L2-025 divide and rule (25 points)

The title of this brush is 2021 Nian TIANTI

L2-025 Divide and rule (25 branch )

Divide and rule , It is one of the common strategies of the strategists to break each other . In the war , We hope to capture some of the enemy's cities first , Leave the rest of the city alone , Then break them separately . To this end, the staff has provided a number of strike programs . Please write a program for this question , Judge the feasibility of each plan .

Input format ：

Enter two positive integers on the first line N and M（ Not more than 10 000）, Number of enemy cities （ So the default city is from 1 To N Number ） And the number of roads connecting the two cities . And then M That's ok , Each line gives the number of the two cities connected by a road , Separated by a space . After the city information, give a series of plans of the staff , It's a positive integer K （≤ 100） And then K Line plan , Each line is given in the following format ：

Np v[1] v[2] ... v[Np]

among Np It's the number of cities planned to be conquered in this plan , Later series v[i] It's the number of the planned city .

Output format ：

For each package , Output if possible YES, Otherwise output NO.

sample input ：

10 11
8 7
6 8
4 5
8 4
8 1
1 2
1 4
9 8
9 1
1 10
2 4
5
4 10 3 8 4
6 6 1 7 5 4 9
3 1 8 4
2 2 8
7 9 8 7 6 5 4 2

sample output ：

NO
YES
YES
NO
NO

Answer key

Just need to beat the point Mark it

Then traverse 1…N Again , Look at their neighbors , If there are some adjacent points, you can go .

That is, this point can go to the next . So no

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int n = 10005;
vector<int> v[n];
int main()
{
    int N, M;
    cin >> N >> M;
    int x, y;
    while (M--)
    {
        cin >> x >> y;
        v[x].push_back(y);
        v[y].push_back(x);
    }
    cin >> M;
    while (M--)
    {
        int K;
        cin >> K;
        int dp[n] = {0};
        for (int i = 0; i < K; i++)
            cin >> x, dp[x]++;
        int t = 0;
        for (int i = 1; i <= N; i++)
            if (!dp[i])
                for (auto &e : v[i])
                    if (dp[e] == 0)
                    {
                        t = 1;
                        break;
                    }
        printf("%s", t == 0 ? "YES\n" : "NO\n");
    }
    return 0;
}