hdu 5952 Counting Cliques_hdu counting cliques-优快云博客

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Counting Cliques

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2630 Accepted Submission(s): 966

Problem Description

A clique is a complete graph, in which there is an edge between every pair of the vertices. Given a graph with N vertices and M edges, your task is to count the number of cliques with a specific size S in the graph.

Input

The first line is the number of test cases. For each test case, the first line contains 3 integers N,M and S (N ≤ 100,M ≤ 1000,2 ≤ S ≤ 10), each of the following M lines contains 2 integers u and v (1 ≤ u < v ≤ N), which means there is an edge between vertices u and v. It is guaranteed that the maximum degree of the vertices is no larger than 20.

Output

For each test case, output the number of cliques with size S in the graph.

Sample Input

Sample Output

Source

2016ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

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jiangzijing2015

Statistic | Submit | Discuss | Note

因为是稀疏图所以对边进行暴力就行。

#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
using namespace std;
int mark[105][105];
int stack[105];
int u=0,v=0;
int res=0;
int n,m,k;
vector<int>G[105];
void dfs(int p)
{
    if(v==k)
    {
        res++;
        return;
    }
    for(int i=0;i<G[p].size();i++)
    {
        int tmp = G[p][i];
        //if(vis[tmp])continue;
        bool Flag=true;
        for(int j=0;j<v;j++)
        {
            if(!mark[tmp][stack[j]])
            {
                Flag=false;
                break;
            }
        }
        if(Flag)
        {
            stack[v++]=tmp;
            dfs(tmp);
            v--;
        }
    }
    return;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        u=0,v=0;
        res=0;
        for(int i=0;i<100;i++)
            G[i].clear();
        scanf("%d%d%d",&n,&m,&k);
        int x,y;
        memset(mark,0,sizeof(mark));
        memset(vis,0,sizeof(vis));
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&x,&y);
            if(y<x)
            {
                G[y].push_back(x);
            }
            else G[x].push_back(y);
            mark[y][x]=mark[x][y]=1;
        }
        for(int i=1;i<=n;i++)
        {
            stack[v++]=i;
            dfs(i);
            v--;
        }
        printf("%d\n",res);
    }
}