nyoj 游戏高手的烦恼 (二分图最小点覆盖)
还是想半天都没想明白。。 做得不多不熟，所以也联系不起来。
二分图最小点覆盖= 二分图的匹配数 详细请看某周的hihocoder
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<string> 6 #include<queue> 7 #include<algorithm> 8 #include<map> 9 #include<iomanip> 10 #include<climits> 11 #include<string.h> 12 #include<cmath> 13 #include<stdlib.h> 14 #include<vector> 15 #include<set> 16 #define INF 1e7 17 #define MAXN 100010 18 #define maxn 50 19 #define maxm 1000 20 #define Mod 1000007 21 using namespace std; 22 typedef long long LL; 23 24 25 int T; 26 int u, v; 27 int n, m; 28 int ans; 29 vector<int> G[555]; 30 int vis[10010]; 31 int match[10010]; 32 33 bool path(int u) 34 { 35 for (int i = 0; i < G[u].size(); ++i) { 36 int v = G[u][i]; 37 if (true == vis[v]) continue; 38 vis[v] = true; 39 if (match[v] == -1 || path(match[v])) { 40 match[v] = u; 41 return true; 42 } 43 } 44 return false; 45 } 46 47 void hungarian() 48 { 49 ans = 0; 50 memset(match,-1,sizeof(match)); 51 for (int i = 1; i <= n; ++i) { 52 memset(vis,0,sizeof(vis)); 53 if (path(i)) ans++; 54 } 55 } 56 57 void run() 58 { 59 scanf("%d%d",&n,&m); 60 for (int i = 0; i <= n; ++i) { 61 G[i].clear(); 62 } 63 for (int i = 0; i < m; ++i) { 64 scanf("%d%d",&u,&v); 65 G[u].push_back(v); 66 } 67 hungarian(); 68 printf("%d\n",ans); 69 70 } 71 int main() 72 { 73 scanf("%d", &T); 74 while (T--) 75 run(); 76 return 0; 77 }
