PAT（甲级）2019年冬季考试7-3 Summit (25 分)

7-3 Summit (25 分)

A summit (峰会) is a meeting of heads of state or government. Arranging the rest areas for the summit is not a simple job. The ideal arrangement of one area is to invite those heads so that everyone is a direct friend of everyone.

Now given a set of tentative arrangements, your job is to tell the organizers whether or not each area is all set.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N (≤ 200), the number of heads in the summit, and M, the number of friendship relations. Then M lines follow, each gives a pair of indices of the heads who are friends to each other. The heads are indexed from 1 to N.

Then there is another positive integer K (≤ 100), and K lines of tentative arrangement of rest areas follow, each first gives a positive number L (≤ N), then followed by a sequence of L distinct indices of the heads. All the numbers in a line are separated by a space.

Output Specification:

For each of the K areas, print in a line your advice in the following format:

if in this area everyone is a direct friend of everyone, and no friend is missing (that is, no one else is a direct friend of everyone in this area), print Area X is OK…

if in this area everyone is a direct friend of everyone, yet there are some other heads who may also be invited without breaking the ideal arrangement, print Area X may invite more people, such as H. where H is the smallest index of the head who may be invited.

if in this area the arrangement is not an ideal one, then print Area X needs help. so the host can provide some special service to help the heads get to know each other.

Here X is the index of an area, starting from 1 to K.

Sample Input:

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

Sample Output:

Area 1 is OK.
Area 2 is OK.
Area 3 is OK.
Area 4 is OK.
Area 5 may invite more people, such as 3.
Area 6 needs help.

就图的遍历，先判断给的set中的节点是否都相邻,再判断没给出的节点中是否有节点和set中的所有节点都是相邻的，是的话输出这set需要补充，不是的话输出OK

#include<bits/stdc++.h>
using namespace std;
int N, M, e[205][205], K;
int main() {
	scanf("%d %d", &N, &M);
	int a, b;
	for (int i = 0; i < M; i++) {
		scanf("%d %d", &a, &b);
		e[a][b] = e[b][a] = 1;
	}
	scanf("%d", &K);
	for (int i = 1; i <= K; i++) {
		int cnt, flag1 = 0, flag2 = 0, id;
		bool visit[210] = {false};
		scanf("%d", &cnt);
		vector<int> v(cnt + 1);
		for (int j = 1; j <= cnt; j++) {
			scanf("%d", &v[j]);
			visit[v[j]] = true;
		}
		for (int j = 1; j < cnt; j++) {
			if (flag1) break;
			for (int t = j + 1; t <= cnt; t++) {
				if (e[v[j]][v[t]] != 1) {
					flag1 = 1;
					break;
				}
			}
		}
		if (flag1) {
			printf("Area %d needs help.\n", i);
			continue;
		}
		for (int j = 1; j <= N; j++) {
			if (!visit[j]) {
				int t;
				for (t = 1; t <= cnt; t++) {
					if (e[j][v[t]] != 1) {
						break;
					}
				}
				if (t == cnt + 1) {
					flag2 = 1;
					id = j;
					break;
				}
			}
		}
		if (flag2 == 1) printf("Area %d may invite more people, such as %d.\n", i, id);
		else {
			printf("Area %d is OK.\n", i);
		}
	}
	return 0;
}