• PAT Advanced 1142 Maximal Clique (25) [图论,⽆向完全图,团,极大团]


    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. Afer 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

    题目分析

    团clique(clique)是一个无向图(undirected graph )的子图,该子图中任意两个顶点之间均存在一条边。
    极大团maximal clique是一个团,该团不能被更大的团所包含,换句话说,图中再也不存在一个点与该团中的任意顶点之间存在一条边。
    已知一系列测试样例,判断每个测试样例中的顶点是否为可组成团,若是团,判断是否为极大团
    极大图条件:

    1. 极大团中所有顶点在图中两两间有边
    2. 图中再没有一个顶点与极大团顶点间存在边

    解题思路

    1. 存储图(邻接矩阵)
    2. 判断每个顶点集是否为极大团,若是团,判断是否为极大团

    Code

    #include <iostream>
    using namespace std;
    const int maxn=210;
    int nv,ne,e[maxn][maxn];
    int main(int argc,char * argv[]) {
    	scanf("%d %d",&nv,&ne);
    	int a,b,m,k;
    	for(int i=1; i<=ne; i++) {
    		scanf("%d %d",&a,&b);
    		e[a][b]=e[b][a]=1;
    	}
    	scanf("%d",&m);
    	for(int i=0; i<m; i++) {
    		scanf("%d", &k);
    		int cv[maxn]= {0},incv[maxn]= {0};
    		for(int j=0; j<k; j++) {
    			scanf("%d", &cv[j]);
    			incv[cv[j]]=1;
    		}
    		// 判断是否为clique
    		bool isClique=true,isMaximal=true;
    		for(int j=0; j<k; j++) {
    			if(isClique==false)break;
    			for(int r=j+1;r<k;r++){ //0~j已经跟子顶点集合中任意顶点匹配过,无需再匹配 
    				if(e[cv[j]][cv[r]]==0){
    					isClique=false;
    					printf("Not a Clique
    ");
    					break;
    				} 
    			}
    		}
    		if(isClique==false)continue;
    		// 判断是否为 maximal clique
    		for(int j=1;j<=nv;j++){
    			if(incv[j]==1)continue; //在子顶点集合中的点跳过 
    			for(int r=0;r<k;r++){
    				if(e[j][cv[r]]==0)break; //如果边不存在,退出循环,继续下个剩余顶点校验 
    				if(r==k-1)isMaximal=false; //如果存在顶点与Clique中所有顶点有相邻有边,则可将该节点加入Clique中,所以原Clique不是最大Clique 
    			}
    			if(isMaximal==false){
    				printf("Not Maximal
    ");
    				break;
    			}
    		} 
    		if(isMaximal&&isClique){
    			printf("Yes
    ");
    		}
    	}
    	return 0;
    }
    

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  • 原文地址:https://www.cnblogs.com/houzm/p/12384638.html
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