hdu1281---棋盘游戏

以每一个可以放车的点的行列坐标为点，建图，然后求最大匹配
接下来，删边，观察最大匹配是否改变，如果改变，这个点就是关键点

/*************************************************************************
    > File Name: hdu1281.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月16日 星期一 18时01分44秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

const int N = 110;
int edge[N][N];
int un, vn, n, m;
int match[N];
bool used[N];
struct node
{
    int x, y;
}imp[N * N];

bool dfs (int u)
{
    for (int i = 1; i <= vn; ++i)
    {
        if (edge[u][i] && !used[i])
        {
            used[i] = 1;
            if (match[i] == -1 || dfs (match[i]))
            {
                match[i] = u;
                return 1;
            }
        }
    }
    return 0;
}

int hungry ()
{
    memset (match, -1, sizeof(match));
    int ans = 0;
    for (int i = 1; i <= un; ++i)
    {
        memset (used, 0, sizeof(used));
        if (dfs(i))
        {
            ++ans;
        }
    }
    return ans;
}

int main ()
{
    int k;
    int icase = 1;
    while (~scanf("%d%d%d", &n, &m, &k))
    {
        memset (edge, 0, sizeof(edge));
        for (int i = 1; i <= k; ++i)
        {
            scanf("%d%d", &imp[i].x, &imp[i].y);
            edge[imp[i].x][imp[i].y] = 1;
        }
        un = n;
        vn = m;
        int ans = hungry();
        int ret = 0;
        for (int i = 1; i <= k; ++i)
        {
            edge[imp[i].x][imp[i].y] = 0;
            if (hungry() != ans)
            {
                ++ret;
            }
            edge[imp[i].x][imp[i].y] = 1;
        }
        printf("Board %d have %d important blanks for %d chessmen.\n", icase++, ret, ans);
    }
    return 0;
}