PAT A1142 Maximal Clique （25 分）——图

A clique 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

 

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <map>
#include <string>
#include <vector>
#include <set>
#include <cctype>
using namespace std;
const int maxn=210;
int n,m,k;
set<int> adj[maxn];
int g[maxn][maxn];
int main(){
    fill(g[0],g[0]+maxn*maxn,0);
    scanf("%d %d",&n,&m);
    for(int i=0;i<m;i++){
        int c1,c2;
        scanf("%d %d",&c1,&c2);
        g[c1][c2]=1;
        g[c2][c1]=1;
        adj[c1].insert(c2);
        adj[c2].insert(c1);
    }
    scanf("%d",&k);
    for(int i=0;i<k;i++){
        int x;
        scanf("%d",&x);
        int que[maxn]={0};
        int vis[maxn]={0};
        for(int j=0;j<x;j++){
            int tmp;
            scanf("%d",&tmp);
            que[j]=tmp;
            vis[tmp]=1;
        }
        int flag=0;
        for(int j=0;j<x;j++){
            if(flag==0){
                for(int q=j+1;q<x;q++){
                    if(g[que[j]][que[q]]==0){
                        flag=1;
                        printf("Not a Clique
");
                        break;
                    }
                }
            }
            else break;
        }
        if(flag==0){
            int flag1=0,flag2=0;
            for(int j=1;j<=n;j++){
                flag2=0;
                if(vis[j]==1) continue;
                for(int q=0;q<x;q++){
                    if(g[que[q]][j]==0){
                        flag2=1;
                        break;
                    }
                }
                if(flag2==0){
                    flag1=1;
                    printf("Not Maximal
");
                    break;
                }
            }
            if(flag1==0)printf("Yes
");
        }
    }
}

注意点：题目意思就是找一个团，里面每个元素之间都两两相连，如果没有另外的点可以加进去后还满足这个条件就是最大团。一开始一直在想把给定序列的元素每个遍历，找到每个元素相连的元素，再去对他们做交集，看看和给定序列一不一样。做交集这个就很麻烦了，看了大佬的思路，原来不用纠结在给定的点上，要看另外的点能不能加进来，其实就是题目里给的意思，我自己转弯了。

这种多个判断的题已经不是第一次做到了，类似的还有1150 Travelling Salesman Problem （25 分），也是和图有关，判断各种东西