标签：cas back put cto clu between 引入 eve void
InputThe 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.OutputFor 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
思路：
如何找到一个k阶的完全图？如果一个图是完全图，那么引入一个新的点，这个点与原图中的每一点都有边相连，新图还是完全图。用了num数组来记录原图上的点。建图时，、只建了从编号小的点到编号大的点之间的边。
这是由于，每次没有必要建立反向边。反而不建反向边的话，会少了去重的过程。
代码：
#include<iostream> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<map> #include<set> #include<cstdio> #include<cstring> #include<cmath> #include<ctime> #define fuck(x) cout<<#x<<" = "<<x<<endl; #define ls (t<<1) #define rs ((t<<1)+1) using namespace std; typedef long long ll; typedef unsigned long long ull; const int maxn = 108; const int inf = 2.1e9; const ll Inf = 999999999999999999; const int mod = 1000000007; const double eps = 1e-6; const double pi = acos(-1); vector<int>u[maxn]; int n,m,k; int ans; int top; int num[maxn]; bool mp[maxn][maxn]; void dfs(int t,int d) { if(d==k){ans++;return;} int siz = u[t].size(); bool flag = false; for(int i=0;i<siz;i++){ int cnt = u[t][i]; flag = false; for(int j=1;j<=top;j++){ if(!mp[cnt][num[j]]){flag = true;break;} } if(flag){continue;} num[++top]=cnt; dfs(cnt,d+1); top--; } } int main() { int T; scanf("%d",&T); while(T--){ ans = 0; scanf("%d%d%d",&n,&m,&k); memset(mp,0,sizeof(mp)); for(int i=1;i<=n;i++){ u[i].clear(); } int x,y; for(int i=1;i<=m;i++){ scanf("%d%d",&x,&y); if(x>y){swap(x,y);} u[x].push_back(y); mp[x][y]=mp[y][x]=true; } for(int i=1;i<=n;i++){ num[++top]=i; dfs(i,1); top--; } printf("%d\n",ans); } return 0; }
HDU - 5952 Counting Cliques(DFS)
标签：cas back put cto clu between 引入 eve void
原文地址：https://www.cnblogs.com/ZGQblogs/p/9890878.html
