A1142 Maximal Clique（25分）PAT 甲级(Advanced Level) Practice（C++）满分题解【图+极大团】

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

题意分析：

这道题首先在于看懂题目，判断输入的图是否是一个两两连通的团，或者是否是极大团，最后总结归纳为以下两个条件：

1. 是否两两相连
2. 是否存在每个人都有的共同邻居

然后分别定义两个判断上述条件的bool类型函数，再根据要求在输出时进行判断 

代码如下：

#include
using namespace std;
const int maxn = 210;
int Nv, Ne, v1, v2, m;
int edge[maxn][maxn] = { 0 };//邻接矩阵
bool isconnected(vector<int> vlist) {
	//判断是否两两相连
	bool flag = true;
	for (int i = 0; i < vlist.size()-1; i++) {
		for (int j = i + 1; j < vlist.size(); j++) {
			if (edge[vlist[i]][vlist[j]] != 1) {
				flag = false;
				return flag;
			}
		}
	}
	return flag;
}
bool addNeighbour(vector<int> vlist) {
	bool flag = false;//true可以加邻居
	for (int i = 1; i <= Nv; i++) {
		if (find(vlist.begin(), vlist.end(), i) == vlist.end()) {//该节点不在查询队列里
			for (int j = 0; j < vlist.size(); j++) {
				if (edge[vlist[j]][i] != 1) {
					flag = false;
					break;
				}
				else flag = true;
			}
		}
		if (flag)break;
	}
	return flag;
}
int main()
{
	cin >> Nv >> Ne;
	//输入
	for (int i = 0; i < Ne; i++) {
		cin >> v1 >> v2;
		edge[v1][v2] = edge[v2][v1] = 1;
	}
	//查询，判断是否是极大团
	//1、是否两两相连
	//2、是否存在每个人都有的共同邻居
	cin >> m;
	int num,v;
	for (int i = 0; i < m; i++) {
		cin >> num;
		vector<int> querylist(num);
		for (int j = 0; j < num; j++) {
			cin >> querylist[j];
		}
		sort(querylist.begin(), querylist.end());
		if (isconnected(querylist)) {
			if(!addNeighbour(querylist))cout << "Yes" << endl;
			else cout << "Not Maximal" << endl;;
		}
		else cout << "Not a Clique" << endl;
	}
	return 0;
}

运行结果如下：