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

àì夳堔傛蜴生んèń 2023-02-20 02:27 6阅读 0赞

[算法笔记总目录][Link 1]  
[关键英语单词解释][Link 2]  
本题代码与[PAT甲级2017冬7-3 1142 Maximal Clique (25分)][PAT_2017_7-3 1142 Maximal Clique _25]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][Link 3]

#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; 
    }

[代码二来源地址][Link 4]

#include<iostream>
    #include<climits>
    #include<cctype>
    #include<map>
    #include<unordered_map>
    #include<set>
    #include<vector>
    #include<algorithm>
    #include<string>
    #include<cstring>
    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;
    }

[Link 1]: https://blog.csdn.net/qq_21480607/article/details/106660085
[Link 2]: https://blog.csdn.net/qq_21480607/article/details/106987591
[PAT_2017_7-3 1142 Maximal Clique _25]: https://blog.csdn.net/qq_21480607/article/details/106896828
[Link 3]: https://www.cnblogs.com/caiyishuai/p/12005418.html
[Link 4]: https://blog.csdn.net/qq_36993032/article/details/103823877