HihoCoder1087Hamiltonian Cycle(DP状态压缩)

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

Given a directed graph containing n vertice (numbered from 1 to n) and m edges. Can you tell us how many different Hamiltonian Cycles are there in this graph?

A Hamiltonian Cycle is a cycle that starts from some vertex, visits each vertex (except for the start vertex) exactly once, and finally ends at the start vertex.

Two Hamiltonian Cycles C1, C2 are different if and only if there exists some vertex i that, the next vertex of vertex i in C1 is different from the next vertex of vertex i in C2.

输入

The first line contains two integers n and m. 2 <= n <= 12, 1 <= m <= 200.

Then follows m line. Each line contains two different integers a and b, indicating there is an directed edge from vertex a to vertex b.

输出

Output an integer in a single line -- the number of different Hamiltonian Cycles in this graph.

提示

额外的样例:

样例输入	样例输出
3 3 1 2                2 1               1 3	0

样例输入

4 7
1 2
2 3
3 4
4 1
1 3
4 2
2 1

样例输出

2

搜索大概也可以搞定。

• 求哈密顿环的数目
• 既然是环，且每个点都经过，我们假定一个起点，得到的结果是一样的，我的代码假定的是1为起点。
• 这题有重边，但是必须两个点之间重边只看成一条边才能AC

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
using namespace std;
int Mp[][];
int dp[<<][];//vis,now
int main()
{
    int n,m,x,y,i,k,p,ans=;
    scanf("%d%d",&n,&m);
    for(i=;i<=m;i++){
        scanf("%d%d",&x,&y);
        Mp[x][y]=;
    }
    for(i=;i<=n;i++) dp[][]=;
    for(i=;i<(<<n);i++)
    {
         for(k=;k<=n;k++)//now
         {
            if(!(i&(<<(k-)))) continue;
            for(p=;p<=n;p++)//pre
            {
               if(!(i&(<<(p-)))||k==p) continue;
               dp[i][k]=dp[i][k]+dp[i^(<<(k-))][p]*Mp[p][k];
            }
         }
    }
    for(i=;i<=n;i++) ans+=dp[(<<n)-][i]*Mp[i][];
    printf("%d\n",ans);
    return ;
}