1142 Maximal Clique (25 point(s))

1142 Maximal Clique (25 point(s))

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：

在一个无向图中，如果在给定的顶点集中任意两个不同的点之间都有一条边，那么我们称这样的点集为Clique。如果一个Clique点集不可以再加入任何一个新的结点构成新的Clique（即是最大的），我们称这样的Clique为Maximal Clique。给定无向图和子集，进行判断。

使用邻接矩阵存储无向图，将给出的子集存储在set<int> s中，按照定义暴力求解。

1. 先判断是不是Clique。遍历s中任意两个元素，如果存在graph[i][j]==false则不是；

2. 再判断是不是Maximal Clique。遍历每一个不在s的元素i，如果存在对于s中所有的j有graph[i][j]==true，则不是；

3. 否则输出“Yes”。

完成本题后的反思：STL中set/vector等查找元素是否存在、个数等库函数

#include<iostream>
#include<cstring>
#include<vector>
#include<set>
using namespace std;
const int N = 207;
int nv,ne;
bool graph[N][N]={false};
set<int> s;
bool isClique(){
	set<int>::iterator i;
	for(i=s.begin();i!=s.end();i++){
		set<int>::iterator j=i;
		for( j++;j!=s.end();j++){
			if(!graph[(*i)][(*j)]) return false;
		}
	}
	return true;
}
bool isMax(){
	for(int i=1;i<=nv;i++){
		if(s.find(i)==s.end()){//如果不在这个集合
			bool flag = false; 
			set<int>::iterator it;
			for(it=s.begin();it!=s.end();it++){
				if(!graph[i][*it]){
					flag =true;break;
				}
			} 
			if(!flag) return false;
		}
	}
	return true;
}
int main(void){
	int a,b,M;cin>>nv>>ne;
	while(ne--){
		cin>>a>>b;
		graph[a][b]=graph[b][a]=true;
	}
	int K,x;cin>>M;
	while(M--){
		s.clear();
		cin>>K;
		while(K--){
			cin>>a;
			s.insert(a); 
		}
		if(!isClique()) cout<<"Not a Clique"<<endl;
		else if(!isMax()) cout<<"Not Maximal"<<endl;
		else cout<<"Yes"<<endl;
	}
	return 0;
}

此题的主要操作是要对特定的边是否存在进行判断，因此使用邻接矩阵更加简单。

大神用向量数组存储的代码：

https://blog.youkuaiyun.com/richenyunqi/article/details/79837901