D - A Simple Task CodeForces - 11D

题目链接：http://codeforces.com/problemset/problem/11/D

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 m lines contains two integers a and b, (1 ≤ a, b ≤ n, a ≠ b) indicating that vertices a and 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.

Examples

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.

题目大意：给定一个简单图，输出其中的简单环的数目。简单环的含义是，不包含重复顶点、重复边的环。
代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<stack>
#include<map>#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<cmath>
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
#define sc1(a) scanf("%lld",&a)
#define pf1(a) printf("%lld\n",a)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const LL INF=1e18;
const ull base=2333;
const int maxn=1e1+50;
const int maxm=1e3+50;
const int maxv=1e6+5;
const int mod=1e9+7;
const int ba=3e5;
LL N,M;
LL e[maxn][maxn];
LL dp[1<<20][25];//dp[S][i]  经过集合S 当前在i的总方案数
LL ans=0;
LL lowbit(LL x)
{
    for(LL i=0;i<N;i++)
    {
        if(x&(1<<i)) return i;
    }
}
void solve()
{
    for(LL i=0;i<N;i++) dp[1<<i][i]=1;
    for(LL s=0;s<(1<<N);s++)
    {
        for(LL i=0;i<N;i++)
        {
            if(s&(1<<i)) //已经走过当前点
            {
                LL x=lowbit(s);
                for(LL j=x;j<N;j++)//当前集合最小的点
                {
                    if(e[i][j]) //有边
                    {
                        if((s&(1<<j))==0) //没有走过当前点
                        {
                            dp[s|(1<<j)][j]+=dp[s][i];
                        }
                        else //已经走过了 再次走到 说明成环了
                        {
                            if(j==x)
                            {
                                ans+=dp[s][i];
                            }
                        }
                    }
                }
            }
        }
    }
    pf1((ans-M)/2);
}
int main()
{
    sc1(N);sc1(M);
    LL u,v;
    for(int i=1;i<=M;i++)
    {
        sc1(u);sc1(v);
        u--;v--;
        e[u][v]=e[v][u]=1;
    }
    solve();
    return 0;
}
/**

*/