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\n",x)
 #define pd(x) printf("%f\n",x)
 #define pl(x) printf("%lld\n",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+;
 const int mod=1e9+;
 const int MOD=mod-;
 const db  eps=1e-;
 using namespace std;
 bool d[][];
 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=;i<g[u].size();i++)
     {
         ok=;
         int v=g[u][i];
         for(int j=;j<=cnt;j++){
             if(!d[a[j]][v]) {ok=;break;}
         }
         if(ok)
         {
             a[++cnt]=v;//加点
             dfs(v,a,cnt);//继续搜
             a[cnt--]=;
         }
     }
 }
 int main(){
 //freopen("data.in","r",stdin);
 //freopen("data.out","w",stdout);
     ci(t);
     while(t--)
     {
         ci(n),ci(m),ci(s);
         ans=;
         for(int i=;i<=n;i++) g[i].clear();
         memset(d,,sizeof(d));
         for(int i=;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]=;
         }
         for(int i=;i<=n;i++){
             if(g[i].size()>=s-){
                 int a[];
                 a[]=i;//构建团
                 int cnt=;
                 dfs(i,a,cnt);
             }
         }
         pi(ans);
     }
     return ;
 }