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）
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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\n", 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 }