hdu3001Travelling

参考了别人的代码   https://blog.csdn.net/u010372095/article/details/38474721

深感自己的弱小

 

这是tsp问题，和基本的tsp问题没什么大的区别，唯一的区别就是采用了三进制

原来的二进制的某一位只能表示到达或没到达过，现在加了三进制，就能表示到达过几次了

tsp问题网上的讲解我推荐这一篇    http://www.360doc.com/content/17/0826/11/36546539_682232069.shtml

我个人感觉讲的算是很详细了，剩下的就是代码实现了。

 

#include<iostream>
#include<queue>
#include<cstring>
#include<vector>
#include<cstdio>
#include<cmath>
#include<map>
#include<string>
using namespace std;
#define ll long long
#define se second
#define fi first
#define oo 0x3fffffff
int n,m;
int dp[15][60000];
int dis[15][15];
int three[12];
int in[12];
int tothree(int n);
int main()
{
    in[0] = 1;
    for(int i = 1; i <= 10; ++i)
        in[i] = in[i-1]*3;

    while(scanf("%d%d",&n,&m) != EOF)
    {
        for(int i = 0; i < n; ++i)
        {
            for(int j = 0; j < in[n]; ++j)
                dp[i][j] = -1;
            for(int j = 0; j < n; ++j)
                dis[i][j] = -1;
        }
        //cout << 1 << endl;
        for(int i = 0; i < m; ++i)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            a --; b --;
            if(dis[a][b] == -1)
                dis[a][b] = dis[b][a] = c;
            else
                dis[a][b] = dis[b][a] = min(dis[a][b],c);
        }

        int minn = -1;
        for(int l = 1; l < in[n]; ++l)
        {
            int k = tothree(l);
            for(int i = 0; i < n; ++i)
            {
                if(three[i])
                {
                    if(k == 1)
                        dp[i][l] = 0;
                    if(dp[i][l] == -1)
                        continue;
                    if(k == n)
                    {
                        if(minn == -1) minn = dp[i][l];
                        else minn = min(minn,dp[i][l]);
                    }

                    for(int j = 0; j < n; ++j)
                    if(i != j && three[j] < 2 && dis[i][j] != -1)
                    {
                        int tog = l + in[j];
                        if(dp[j][tog] == -1)
                            dp[j][tog] = dp[i][l] + dis[i][j];
                        else
                            dp[j][tog] = min(dp[j][tog],dp[i][l] + dis[i][j]);
                    }
                }
            }
        }
        printf("%d\n",minn);
    }
}
int tothree(int n)
{
    int k = 0;
    for(int i = 0; i < 10; ++i)
    {
        three[i] = n%3; n/=3; if(three[i]) k++;
    }
    return k;
}