我是用并查集，同时维护每个节点到根节点的奇偶性，2表示既可以奇数步到也可以偶数步到
#include <iostream>
#include <fstream>
using namespace std;
const int N = 1000001;
int case_num, n, k, a, b, c, ra;
int root[N];
int type_to_root[N];

int Find(int x)
{
	if(root[x] != root[root[x]])
	{
		int temp = root[x];
		root[x] = Find(root[x]);
		if(type_to_root[root[x]] == 2)
		{
			type_to_root[x] = 2;
			type_to_root[temp] = 2;
		}
		else
			type_to_root[x] = (type_to_root[x]+type_to_root[temp])%2;
	}
	return root[x];
}
void Union(int x, int y)
{
	int rx = Find(x); int ry = Find(y);
	if(rx != ry)
	{
		root[rx] = ry;
		if(type_to_root[rx] == 2 || type_to_root[ry] == 2)
		{
			type_to_root[rx] = 2;
			type_to_root[ry] = 2;
		}
		else
			type_to_root[rx] = (1+type_to_root[x]+type_to_root[y])%2;
	}
	else
	{
		if(type_to_root[x] == type_to_root[y])
		{
			type_to_root[rx] = 2;
		}
	}
}
int main()
{
	fstream fin("1227.txt");
	cin>>case_num;
	while(case_num--)
	{
		cin>>n>>k;
		for(int i = 0; i < N; i++)
		{
			root[i] = i;
			type_to_root[i] = 0;
		}
		for(int i = 0; i < n; i++)
		{
			cin>>a>>c;
			for(int j = 0; j < c; j++)
			{
				cin>>b;
				Union(a, b);
			}
		}
		if(k == 1)
		{
			cout<<"YES"<<endl;
		}
		else
		{
			bool test = true;
			cin>>a;
			ra = Find(a);
			for(int i = 1; i < k; i++)
			{
				cin>>b;
				if(ra != Find(b) || (type_to_root[ra] != 2 && (type_to_root[a]+type_to_root[b]) == 1) )
				{
					test = false;
					break;
				}
			}
			if(test)
				cout<<"YES"<<endl;
			else
				cout<<"NO"<<endl;
		}
	}
	return 0;
}