PAT甲级2019冬7-3 Summit（1166）(邻接矩阵存储，直接暴力)

算法笔记总目录
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本题代码与PAT甲级2017冬7-3 1142 Maximal Clique (25分)除输出不一样外，其他不变。

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：

7-3首脑会议（25分）

首脑会议是国家元首或政府首脑的会议。安排峰会的休息区并不是一件简单的工作。一个地区的理想安排是邀请那些头儿，这样每个人都是每个人的直接朋友。

现在给出一组暂定的安排，你的工作是告诉组织者每个区域是否都已设置好。

输入规格：

每个输入文件包含一个测试用例。对于每一种情况，第一行给出两个正整数N（≤200），表示首脑人数，M表示友好关系人数。接着M行，每行给出一对彼此是朋友的头部的索引。头部的索引范围是从1到N。

然后是另一个正整数K（≤100），接下来是K行暂定的休息区排列，每行先给出一个正数L（≤N），然后是一系列L个不同的头部指数。一行中的所有数字都用空格隔开。

输出规格：

对于每个K区域，请按以下格式在一行中打印您的建议：

如果在这个区域中每个人都是每个人的直接朋友，并且没有朋友丢失（也就是说，没有其他人是这个区域中每个人的直接朋友），那么打印区域X就可以了。。

如果在这方面，每个人都是每个人的直接朋友，但也有一些其他的头像可能会被邀请，而不会破坏理想的安排，那么打印区域X可能会邀请更多的人，例如H，其中H是可能被邀请的头像的最小索引。

如果在这个区域的排列不理想，则打印区域X需要帮助。因此，主人可以提供一些特殊的服务，以帮助头部相互了解。

这里X是一个区域的索引，从1到K。

思路：

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

代码一来源地址

#include
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; 
}

代码二来源地址

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
bool e[205][205];
int main(){
    int n, m , k, a ,b;
    cin >> n >> m;
    fill(e[0], e[0] + 205*205, false);
    for(int i = 0; i < m; i++){
        cin >> a >> b;
        e[a][b] = true;
        e[b][a] = true;
    }
    cin >> k;
    for(int i = 1; i <= k; i++){
        bool flag1 = true;
        cin >> a;
        vector<int> v(a);
        vector<bool> exist(n + 1, false);
        for(int j = 0; j < a; j++){
            scanf("%d", &v[j]);
            exist[v[j]] = true;
        }
        for(int j = 0; j < a && flag1; j++){
            for(int l = j + 1; l < a && flag1; l++)
                if(!e[v[j]][v[l]]){
                    flag1 = false;
                    break;
                }
        }
        if(!flag1) printf("Area %d needs help.\n", i);
        else{
    bool flag3= true;
            for(int j = 1; j <= n; j++){
                if(!exist[j]){
                            int flag2 = true;
                    for(int l = 0; l <a; l++){
                            if(!e[j][v[l]]){
                                flag2= false;
                                break;
                            }
                    }
                    if(flag2) {
                        printf("Area %d may invite more people, such as %d.\n", i, j);
                        flag3 = false;
                        break;
                    }
                }
            }
            if(flag3)   printf("Area %d is OK.\n", i);
        }
    }
    return 0;
}