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.
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.
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.
题意：给出顶点数和边数，再给出查询的次数，每次查询中都会列几个顶点，查子集两两相连能不能扩充。 思路：先用二维矩阵存储顶点和边之间的关系，先判断子集的顶点之间是不是两两相连的，不能就直接输出Not a Clique，能的话再判断其他顶点能不能对这个子集中的点全部都连通，如果都不能的话输出Yes,能的话输出Not Maximal。判断子集外的点对子集上顶点的连通性就直接先用hash存储之前子集上的点，将总体集合中不在hash上的值遍历过去，与子集中的顶点进行判断有无连通性。
#include<iostream> #include<cstdio> #include<stack> #include<queue> #include<vector> #include<algorithm> #include<cstring> #include<string> #include<cmath> #include<set> #include<map> #include<cctype> #include<cstdlib> #include<ctime> #include<unordered_map> using namespace std; int e[210][210]; int main() { int nv,ne; scanf("%d%d",&nv,&ne); for(int i = 0;i < ne;i++) { int t1,t2; scanf("%d%d",&t1,&t2); e[t1][t2] = e[t2][t1] = 1; } int m; scanf("%d",&m); for(int i = 0;i < m;i++) { int k; int hash1[210] = {0}; int isclique = 1,ismax = 1; scanf("%d",&k); vector<int> v(k); for(int j = 0;j < k;j++) { scanf("%d",&v[j]); hash1[v[j]] = 1; } //字团内能否互相连 for(int j = 0;j < k;j++) { if(isclique == 0) break; for(int l = j + 1;l < k;l++) { //不能互相连 if(e[v[j]][v[l]] == 0) { isclique = 0; printf("Not a Clique\n"); break; } } } if(isclique == 0) continue; //字团外的点能否与之互相连 for(int j = 1;j <= nv;j++) { if(hash1[j] == 0) { for(int l = 0;l < k;l++) { //这个点与字团中的有一点不相连 if(e[v[l]][j] == 0) break; if(l == k - 1) ismax = 0; } } if(ismax == 0) { printf("Not Maximal\n"); break; } } if(ismax == 1) printf("Yes\n"); } return 0; }