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.

题目描述：安排峰会的休息地是一个不简单的活，每个安排的休息地中，需要所有的人都彼此认识。

输入：第一行n，m分别表示有n个人参加峰会，m对好友关系。接下来m行，分别表示这两个人是好友关系。

接下来有一个k值，表示需要查询的休息地个数。往下的k行，第一个值表示该休息地有L个人，接着同一行有L个数。

输出：

1)如果查询的休息地中，有任意两个人不是相互认识的，则输出格式为Area X needs help.，其中X表示第几次查询的编号；

2)如果查询的休息地中，任意两个人都相互认识，并且能够再找出一个人，这人加入进来后，彼此仍旧相互认识，则输出Area X may invite more people, such as other-id.其中X表示第几次查询的编号;若这样的人有多个，输出编号最小一个即可；

3)如果查询的休息地中，任意两个人都相互认识，并且无法再加入其他人。则输出Area X is OK.，其中X表示第几次查询的编号。

解题思路：不需要存图，但需要保存好友关系，方便查询。

#include<bits/stdc++.h>
#include<map>
using namespace std;
#define ll long long
int n, m;
vector<int> v;
unordered_map<int,unordered_map<int,int> > p;
void solve(int index)
{
	bool vis[400];
	memset(vis,false,sizeof(vis));
	int l = v.size();
	bool f1 = true;
	for(int i=0; i<l&&f1; i++){
		vis[i] = true;
		for(int j=i+1; j<l; j++){
			if(p[v[i]][v[j]]==0){
				f1 = false;
				break;
			}
		}
	}
	if(f1==false){
		printf("Area %d needs help.\n", index);
		return ;
	}
	int ans;
	for(int i=1; i<=n&&f1; i++){
		if(vis[i]==false){
			f1 = false;
			for(int j=0; j<l; j++){
				if(p[i][v[j]]==0){
					f1 = true;
					break;
				}
			}
			if(f1==false){
				ans = i;
			}
		}
	}
	if(f1==false){
		printf("Area %d may invite more people, such as %d.\n", index, ans);
	}
	else{
		printf("Area %d is OK.\n", index);
	}
}
int main()
{
	int a, b;
	scanf("%d %d", &n, &m);
	for(int i=0; i<m; i++){
		scanf("%d %d", &a, &b);
		p[a][b] = p[b][a] = 1;
	}
	int k;
	scanf("%d", &k);
	for(int i=0; i<k; i++){
		scanf("%d", &m);
		for(int j=0; j<m; j++){
			scanf("%d", &a);
			v.push_back(a);
		}
		solve(i+1);
		v.clear();
	}
	return 0;
}