HDU 5952 Counting Cliques 【DFS+剪枝】 （2016ACM/ICPC亚洲区沈阳站）

Counting Cliques

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 539    Accepted Submission(s): 204

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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题目链接：

　　http://acm.hdu.edu.cn/showproblem.php?pid=5952

题目大意：

　　给N个点M条边，求大小位S的集合的个数，要求集合内的元素两两有边（完全图）。

题目思路：

　　【DFS+剪枝】

　　每个点最多20条出边。暴力。

　　首先去掉边数<S-1的点，枚举剩下的X，枚举完X可以把X的所有边删掉。加几个小剪枝即可。

 //
 //by coolxxx
 //#include<bits/stdc++.h>
 #include<iostream>
 #include<algorithm>
 #include<string>
 #include<iomanip>
 #include<map>
 #include<stack>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<memory.h>
 #include<time.h>
 #include<stdio.h>
 #include<stdlib.h>
 #include<string.h>
 //#include<stdbool.h>
 #include<math.h>
 #pragma comment(linker,"/STACK:1024000000,1024000000")
 #define min(a,b) ((a)<(b)?(a):(b))
 #define max(a,b) ((a)>(b)?(a):(b))
 #define abs(a) ((a)>0?(a):(-(a)))
 #define lowbit(a) (a&(-a))
 #define sqr(a) ((a)*(a))
 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
 #define mem(a,b) memset(a,b,sizeof(a))
 #define eps (1e-8)
 #define J 10000
 #define mod 2147493647
 #define MAX 0x7f7f7f7f
 #define PI 3.14159265358979323
 #define N 104
 #define M 2004
 using namespace std;
 typedef long long LL;
 double anss;
 LL aans;
 int cas,cass;
 int n,m,lll,ans,sz;
 int last[N];
 struct xxx
 {
     int next,to;
 }a[M];
 int b[N],c[N],in[N];
 void add(int x,int y)
 {
     a[++sz].next=last[x];
     a[sz].to=y;
     last[x]=sz;
 }
 bool ma[N][N];
 void dfs(int top,int now)
 {
     if(top-+lll-now<cas)return;
     if(top==cas+)
     {
         ans++;
         return;
     }
     int i,j,to;
     for(i=last[now];i;i=a[i].next)
     {
         to=a[i].to;
         if(!ma[c[now]][c[to]] || to<now)continue;
         for(j=;j<top;j++)
             if(!ma[c[to]][b[j]])break;
         if(j<top)continue;
         b[top]=to;
         dfs(top+,to);
         b[top]=;
     }
 }
 int main()
 {
     #ifndef ONLINE_JUDGE
     freopen("1.txt","r",stdin);
 //    freopen("2.txt","w",stdout);
     #endif
     int i,j,k;
     int x,y,z;
 //    init();
     for(scanf("%d",&cass);cass;cass--)
 //    for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
 //    while(~scanf("%s",s))
 //    while(~scanf("%d%d",&n,&m))
     {
         mem(ma,);mem(in,);mem(last,);ans=;lll=;sz=;
         scanf("%d%d%d",&n,&m,&cas);
         for(i=;i<=m;i++)
         {
             scanf("%d%d",&x,&y);
             in[x]++,in[y]++;
             ma[x][y]=ma[y][x]=;
         }
         for(i=;i<=n;i++)
             if(in[i]>=cas-)
                 c[++lll]=i;
         for(i=;i<=lll;i++)
             for(j=i+;j<=lll;j++)
                 if(ma[c[i]][c[j]])
                     add(i,j);
         for(i=;i<=lll;i++)
         {
             b[]=c[i];
             dfs(,i);
             for(j=last[c[i]];j;j=a[j].next)
                 ma[c[a[j].to]][c[i]]=ma[c[i]][c[a[j].to]]=;
         }
         printf("%d\n",ans);
     }
     return ;
 }
 /*
 //

 //
 */