http://acm.hdu.edu.cn/showproblem.php?pid=1083

题意：有p门的课，每门课都有若干学生，现在要为每个课程分配一名课代表，每个学生只能担任一门课的课代表，如果每个课都能找到课代表，则输出"YES"，否则"NO"。

题解：二分图匹配

 

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=1000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int mp[N][N];
bool vis[N];
int match[N];
char str;
struct node{};

bool dfs(int u){//cout<<u<<endl;
    for(int i=1;i<=n;i++){
        if(vis[i]==0&&mp[u][i]){
            vis[i]=1;
            if(match[i]==-1||dfs(match[i])){
                match[i]=u;
                return 1;
            }
        }
    }
    return 0;
}

void maxmatch(){
    for(int i=1;i<=p;i++){
        memset(vis,0,sizeof(vis));
        if(!dfs(i))
            break;
        ans++;
    }
}
void init(){
    memset(mp,0,sizeof(mp));
    memset(match,-1,sizeof(match));
    ans=0;
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&p,&n);
        init();
        for(int i=1;i<=p;i++){
            scanf("%d",&k);
            for(int j=1;j<=k;j++){
                scanf("%d",&v);
                mp[i][v]=1;
            }
        }
        maxmatch();
        //cout<<ans<<endl;
        if(ans==p){
            cout<<"YES"<<endl;
        }else{
            cout<<"NO"<<endl;
        }
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}