CF11D：A Simple Task（状压dp & 图）

最新推荐文章于 2023-04-08 11:00:02 发布

junior19

最新推荐文章于 2023-04-08 11:00:02 发布

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分类专栏：状压dp

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本文介绍了一种高效算法，用于计算给定简单图中的简单环数量。通过使用位操作和动态规划技巧，该算法能在有限时间内找出所有不重复顶点和边的环。

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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;
}