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
1 #include <stdio.h> 2 #include <algorithm> 3 #include <iostream> 4 #include <map> 5 #include <string> 6 #include <vector> 7 #include <set> 8 #include <cctype> 9 using namespace std; 10 const int maxn=210; 11 int n,m,k; 12 set<int> adj[maxn]; 13 int g[maxn][maxn]; 14 int main(){ 15 fill(g[0],g[0]+maxn*maxn,0); 16 scanf("%d %d",&n,&m); 17 for(int i=0;i<m;i++){ 18 int c1,c2; 19 scanf("%d %d",&c1,&c2); 20 g[c1][c2]=1; 21 g[c2][c1]=1; 22 adj[c1].insert(c2); 23 adj[c2].insert(c1); 24 } 25 scanf("%d",&k); 26 for(int i=0;i<k;i++){ 27 int x; 28 scanf("%d",&x); 29 int que[maxn]={0}; 30 int vis[maxn]={0}; 31 for(int j=0;j<x;j++){ 32 int tmp; 33 scanf("%d",&tmp); 34 que[j]=tmp; 35 vis[tmp]=1; 36 } 37 int flag=0; 38 for(int j=0;j<x;j++){ 39 if(flag==0){ 40 for(int q=j+1;q<x;q++){ 41 if(g[que[j]][que[q]]==0){ 42 flag=1; 43 printf("Not a Clique "); 44 break; 45 } 46 } 47 } 48 else break; 49 } 50 if(flag==0){ 51 int flag1=0,flag2=0; 52 for(int j=1;j<=n;j++){ 53 flag2=0; 54 if(vis[j]==1) continue; 55 for(int q=0;q<x;q++){ 56 if(g[que[q]][j]==0){ 57 flag2=1; 58 break; 59 } 60 } 61 if(flag2==0){ 62 flag1=1; 63 printf("Not Maximal "); 64 break; 65 } 66 } 67 if(flag1==0)printf("Yes "); 68 } 69 } 70 }
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