1166 Summit

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.

思路

先给出M个关系即m条边，然后给出k个集合，每个集合里面的人必须是朋友关系，如果可以找出不在集合里的一个人并放入集合，依旧满足这个性质，输出Area 5 may invite more people, such as 3.如果找不到这样的人，输出Area 4 is OK.如果原始集合不满足这个性质，输出Area 6 needs help.

cpp代码

#include<iostream>
#include<vector>
#include<unordered_map>
#include<cstring>
using namespace std;
const int N=210;
int g[N][N];
int n,m;
bool st[N];
int back[N];
int main(){
    cin>>n>>m;
    for(int i=0;i<m;i++){
        int a,b;
        cin>>a>>b;
        g[a][b]=1;
        g[b][a]=1;
    }
    int q;
    cin>>q;
    int in=1;
    while(q--){
        int k;
        cin>>k;
        for(int i=0;i<k;i++)cin>>back[i];
        bool is_every=true;
        int id=-1;
        for(int i=0;i<k;i++){
            for(int j=i+1;j<k;j++){
                int a=back[i],b=back[j];
                if(!g[a][b]||!g[b][a]){
                    is_every=false;
                    break;
                }
            }
        }
        if(!is_every){
            printf("Area %d needs help.\n",in);
            in++;
            continue;
        }
        for(int i=0;i<k;i++)st[back[i]]=true;
        for(int i=1;i<=n;i++){
            bool flag=true;
            if(!st[i]){//不在集合里的人
                for(int j=0;j<k;j++){
                    int t=back[j];
                    if(!g[i][t]||!g[t][i]){
                        flag=false;
                        break;
                    }
                }
                if(flag){//集合里的所有人都和这个集合外的人有交集
                    id=i;
                    break;
                }
            }
        }
        memset(st,false,sizeof st);
        if(~id)printf("Area %d may invite more people, such as %d.\n",in,id);
        else printf("Area %d is OK.\n",in);
        in++;
    }
    return 0;
}