POJ-1274 The Perfect Stall 二分匹配

简单二分匹配。

代码如下：

#include <cstring>
#include <cstdlib>
#include <cstdio>
#define MAXN 205
using namespace std;

int N, M, G[MAXN][MAXN];

int visit[MAXN], marry[MAXN];

int path(int u)
{
    for (int i = 1; i <= M; ++i) {
        if (!G[u][i] || visit[i]) {
            continue;
        }
        visit[i] = 1;
        if (!marry[i] || path(marry[i])) {
            marry[i] = u;
            return 1;
        }
    }
    return 0;
}

int main()
{
    int k, c, cnt;
    while (scanf("%d %d", &N, &M) == 2) {
        cnt = 0;
        memset(G, 0, sizeof (G));
        memset(marry, 0, sizeof (marry));
        for (int i = 1; i <= N; ++i) {
            scanf("%d", &k);
            for (int j = 0; j < k; ++j) {
                scanf("%d", &c);
                G[i][c] = 1;
            }
        }
        for (int i = 1; i <= N; ++i) {
            memset(visit, 0, sizeof (visit));
            if (path(i)) {
                ++cnt;
            }
        }
        printf("%d\n", cnt);
    }
    return 0;
}