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：
1 #include<bits/stdc++.h> 2 //#include<regex> 3 #define db double 4 #include<vector> 5 #define ll long long 6 #define vec vector<ll> 7 #define Mt vector<vec> 8 #define ci(x) scanf("%d",&x) 9 #define cd(x) scanf("%lf",&x) 10 #define cl(x) scanf("%lld",&x) 11 #define pi(x) printf("%d ",x) 12 #define pd(x) printf("%f ",x) 13 #define pl(x) printf("%lld ",x) 14 #define MP make_pair 15 #define PB push_back 16 #define inf 0x3f3f3f3f3f3f3f3f 17 #define fr(i,a,b) for(int i=a;i<=b;i++) 18 const int N=1e3+5; 19 const int mod=1e9+7; 20 const int MOD=mod-1; 21 const db eps=1e-18; 22 using namespace std; 23 bool d[105][105]; 24 int n,m,s,t; 25 int ans; 26 vector<int> g[N]; 27 void dfs(int u,int *a,int cnt) 28 { 29 if(cnt==s){ 30 ans++; 31 return; 32 } 33 bool ok; 34 for(int i=0;i<g[u].size();i++) 35 { 36 ok=1; 37 int v=g[u][i]; 38 for(int j=1;j<=cnt;j++){ 39 if(!d[a[j]][v]) {ok=0;break;} 40 } 41 if(ok) 42 { 43 a[++cnt]=v;//加点 44 dfs(v,a,cnt);//继续搜 45 a[cnt--]=0; 46 } 47 } 48 } 49 int main(){ 50 //freopen("data.in","r",stdin); 51 //freopen("data.out","w",stdout); 52 ci(t); 53 while(t--) 54 { 55 ci(n),ci(m),ci(s); 56 ans=0; 57 for(int i=1;i<=n;i++) g[i].clear(); 58 memset(d,0,sizeof(d)); 59 for(int i=0;i<m;i++){ 60 int u,v; 61 ci(u),ci(v); 62 if(u>v) swap(u,v); 63 g[u].push_back(v); 64 d[u][v]=d[v][u]=1; 65 } 66 for(int i=1;i<=n;i++){ 67 if(g[i].size()>=s-1){ 68 int a[105]; 69 a[1]=i;//构建团 70 int cnt=1; 71 dfs(i,a,cnt); 72 } 73 } 74 pi(ans); 75 } 76 return 0; 77 }