PAT-2019年冬季考试-甲级 7-3 Summit (25分) (邻接矩阵存储，直接暴力)

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.

题意：

给N个点，M条边。K个询问。每个询问给出L个点，问这L个点是不是两两相连的。

如果两两相连：

　　存不存在一个其它的点，与这L个点都有连接：

　　　　有：Area i may invite more people, such as 这个点.

　　　　没有：Area i is OK.

不是两两相连：Area i needs help.

 

题解：

 

看懂题意发现挺简单的嘛，不需要并查集，直接邻接矩阵存储，直接暴力就ok了，考试的时候一遍过，惊喜！

 

AC代码：

#include<bits/stdc++.h>
using namespace std;
int e[205][205];
int a[205];
int v[205];
int n,m;
int main(){
    cin>>n>>m;
    memset(e,0,sizeof(e));
    for(int i=1;i<=m;i++){
        int u,v;
        cin>>u>>v;
        e[u][v]=e[v][u]=1;//邻接矩阵存储 
    }
    int k;
    cin>>k;
    int num;
    for(int i=1;i<=k;i++){//k个询问 
        cin>>num;
        memset(v,0,sizeof(v));//v来标记所询问的num个点 
        for(int j=1;j<=num;j++) {
            cin>>a[j];
            v[a[j]]=1;//做上标记 
        }
        int f=1;//是不是两两相连 
        for(int j=1;j<=num;j++){
            for(int p=j+1;p<=num;p++){
                if(e[a[j]][a[p]]!=1) f=0;
                break;
            }
        }
        if(!f) cout<<"Area "<<i<<" needs help.";
        else{//如果是两两相连 
            int ans=-1;//是否存在 
            for(int j=1;j<=n;j++){//查询存不存在一个点与这num个点都相连 
                if(v[j]) continue;//本身是num个点里的不算 
                int ff=1; 
                for(int p=1;p<=num;p++){
                    if(e[a[p]][j]!=1){
                        ff=0;
                        break;    
                    }
                }
                if(ff) {//满足与num中的每个点都相连 
                    ans=j;//存在 
                    break;
                }
            }
            if(ans!=-1){//存在 
                cout<<"Area "<<i<<" may invite more people, such as "<<ans<<".";
            }else{//不存在 
                cout<<"Area "<<i<<" is OK."; 
            }
        }
        if(i!=k) cout<<endl;//行末无空行 
    }
    return 0; 
}