hdu 5952 连通子图

Counting Cliques

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 184    Accepted Submission(s): 56

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

思路：构造一个团，如果一个点与这个团的所有点都有边，则将其加入团中，统计含s个点的团的个数。关于优化，可以建单向边来减少搜索量。

代码：

#include<bits/stdc++.h>
//#include<regex>
#define db double
#include<vector>
#define ll long long
#define vec vector<ll>
#define Mt  vector<vec>
#define ci(x) scanf("%d",&x)
#define cd(x) scanf("%lf",&x)
#define cl(x) scanf("%lld",&x)
#define pi(x) printf("%d
",x)
#define pd(x) printf("%f
",x)
#define pl(x) printf("%lld
",x)
#define MP make_pair
#define PB push_back
#define inf 0x3f3f3f3f3f3f3f3f
#define fr(i,a,b) for(int i=a;i<=b;i++)
const int N=1e3+5;
const int mod=1e9+7;
const int MOD=mod-1;
const db  eps=1e-18;
using namespace std;
bool d[105][105];
int n,m,s,t;
int ans;
vector<int> g[N];
void dfs(int u,int *a,int cnt)
{
    if(cnt==s){
        ans++;
        return;
    }
    bool ok;
    for(int i=0;i<g[u].size();i++)
    {
        ok=1;
        int v=g[u][i];
        for(int j=1;j<=cnt;j++){
            if(!d[a[j]][v]) {ok=0;break;}
        }
        if(ok)
        {
            a[++cnt]=v;//加点
            dfs(v,a,cnt);//继续搜
            a[cnt--]=0;
        }
    }
}
int main(){
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
    ci(t);
    while(t--)
    {
        ci(n),ci(m),ci(s);
        ans=0;
        for(int i=1;i<=n;i++) g[i].clear();
        memset(d,0,sizeof(d));
        for(int i=0;i<m;i++){
            int u,v;
            ci(u),ci(v);
            if(u>v) swap(u,v);
            g[u].push_back(v);
            d[u][v]=d[v][u]=1;
        }
        for(int i=1;i<=n;i++){
            if(g[i].size()>=s-1){
                int a[105];
                a[1]=i;//构建团
                int cnt=1;
                dfs(i,a,cnt);
            }
        }
        pi(ans);
    }
    return 0;
}