PAT (Advanced Level) Practice 1142 Maximal Clique (25分) （算法设计）

1.题目

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

2.题目分析

 A maximal clique就是给的点在图中彼此互相都能联通，而且这些点中的任意一个与其它剩下的所有点不能全联通，

只满足第一点就是Not Maximal

3.代码

#include<iostream>
#include<set>
#include<string>
#include<cstring>
using namespace std;
int nv, ne,m;
int edges[201][201];
int temp[202];
int main()
{
	scanf("%d %d", &nv, &ne);
	set<int>run;
	for (int i = 1; i <= ne; i++)
	{
		int a, b;
		scanf("%d %d", &a, &b);
		edges[a][b] = 1;
		edges[b][a] = 1;
		run.insert(a);
		run.insert(b);
	}
	scanf("%d", &m);
	for (int i = 1; i <= m; i++)
	{
		int k;
		bool max = false, noclique=false;
		int maximal = 0;
		set<int>temprun = run;//temprun中是图中除了所给节点剩下的节点
		scanf("%d", &k);
		fill(temp, temp + nv, 0);
		for (int j = 1; j <= k; j++)
		{
			scanf("%d", &temp[j]);
			temprun.erase(temp[j]);
		}
		for (int j = 1; j <=k; j++)
		{
			for (int s = 1; s <= k; s++)
			{
				if (j!=s&&edges[temp[j]][temp[s]] != 1) { noclique = true; break; }
			}
			if (noclique)break;
		}
		if (noclique){printf("Not a Clique\n"); continue;}
		for (auto it = temprun.begin(); it != temprun.end(); it++)
		{
			maximal = 0;
			for (int j = 1; j <= k; j++)
			{
				if (edges[*it][temp[j]] != 0) maximal++; 
			}
			if (maximal == k) { max = true; break; }
		}
		if (max) { printf("Not Maximal\n"); continue; }
		printf("Yes\n");
	}
}