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=;
 int n,m,k;
 set<int> adj[maxn];
 int g[maxn][maxn];
 int main(){
     fill(g[],g[]+maxn*maxn,);
     scanf("%d %d",&n,&m);
     for(int i=;i<m;i++){
         int c1,c2;
         scanf("%d %d",&c1,&c2);
         g[c1][c2]=;
         g[c2][c1]=;
         adj[c1].insert(c2);
         adj[c2].insert(c1);
     }
     scanf("%d",&k);
     for(int i=;i<k;i++){
         int x;
         scanf("%d",&x);
         int que[maxn]={};
         int vis[maxn]={};
         for(int j=;j<x;j++){
             int tmp;
             scanf("%d",&tmp);
             que[j]=tmp;
             vis[tmp]=;
         }
         int flag=;
         for(int j=;j<x;j++){
             if(flag==){
                 for(int q=j+;q<x;q++){
                     if(g[que[j]][que[q]]==){
                         flag=;
                         printf("Not a Clique\n");
                         break;
                     }
                 }
             }
             else break;
         }
         if(flag==){
             int flag1=,flag2=;
             for(int j=;j<=n;j++){
                 flag2=;
                 if(vis[j]==) continue;
                 for(int q=;q<x;q++){
                     if(g[que[q]][j]==){
                         flag2=;
                         break;
                     }
                 }
                 if(flag2==){
                     flag1=;
                     printf("Not Maximal\n");
                     break;
                 }
             }
             if(flag1==)printf("Yes\n");
         }
     }
 }

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