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-1+lll-now<cas)return;
    if(top==cas+1)
    {
        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=1;j<top;j++)
            if(!ma[c[to]][b[j]])break;
        if(j<top)continue;
        b[top]=to;
        dfs(top+1,to);
        b[top]=0;
    }
}
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,0);mem(in,0);mem(last,0);ans=0;lll=0;sz=0;
        scanf("%d%d%d",&n,&m,&cas);
        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&x,&y);
            in[x]++,in[y]++;
            ma[x][y]=ma[y][x]=1;
        }
        for(i=1;i<=n;i++)
            if(in[i]>=cas-1)
                c[++lll]=i;
        for(i=1;i<=lll;i++)
            for(j=i+1;j<=lll;j++)
                if(ma[c[i]][c[j]])
                    add(i,j);
        for(i=1;i<=lll;i++)
        {
            b[1]=c[i];
            dfs(2,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]]=0;
        }
        printf("%d
",ans);
    }
    return 0;
}
/*
//

//
*/