PAT 1142 Maximal Clique (25分)

原题链接：1142 Maximal Clique (25分)
关键词：图论、枚举
整理时间：2020.7.24

A clique（团） is a subset of vertices of an undirected graph such that every two distinct vertices in the clique are adjacent. A maximal clique is a clique that cannot be extended by including one more adjacent vertex. (Quoted from https://en.wikipedia.org/wiki/Clique_(graph_theory))

Now it is your job to judge if a given subset of vertices can form a maximal clique.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers Nv (≤ 200), the number of vertices in the graph, and Ne, the number of undirected edges. Then Ne lines follow, each gives a pair of vertices of an edge. The vertices are numbered from 1 to Nv.

After the graph, there is another positive integer M (≤ 100). Then M lines of query follow, each first gives a positive number K (≤ Nv), then followed by a sequence of K distinct vertices. All the numbers in a line are separated by a space.

Output Specification:

For each of the M queries, print in a line Yes if the given subset of vertices can form a maximal clique; or if it is a clique but not a maximal clique, print Not Maximal; or if it is not a clique at all, print Not a Clique.
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
3 4 3 6
3 3 2 1

Sample Output:

Yes
Yes
Yes
Yes
Not Maximal
Not a Clique

题目大意： 一个团是指这其中的任意两个顶点都是连通的，最大团是指不能再往里面加顶点构成团。现在给出一个图和若干个询问，问你询问的结点是否是团、或者最大团。

分析：设置两个bool型变量用于判断是否是团和是否是最大团。

1. 如果存在两个顶点不相连：不是团；
2. 是团的话，看图里的其他结点，如果存在一个结点与团中的结点都相连：不是最大团；
3. 否则:最大团

代码：

#include <iostream>
#include <vector>
#include <set>
using namespace std;

const int maxn = 210;

int nv, ne, m, k;   //v无向图中点数量 e边的数量 m询问数量 k子集包含点的数量
int g[maxn][maxn] = {0};

int main(){
    cin >> nv >> ne;
    for(int i = 0; i < ne; i ++ ){
        int a, b;
        cin >> a >> b;
        g[a][b] = g[b][a] = 1;
    }
    
    cin >> m;	//m个询问
    for(int i = 0; i < m; i ++ ){
        cin >> k;	//k个结点
        bool is_cli = true, is_max = true;
        
        vector<int> v(k);
        set<int> vis;  // 出现过的点 
        for(int i = 0; i < k; i++){
            cin >> v[i];
            vis.insert(v[i]); 
        }
        
        for(int i = 0; i < k - 1; i++){
            for(int j = i + 1; j < k; j++){
                if(g[v[i]][v[j]] != 1){	//不连通不是团
                    is_cli = false;
                    puts("Not a Clique");
                    break;  // 跳出内循环了 
                }
            }
            if(is_cli == false)
                break;	//跳出外循环
        }

        if(is_cli == false) continue;	//进行下一组询问
        
        // 查看所有的点，看是否存在一个点满足和集合中所有的结点连接 
        for(int i = 1; i <= nv; i++){
            if(!vis.count(i)){	//i是图里的其他点
                int j;
                for(j = 0; j < k; j++){
                    if(g[i][v[j]] == 0)
                        break;  
                }   
                if(j == k){
                    is_max = false;
                    break;
                }
            } 
        }
        if(is_max) puts("Yes");
        else puts("Not Maximal");
        
    }
    return 0;
}