hdu 1054 Strategic Game 【匈牙利算法】

题目链接：http://acm.acmcoder.com/showproblem.php?pid=1054

题意：求无向图的最小顶点覆盖 = 最大匹配数 / 2;

代码：

#include <stdio.h>
#include <ctime>
#include <math.h>
#include <limits.h>
#include <complex>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <bitset>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <iostream>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <time.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>

using namespace std;

int n, m, k, num;
int p[1510][1510];
int book[1510];
int match[1510];
int t, tt;
int a, b;
char tmp;

bool dfs(int u)
{
    int i;
    for (i = 0; i < n; i++)
    {
        if (book[i] == 0 && p[u][i] == 1)
        {
            book[i] = 1;
            if (match[i] == 0 || dfs(match[i]))
            {
                match[i] = u;
                return true;
            }
        }
    }
    return false;
}

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        memset(p,0,sizeof(p));
        memset(match, 0, sizeof(match));

        for (int i = 0; i < n; i++)
        {
            scanf("%d%c%c%d%c",&m,&tmp,&tmp,&k,&tmp);
            for (int j = 1; j <= k; j++)
            {
                scanf("%d",&num);
                p[m][num] = 1;
                p[num][m] = 1;
            }
        }

        int ans = 0;

        for (int i = 0; i < n; i++)
        {
            memset(book, 0, sizeof(book));
            if (dfs(i))
                ans++;
        }
        cout << ans/2 << endl;
    }
    return 0;
}