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。

 

  1 //
  2 //by coolxxx
  3 //#include<bits/stdc++.h>
  4 #include<iostream>
  5 #include<algorithm>
  6 #include<string>
  7 #include<iomanip>
  8 #include<map>
  9 #include<stack>
 10 #include<queue>
 11 #include<set>
 12 #include<bitset>
 13 #include<memory.h>
 14 #include<time.h>
 15 #include<stdio.h>
 16 #include<stdlib.h>
 17 #include<string.h>
 18 //#include<stdbool.h>
 19 #include<math.h>
 20 #pragma comment(linker,"/STACK:1024000000,1024000000")
 21 #define min(a,b) ((a)<(b)?(a):(b))
 22 #define max(a,b) ((a)>(b)?(a):(b))
 23 #define abs(a) ((a)>0?(a):(-(a)))
 24 #define lowbit(a) (a&(-a))
 25 #define sqr(a) ((a)*(a))
 26 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
 27 #define mem(a,b) memset(a,b,sizeof(a))
 28 #define eps (1e-8)
 29 #define J 10000
 30 #define mod 2147493647
 31 #define MAX 0x7f7f7f7f
 32 #define PI 3.14159265358979323
 33 #define N 104
 34 #define M 2004
 35 using namespace std;
 36 typedef long long LL;
 37 double anss;
 38 LL aans;
 39 int cas,cass;
 40 int n,m,lll,ans,sz;
 41 int last[N];
 42 struct xxx
 43 {
 44     int next,to;
 45 }a[M];
 46 int b[N],c[N],in[N];
 47 void add(int x,int y)
 48 {
 49     a[++sz].next=last[x];
 50     a[sz].to=y;
 51     last[x]=sz;
 52 }
 53 bool ma[N][N];
 54 void dfs(int top,int now)
 55 {
 56     if(top-1+lll-now<cas)return;
 57     if(top==cas+1)
 58     {
 59         ans++;
 60         return;
 61     }
 62     int i,j,to;
 63     for(i=last[now];i;i=a[i].next)
 64     {
 65         to=a[i].to;
 66         if(!ma[c[now]][c[to]] || to<now)continue;
 67         for(j=1;j<top;j++)
 68             if(!ma[c[to]][b[j]])break;
 69         if(j<top)continue;
 70         b[top]=to;
 71         dfs(top+1,to);
 72         b[top]=0;
 73     }
 74 }
 75 int main()
 76 {
 77     #ifndef ONLINE_JUDGE
 78     freopen("1.txt","r",stdin);
 79 //    freopen("2.txt","w",stdout);
 80     #endif
 81     int i,j,k;
 82     int x,y,z;
 83 //    init();
 84     for(scanf("%d",&cass);cass;cass--)
 85 //    for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
 86 //    while(~scanf("%s",s))
 87 //    while(~scanf("%d%d",&n,&m))
 88     {
 89         mem(ma,0);mem(in,0);mem(last,0);ans=0;lll=0;sz=0;
 90         scanf("%d%d%d",&n,&m,&cas);
 91         for(i=1;i<=m;i++)
 92         {
 93             scanf("%d%d",&x,&y);
 94             in[x]++,in[y]++;
 95             ma[x][y]=ma[y][x]=1;
 96         }
 97         for(i=1;i<=n;i++)
 98             if(in[i]>=cas-1)
 99                 c[++lll]=i;
100         for(i=1;i<=lll;i++)
101             for(j=i+1;j<=lll;j++)
102                 if(ma[c[i]][c[j]])
103                     add(i,j);
104         for(i=1;i<=lll;i++)
105         {
106             b[1]=c[i];
107             dfs(2,i);
108             for(j=last[c[i]];j;j=a[j].next)
109                 ma[c[a[j].to]][c[i]]=ma[c[i]][c[a[j].to]]=0;
110         }
111         printf("%d\n",ans);
112     }
113     return 0;
114 }
115 /*
116 //
117 
118 //
119 */
View Code

 

posted @ 2016-10-31 17:00  Cool639zhu  阅读(409)  评论(0)  编辑  收藏  举报