Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1800 Accepted Submission(s): 606
Special Judge
/** 题意:给出一个n*n的矩阵,然后进行行列变换,使得矩形的对角线上都是1 做法:二分图匹配匈牙利 **/ #include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> #include<cmath> #include<vector> using namespace std; #define maxn 110 int g[maxn][maxn]; int linker[maxn]; bool vis[maxn]; vector<int>G[maxn]; int n; int dfs(int u) { for(int i=0;i<(int)G[u].size();i++) { int v = G[u][i]; if(!vis[v]) { vis[v] = true; if(linker[v] == -1 || dfs(linker[v])) { linker[v] = u; return 1; } } } return 0; } int hungary() { int res =0; memset(linker,-1,sizeof(linker)); for(int i=0;i<n;i++) { memset(vis,false,sizeof(vis)); res += dfs(i); } return res; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif // ONLINE_JUDGE while(~scanf("%d",&n)) { for (int i = 0; i <= n; i++) G[i].clear(); for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { scanf("%d", &g[i][j]); if (g[i][j]) G[i].push_back(j); } } int res = hungary(); if(res == n) { printf("%d\n",res); for(int i=0;i<n;i++) { printf("R %d %d\n",linker[i]+1,i+1); for(int j=0;j<n;j++) { if(linker[j] == i ) linker[j] = linker[i]; } } } else printf("-1\n"); } return 0; }
原文:http://www.cnblogs.com/chenyang920/p/4392665.html