hdu 1281 二分图最大匹配

很容易想到是最大匹配：行和列看做点，每一个可放置的位置看做边，先求一遍最大匹配，然后枚举删除每一条边，如果删除了某一条边发现求出来的最大匹配比之前少，则这条边对应的格点是”重要点“。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 101;
bool mp[N][N];
int mark[N];
bool visit[N];
int n, m, k;

int dfs( int u )
{
    for ( int i = 1; i <= m; i++ )
    {
        if ( !visit[i] && mp[u][i] )
        {
            visit[i] = 1;
            if ( mark[i] == -1 || dfs(mark[i]) )
            {
                mark[i] = u;
                return 1;
            }
        }
    }
    return 0;
}

inline int hungary()
{
    int res = 0;
    memset( mark, -1, sizeof(mark) );
    for ( int i = 1; i <= n; i++ )
    {
        memset( visit, 0, sizeof(visit) );
        res += dfs(i);
    }
    return res;
}

int main()
{
    int _case = 1;
    while ( scanf("%d%d%d", &n, &m, &k) != EOF )
    {
        memset( mp, 0, sizeof(mp) );
        while ( k-- )
        {
            int x, y;
            scanf("%d%d", &x, &y);
            mp[x][y] = 1;
        }
        int cnt = 0, ans = hungary();
        for ( int i = 1; i <= n; i++ )
        {
            for ( int j = 1; j <= m; j++ )
            {
                if ( !mp[i][j] ) continue;
                mp[i][j] = 0;
                if ( hungary() < ans ) cnt++;
                mp[i][j] = 1;
            }
        }
        printf("Board %d have %d important blanks for %d chessmen.
", _case++, cnt, ans);
    }
    return 0;
}