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

解题思路:

题目的意思是.给定图中的一个子图,如果子图中的任意两个顶点相邻的话,那么这个子图为clique

题目给出若干子图顶点,要求判断这个子图是不是clique,如果是,它是不是最大 clique,也就是说,图中还有顶点与该子图中的顶点都有相邻关系

这题我用到了邻接矩阵和邻接表,邻接矩阵用来判断两个顶点矩阵之间是否有相邻关系

邻接表用来判断子图外的顶点是否和子图都有相邻关系

首先在子图内进行判断,也就是子图顶点之间两两比较,排除掉非clique的情况

确认是clique后,判断是否是max clique

我们利用哈希表把子图内的顶点标记,然后把子图所有顶点的邻接顶点存到一个set中,然后将这些结点与子图内的顶点二二比较,如果存在一个顶点与子图中的顶点都有邻接关系,那么这个子图就不是maxclique

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <string.h>
#include <set>
using namespace std;
const int MANX = 210;
int N, M, K;
int BBedges[MANX][MANX];
vector<int> EDGEs[MANX];
bool hashTAble[MANX] = { false };

int main() {
	cin >> N >> M;
	int nx, ny;
	for (int i = 0; i < M; ++i) {
		cin >> nx >> ny;  //无向边
		EDGEs[nx].push_back(ny); 
		EDGEs[ny].push_back(nx);
		BBedges[nx][ny] = 1;
		BBedges[ny][nx] = 1;
	}
	cin >> K;
	for (int i = 0; i < K; ++i) {
		vector<int> nodesTemp;
		int nums,xnode;
		cin >> nums;
		for (int j = 0; j < nums; ++j) {
			cin >> xnode;
			nodesTemp.push_back(xnode);
			hashTAble[xnode] = true;
		}
		//第一轮:是不是clique
		bool isclque = true;
		for (auto x : nodesTemp) {
			for (auto y :nodesTemp) {
				if(x == y) continue;
				if (BBedges[x][y] != 1) {
					isclque = false;
					break;
				}
			}
			if(!isclque) break;
		}
		//第二轮:是不是maxclique
		if (isclque) {
			set<int> nodesSet;
			for (auto x : nodesTemp) {
				for (auto y : EDGEs[x]) {
					nodesSet.insert(y);
				}
			}
			bool notMax = true;
			for (auto z : nodesSet) {
				if (!hashTAble[z]) {
					bool ismax = true;
					for (auto a : nodesTemp) {
						if(a == z) continue;
						if (BBedges[z][a] != 1) {
							ismax = false;
							break;
						}
					}
					if (ismax) {
						cout << "Not Maximal" << endl;
						notMax = false;
						break;
					}
				}
			}
			if (notMax) {
				cout << "Yes" << endl;
			}
		}
		else {
			cout << "Not a Clique" << endl;
		}
		memset(hashTAble, 0, MANX);
	}
	system("PAUSE");
	return 0;
}