HDU 5952 Counting Cliques（dfs）

Counting Cliques

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1855    Accepted Submission(s): 735

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）

 

Recommend

jiangzijing2015

题意：

　　给你n个点，m条边，要你在这些点里面找大小为s 的完全图（完全图是指这个图里任意两点都有一条边）。

　　因为 N 只有100个。S最大也才10。所以我们可以爆搜来解决。然后我就T了  :(

　　因为 1 2 3 如果是完全图的话，那么  2 1 3 也是。所以我们多搜了很多次。

　　如果我们建边的时候是按照 小的指向 大的点来建，就可以避免了很多情况。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <cmath>
 #include <vector>
 #include <queue>
 #include <map>
 #include <stack>
 #include <set>
 using namespace std;
 typedef long long LL;
 #define ms(a, b) memset(a, b, sizeof(a))
 #define pb push_back
 #define mp make_pair
 const LL INF = 0x7fffffff;
 const int inf = 0x3f3f3f3f;
 const int mod = 1e9+;
 const int maxn = +;
 bool edge[maxn][maxn];
 vector <int> E[maxn];
 int num[maxn];
 int p[];
 int ans, n, m, s, u, v;
 void init() {
     ms(edge, );
     ms(num, );
     for(int i = ;i<maxn;i++)   E[i].clear();
     ans = ;
 }
 void dfs(int x, int len){
     int flag = ;
     for(int i=;i<len;i++){
         if(!edge[x][p[i]]){
             flag = ;
             break;
         }
     }
     if(!flag){
         return;
     }
     p[len] = x;
     if(len+==s){
         ans++;
         return;
     }
     for(int i = ;i<E[x].size();i++){
         dfs(E[x][i], len+);
     }
 }
 void solve() {
     scanf("%d%d%d", &n, &m, &s);
     for(int i = ;i<m;i++){
         scanf("%d%d", &u, &v);
         E[min(u, v)].pb(max(u, v));//小的点指向大的点
         edge[u][v] = edge[v][u] = ;
         num[u]++;
         num[v]++;
     }
     for(int i = ;i<=n;i++){
         if(num[i]<s-)  continue;
         dfs(i, );
     }
     printf("%d\n", ans);
 }
 int main() {
 #ifdef LOCAL
     freopen("input.txt", "r", stdin);
 //        freopen("output.txt", "w", stdout);
 #endif
     int T;
     scanf("%d", &T);
     while(T--){
         init();
         solve();
     }
     return ;
 }