(PAT 1142) Maximal Clique (图中顶点与顶点之间的关系)

分手后的思念是犯贱 2022-04-23 09:48 175阅读 0赞

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][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

解题思路:

题目的意思是.给定图中的一个子图,如果子图中的任意两个顶点相邻的话,那么这个子图为clique

题目给出若干子图顶点,要求判断这个子图是不是clique,如果是,它是不是最大 clique,也就是说,图中还有顶点与该子图中的顶点都有相邻关系

这题我用到了邻接矩阵和邻接表,邻接矩阵用来判断两个顶点矩阵之间是否有相邻关系

邻接表用来判断子图外的顶点是否和子图都有相邻关系

首先在子图内进行判断,也就是子图顶点之间两两比较,排除掉非clique的情况

确认是clique后,判断是否是max clique

我们利用哈希表把子图内的顶点标记,然后把子图所有顶点的邻接顶点存到一个set中,然后将这些结点与子图内的顶点二二比较,如果存在一个顶点与子图中的顶点都有邻接关系,那么这个子图就不是maxclique

#include <iostream>
    #include <vector>
    #include <algorithm>
    #include <queue>
    #include <string.h>
    #include <set>
    using namespace std;
    const int MANX = 210;
    int N, M, K;
    int BBedges[MANX][MANX];
    vector<int> EDGEs[MANX];
    bool hashTAble[MANX] = { false };
    
    int main() {
    	cin >> N >> M;
    	int nx, ny;
    	for (int i = 0; i < M; ++i) {
    		cin >> nx >> ny;  //无向边
    		EDGEs[nx].push_back(ny); 
    		EDGEs[ny].push_back(nx);
    		BBedges[nx][ny] = 1;
    		BBedges[ny][nx] = 1;
    	}
    	cin >> K;
    	for (int i = 0; i < K; ++i) {
    		vector<int> nodesTemp;
    		int nums,xnode;
    		cin >> nums;
    		for (int j = 0; j < nums; ++j) {
    			cin >> xnode;
    			nodesTemp.push_back(xnode);
    			hashTAble[xnode] = true;
    		}
    		//第一轮:是不是clique
    		bool isclque = true;
    		for (auto x : nodesTemp) {
    			for (auto y :nodesTemp) {
    				if(x == y) continue;
    				if (BBedges[x][y] != 1) {
    					isclque = false;
    					break;
    				}
    			}
    			if(!isclque) break;
    		}
    		//第二轮:是不是maxclique
    		if (isclque) {
    			set<int> nodesSet;
    			for (auto x : nodesTemp) {
    				for (auto y : EDGEs[x]) {
    					nodesSet.insert(y);
    				}
    			}
    			bool notMax = true;
    			for (auto z : nodesSet) {
    				if (!hashTAble[z]) {
    					bool ismax = true;
    					for (auto a : nodesTemp) {
    						if(a == z) continue;
    						if (BBedges[z][a] != 1) {
    							ismax = false;
    							break;
    						}
    					}
    					if (ismax) {
    						cout << "Not Maximal" << endl;
    						notMax = false;
    						break;
    					}
    				}
    			}
    			if (notMax) {
    				cout << "Yes" << endl;
    			}
    		}
    		else {
    			cout << "Not a Clique" << endl;
    		}
    		memset(hashTAble, 0, MANX);
    	}
    	system("PAUSE");
    	return 0;
    }

[https_en.wikipedia.org_wiki_Clique_graph_theory]: https://en.wikipedia.org/wiki/Clique_%28graph_theory