PAT 甲级 1142 Maximal Clique (25分)

原创于 2020-04-06 12:09:05 发布 · 229 阅读

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本文介绍了一种算法，用于判断给定子集的顶点是否能形成一个最大团子图，即所有顶点两两相邻且无法通过添加更多相邻顶点来扩展。算法首先检查子集是否构成团子图，再逐个尝试添加未包含的顶点以验证其最大性。

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
题目大意：
max_clique：集合内的所有结点都两两领接且不能往这个集合内添加任何结点继续满足这个条件；
clique:结合内的所有结点都两两领接
故本题分为两个步骤先判断是否满足clique条件，然后将所有未包括在判断点集合的点依次加入判断点集合看是否还是clique只要有一个点能满足clique条件那么判断点集合就不是Max_clique；
AC代码：

#include<iostream>
#include<vector>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=201;
bool adjacent[maxn][maxn];//记录临界顶点
int n;
bool is_clique(vector<int>v){
for(int i=0;i<v.size();i++){
for(int j=i+1;j<v.size();j++){
if(adjacent[v[i]][v[j]]!=true){
    return false;}
}
}
return true;}
bool is_Maximal(vector<int>v){
    if(!is_clique(v))
        return false;
for(int i=1;i<=n;i++){
if(find(v.begin(),v.end(),i)==v.end()){//说明该节点不在；
bool flag=true;
for(int j=0;j<v.size();j++){
if(adjacent[i][v[j]]==false){
    flag=false;}
                           }
if(flag==true)
return false;                         }
                    }
return true;
}
int main(){
int m,k;
scanf("%d %d",&n,&m);
for(int i=0;i<m;i++){
int a,b;
scanf("%d%d",&a,&b);
adjacent[a][b]=adjacent[b][a]=true;
}
scanf("%d",&k);
for(int i=0;i<k;i++){
    int temp1,temp2;
    scanf("%d",&temp1);
    vector<int>v;
        for(int j=0;j<temp1;j++){
          scanf("%d",&temp2);
          v.push_back(temp2)  ;
    }
    if(is_Maximal(v)){
        printf("Yes\n"); }
    else if(is_clique(v))
{
    printf("Not Maximal\n");}
else{
    printf("Not a Clique\n");}
}
system("pause");
return 0;
}