POJ 2594 Treasure Exploration（最小路径覆盖 传递闭包）

版权声明：get busy living...or get busy dying https://blog.csdn.net/qq_41444888/article/details/89301696

最小路径覆盖 == 顶点数 - 配对数，此时每个顶点只能经过一次，若是经过一个点多次，例如下面博客所介绍的，当路径相交的时候，应该进行一次传递闭包处理，也就是跑一遍Floyd再进行最大匹配

讲解博客：http://www.cnblogs.com/ka200812/archive/2011/07/31/2122641.html

题目链接：https://vjudge.net/problem/POJ-2594

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
#define ll long long
#define mod 1000000007
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 505;
int boy[maxn];//记录被选着的编号
bool vis[maxn];
int g[maxn][maxn];
int n, m, k;
int dfs(int u)
{
    for(int i = 1; i <= n; i ++)//枚举被选者
    {
        if(g[u][i] && ! vis[i])
        {
            vis[i] = 1;
            if(boy[i] == 0 || dfs(boy[i]))
            {
                boy[i] = u;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    while(scanf("%d%d", &n, &m) != EOF && (n + m))
    {
        memset(boy, 0, sizeof(boy));
        memset(vis, 0, sizeof(vis));
        memset(g, 0, sizeof(g));
        int u, v;
        for(int i = 1; i <= m; i ++)
        {
            scanf("%d%d", &u, &v);
            g[u][v] = 1;
        }
        for(int k = 1; k <= n; k ++)
            for(int i = 1; i <= n; i ++)
                for(int j = 1; j <= n; j ++)
                    if(g[i][k] && g[k][j])
                        g[i][j] = 1;
        int ans = 0;
        for(int i = 1; i <= n; i ++)
        {
            memset(vis, 0, sizeof(vis));
            if(dfs(i)) ans ++;
        }
        printf("%d\n", n - ans);
    }
    return 0;
}