二分匹配

#include<iostream>
using namespace std;
int g[333][333], vis[333], lin[333], n;

int find(int node)
{
      for (int i = 1; i <= n; i++)
         if (!vis[i] && g[node][i])
         {
            vis[i] = 1;
             if (!lin[i] || find(lin[i])) 
             {
                 lin[i] = node;
                  return 1;             
             }
        }
    return 0; 
}

int main()
{
    int cas, i, j, x, p;
    //freopen("D:\\input.txt","r",stdin);
    cin >> cas;
    while (cas--) 
    {        
        memset(g, 0, sizeof(g));
        cin >> p >> n;        
        for (i = 1; i <= p; i++) 
        {
            int m;
            cin >> m;
            for (j = 1, x; j <= m; j++)
                  cin >> x, g[i][x] = 1;
        }
        memset(lin, 0, sizeof lin);
        for (i = 1; i <= p; i++) 
        {
              memset(vis, 0, sizeof (vis));
             if (!find(i))
                 break;
         }
        if (i <= p)cout << "NO" << endl;
        else 
            cout << "YES" << endl;
    }
    return 0;
}