CodeForces 11D(状压DP 求图中环的个数)

Given a simple graph, output the number of simple cycles in it. A simple cycle is a cycle with no repeated vertices or edges.

Input

The first line of input contains two integers n and m (1 ≤ n ≤ 19, 0 ≤ m) – respectively the number of vertices and edges of the graph. Each of the subsequent mlines contains two integers a and b, (1 ≤ a, b ≤ n, a ≠ b) indicating that vertices aand b are connected by an undirected edge. There is no more than one edge connecting any pair of vertices.

Output

Output the number of cycles in the given graph.

Example

Input

4 6
1 2
1 3
1 4
2 3
2 4
3 4

Output

7

Note

The example graph is a clique and contains four cycles of length 3 and three cycles of length 4.

 1-2-3 2-3-4 1-3-4 1-2-4 1-2-3-4 1-2-4-3 1-4-2-3

题意:给出一个图的点数和边数输出这个图中有几个环.

题解:这道题可以很容易想到状压,因为数据也只有19(orz).用sta的二进制表示已经有几个点在这个状态中,那怎么表示形成环呢?只需要找到一个当前状态中的点,它神奇的与前面的某一个点有一条边连着,那么说明肯定能构成环,可以为总答案做贡献.如果不能,那么就为下一个状态提供个数.不过由于是无向图.所以两个点也会被认为成环(两个点构成环的个数为边数),并且每个更大的环都会被计算两次,所以最后要减掉.然后就搞定了.

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;

long long dp[1<<20][20],ans=0;
vector<int> e[20];

int lowbit(int x)
{
    return x&(-x);
}

int main()
{
    int n,m,f,t;
    scanf("%d%d",&n,&m);
    
    for(int i=0;i<m;i++)
    {
        scanf("%d%d",&f,&t);
        e[f-1].push_back(t-1);
        e[t-1].push_back(f-1);
    }
    
    for(int i=0;i<n;i++)
    {
        dp[1<<i][i]=1;
    }
    
    for(int sta=1;sta<(1<<n);sta++)
    {
        for(int i=0;i<n;i++)
        {
            if(dp[sta][i])
            {
                for(int k=0;k<e[i].size();k++)
                {
                    int j=e[i][k];
                    if(lowbit(sta)>(1<<j))
                    {
                        continue;
                    }
                    if(sta&(1<<j))
                    {
                        if(lowbit(sta)==(1<<j))
                        {
                            ans+=dp[sta][i];
                        }
                    }
                    else
                    {
                        dp[sta|(1<<j)][j]+=dp[sta][i];          
                }
            }
        }
    }
    
    ans=(ans-m)/2;
    printf("%lld
",ans);
    return 0;
}

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