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
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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。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <string> 5 #include <algorithm> 6 #include <cmath> 7 #include <vector> 8 #include <queue> 9 #include <map> 10 #include <stack> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 #define ms(a, b) memset(a, b, sizeof(a)) 15 #define pb push_back 16 #define mp make_pair 17 const LL INF = 0x7fffffff; 18 const int inf = 0x3f3f3f3f; 19 const int mod = 1e9+7; 20 const int maxn = 100+10; 21 bool edge[maxn][maxn]; 22 vector <int> E[maxn]; 23 int num[maxn]; 24 int p[20]; 25 int ans, n, m, s, u, v; 26 void init() { 27 ms(edge, 0); 28 ms(num, 0); 29 for(int i = 0;i<maxn;i++) E[i].clear(); 30 ans = 0; 31 } 32 void dfs(int x, int len){ 33 int flag = 1; 34 for(int i=0;i<len;i++){ 35 if(!edge[x][p[i]]){ 36 flag = 0; 37 break; 38 } 39 } 40 if(!flag){ 41 return; 42 } 43 p[len] = x; 44 if(len+1==s){ 45 ans++; 46 return; 47 } 48 for(int i = 0;i<E[x].size();i++){ 49 dfs(E[x][i], len+1); 50 } 51 } 52 void solve() { 53 scanf("%d%d%d", &n, &m, &s); 54 for(int i = 0;i<m;i++){ 55 scanf("%d%d", &u, &v); 56 E[min(u, v)].pb(max(u, v));//小的点指向大的点 57 edge[u][v] = edge[v][u] = 1; 58 num[u]++; 59 num[v]++; 60 } 61 for(int i = 1;i<=n;i++){ 62 if(num[i]<s-1) continue; 63 dfs(i, 0); 64 } 65 printf("%d ", ans); 66 } 67 int main() { 68 #ifdef LOCAL 69 freopen("input.txt", "r", stdin); 70 // freopen("output.txt", "w", stdout); 71 #endif 72 int T; 73 scanf("%d", &T); 74 while(T--){ 75 init(); 76 solve(); 77 } 78 return 0; 79 }