hdu 1054 Strategic Game 最小点覆盖 = 最大二分匹配

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1054

 

简单二分匹配，根据题意构造一个无向图。然后求最小点覆盖，然后扫描mark数组将曾经匹配的点所匹配的边消去。

最小点覆盖 = 最大二分匹配

#include<stdio.h>
#include<string.h>
#define maxn 1505
struct node
{int v;node *next;
}*head[maxn],edge[maxn*maxn],*p;
int n,vis[maxn],mark[maxn];
bool dfs(int v)
{for(node *p = head[v]; p ; p = p->next){if(vis[p->v])continue;vis[p->v] = 1;if(mark[p->v] == -1 || dfs(mark[p->v])){mark[p->v] = v;return true;}}return false;
}
int main()
{int num,v,u,i,j;while(scanf("%d",&n) != EOF){memset(head,NULL,sizeof(head));p = edge;for(i = 0; i < n; i++){scanf("%d:(%d)",&u,&num);for(j = 0; j < num; j++){scanf("%d",&v);p->v = v;p->next = head[u];head[u] = p++;p->v = u;p->next = head[v];head[v] = p++;}}memset(mark,-1,sizeof(mark));int ans = 0;for(i = 0; i < n; i++){memset(vis,0,sizeof(vis));if(dfs(i))ans++;}//减去已有的边。。。for(i = 0; i < n; i++){if(mark[i] != -1){for(j = i+1; j < n; j++)if(mark[j] == i){mark[j] = -1;ans--;}}}printf("%d\n",ans);}return 0;
}