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

Output

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\n",ans);return 0;
}