POJ-2239 Selecting Courses 二分匹配

简单二分匹配

代码如下：

#include <stdio.h>
#include <string.h>
#define MAXN 100

int G[MAXN][305], N;
int marry[305], visit[305];

int path(int u)
{
for (int i = 1; i <= N; ++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 t, d, h, count = 0;
while (scanf("%d", &N) == 1) {
        count = 0;
        memset(G, 0, sizeof (G));
        memset(marry, 0, sizeof (marry));
for (int i = 1; i <= N; ++i) {
            scanf("%d", &t);
for (int j = 0; j < t; ++j) {
                scanf("%d %d", &d, &h);
                G[(d-1)*12+h][i] = 1;
            }
        }
for (int i = 1; i <= 84; ++i) {
            memset(visit, 0, sizeof (visit));
if (path(i))
                ++count;
        }
        printf("%d\n", count);
    }
return 0;
}