HDU 5952 - Counting Cliques （DFS）

Counting Cliques

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2717    Accepted Submission(s): 984

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

  

   3
4 3 2
1 2
2 3
3 4
5 9 3
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5
6 15 4
1 2
1 3
1 4
1 5
1 6
2 3
2 4
2 5
2 6
3 4
3 5
3 6
4 5
4 6
5 6
  

 

Sample Output

  

   3
7
15
  

 

Source

2016ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学） 

 

题意：

求顶点数为s的完全图的个数。

POINT：

单向建边，由小往大连，从1开始找模拟，然后从2，从3。就ok了。

用邻接表。

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
#define LL long long
const int maxn = 111;
int n,m,s;
vector<int>G[maxn];
int flag[maxn];
int pre[maxn];
int sz;
int ans;
int mp[maxn][maxn];
int check(int now){
    for(int i=1;i<=sz;i++){
        if(mp[now][pre[i]]!=1) return 0;
    }
    return 1;
}
void dfs(int now)
{
    if(sz==s){
        ans++;
        return;
    }
    for(int i=0;i<G[now].size();i++){
        int v=G[now][i];
        if(check(v)){
            pre[++sz]=v;
            dfs(v);
            sz--;
        }
        
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        ans=0;
        memset(flag,0,sizeof flag);
        memset(mp,0,sizeof mp);
        scanf("%d %d %d",&n,&m,&s);
        for(int i=1;i<=n;i++) G[i].clear();
        for(int i=1;i<=m;i++){
            int u,v;
            scanf("%d %d",&u,&v);
            mp[u][v]=mp[v][u]=1;
            int a=max(u,v);
            int b=u+v-a;
            G[b].push_back(a);
        }
        for(int i=1;i<=n;i++){
            pre[sz=1]=i;
            dfs(i);
        }
        printf("%d\n",ans);
    }
}