L2-1 分而治之 (25分)

L2-1 分而治之 (25分)

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<iomanip>
#include<stack>
using namespace std;
#define STDIN freopen("in.in", "r", stdin);freopen("out.out", "w", stdout);


const int N = 10000;
int h[N], e[N<<1], ne[N<<1], idx;

int n, m;
void add(int a, int b)
{
    e[idx] = b;
    ne[idx] = h[a];
    h[a] = idx++;
}

bool solve(set<int> &se)
{
    for (int k = 1; k <= n; k++)
    {
        if (se.find(k) == se.end())
            for (int i = h[k]; ~i; i = ne[i])
            {
                int j = e[i];
                if (se.find(j) == se.end()){
                    return false;
                }
            }
    }
    return true;
}

int main()
{
//     STDIN
    cin >> n >> m;
    for (register int i = 0; i <= n; i++) h[i] = -1;
    for (register int i = 1; i <= m; i++)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        add(a, b), add(b, a);
    }
    int k; 
    cin >> k;
    for (register int i = 1; i <= k; i++)
    {
        set<int> se;
        int t; cin >> t;
        for (register int i=1; i<= t;i++)
        {
            int x;
            scanf("%d", &x);
            se.insert(x);
        }
        if (solve(se)) puts("YES");
            else puts("NO");

    }

}