CF11D：A Simple Task（状压dp & 图）

D. A Simple Task

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 61 21 31 42 32 43 4

output

7

Note

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

题意：给N个点，M条边，计算有几个简单环。

思路：转成计算有几条链能成环，限定链上最小的数字为起点，dp[i][j]为i状态下，以j为终点的链有几条，最后答案除二即可。

# include <bits/stdc++.h>using namespace std;int v[20][20]={0};long long dp[1<<20][20]={0};int main(){    int n, m, st=0;    long long ans=0;    scanf("%d%d",&n,&m);    while(m--)    {        int a, b;        scanf("%d%d",&a,&b);        --a;--b;        v[a][b] = v[b][a] = 1;    }    for(int i=0; i<n; ++i) dp[1<<i][i] = 1;    for(int i=1; i<1<<n; ++i)    {        for(int j=0; j<n; ++j)        {            if(dp[i][j] == 0) continue;            for(int k=0; k<n; ++k)            {                if(i&(1<<k))                {                    st = k;                    break;                }            }            if(v[j][st] && __builtin_popcount(i)>2) ans += dp[i][j];            for(int k=st+1; k<n; ++k)                if(!(i&(1<<k)) && v[j][k]) dp[i|(1<<k)][k] += dp[i][j];        }    }    printf("%lld\n",ans/2);    return 0;}