 bool dfs(int x) { 17 visx[x] = true; 18 for(int y = 0; y < ny; y++) { 19 if(visy[y])continue; 20 int tmp = lx[x] + ly[y] - G[x][y]; 21 if(tmp == 0) { 22 visy[y] = true; 23 if(linker[y] == -1 || dfs(linker[y])) { 24 linker[y] = x; 25 return true; 26 } 27 } 28 else if(slack[y] > tmp) 29 slack[y] = tmp; 30 } 31 return false; 32 } 33 int km() { 34 memset(linker,-1,sizeof(linker)); 35 memset(ly,0,sizeof(ly)); 36 for(int i = 0;i < nx;i++) { 37 lx[i] = -inf; 38 for(int j = 0;j < ny;j++) 39 if(G[i][j] > lx[i]) 40 lx[i] = G[i][j]; 41 } 42 for(int x = 0;x < nx;x++) { 43 for(int i = 0;i < ny;i++) 44 slack[i] = inf; 45 while(true) { 46 memset(visx,false,sizeof(visx)); 47 memset(visy,false,sizeof(visy)); 48 if(dfs(x))break; 49 int d = inf; 50 for(int i = 0;i < ny;i++) 51 if(!visy[i] && d > slack[i]) 52 d = slack[i]; 53 for(int i = 0;i < nx;i++) 54 if(visx[i]) 55 lx[i] -= d; 56 for(int i = 0;i < ny;i++) { 57 if(visy[i])ly[i] += d; 58 else slack[i] -= d; 59 } 60 } 61 } 62 int res = 0; 63 for(int i = 0;i < ny;i++) 64 if(linker[i] != -1) 65 res += G[linker[i]][i]; 66 return res; 67 } 68 69 int main() { 70 // freopen("in", "r", stdin); 71 int T; 72 scanf("%d", &T); 73 nx = ny = 26; 74 while(T--) { 75 scanf("%d%d%d",&n,&k,&m); 76 for(int i = 0; i < n; i++) { 77 cin >> ans[i]; 78 } 79 for(int _ = 0; _ < m; _++) { 80 for(int i = 0; i < n; i++) { 81 cin >> stu[i]; 82 } 83 memset(G, 0, sizeof(G)); 84 for(int i = 0; i < n; i++) { 85 G[stu[i]-'A'][ans[i]-'A']++; 86 } 87 printf("%.4lf ", (double)km()/(double)n); 88 } 89 } 90 91 return 0; 92 }