hdu3001 状态压缩dp+三进制

http://acm.hdu.edu.cn/showproblem.php?pid=3001

Problem Description

After coding so many days,Mr Acmer wants to have a good rest.So travelling is the best choice!He has decided to visit n cities(he insists on seeing all the cities!And he does not mind which city being his start station because superman can bring him to any city at first but only once.), and of course there are m roads here,following a fee as usual.But Mr Acmer gets bored so easily that he doesn't want to visit a city more than twice!And he is so mean that he wants to minimize the total fee!He is lazy you see.So he turns to you for help.

 

Input

There are several test cases,the first line is two intergers n(1<=n<=10) and m,which means he needs to visit n cities and there are m roads he can choose,then m lines follow,each line will include three intergers a,b and c(1<=a,b<=n),means there is a road between a and b and the cost is of course c.Input to the End Of File.

 

Output

Output the minimum fee that he should pay,or -1 if he can't find such a route.

 

Sample Input

   
   
   
   

    
    
    
    2 1
1 2 100
3 2
1 2 40
2 3 50
3 3
1 2 3
1 3 4
2 3 10
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    100
90
7 
   
   
   
   

/**
hdu3001  状态压缩dp+三进制。
每个顶点经过最多2次，也就是说有0,1,2三总状态，我们状态转移的时候要用三进制。另外起点任意，
所以dp[bit[i]][i]=0,剩下的初始化为INF。状态转移方程为：dp[next][l] = min(dp[next][l], dp[i][j] + load[j][l]);
其中（ next = i + bit[l]）。
三进制没有对应的移位操作因此我们要模拟实现：首先把0~3^10之间的所有数都用一个数组num[i][j]表示出来，
1~10位分别代表i化为三进制后每个位上对应的值（第一位对应个位，第10位对应最高位），然后状态转移就可以了
*/

#include<stdio.h>
#include<string.h>
#include<iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;

int n, m, minn;
int load[12][12];
int bit[11] = {1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049};
int dp[60000][12];
int num[60000][12];//记录每个数的三进制数

void make_trb()//计算所有数的三进制表示
{
    for(int i = 0; i < bit[10]; i++)
    {
        int b = i;
        for(int j = 0; j < 10; j++)
        {
            num[i][j] = b % 3;
            b /= 3;
        }
    }
}

int main()
{
    make_trb();//计算所有数的三进制表示
    while(~scanf("%d%d", &n, &m))
    {
        memset(load, -1, sizeof(load));
        for(int i = 0; i < m; i++)
        {
            int a,b,c;
            scanf("%d%d%d", &a, &b, &c);
            if(load[a-1][b-1] == -1)//去重边
                load[b-1][a-1] = load[a-1][b-1] = c;
            else
                load[b-1][a-1] = load[a-1][b-1] = min(load[a-1][b-1], c);
        }
        memset(dp,0x3f3f3f3f,sizeof(dp));
        for(int j = 0; j < n; j++)
            dp[ bit[j] ][j] = 0;//对每个点定为初始点0
        int flag,next;
        minn = INF;
        for(int i = 0; i < bit[n]; i++)
        {
            flag = 1;//表示所有位都为1，即所有的城市都遍历过1次以上
            for(int j = 0; j < n; j++)//跟二进制一样遍历所有终点
            {
                if(num[i][j] == 0)
                    flag = 0;
                if(dp[i][j] == INF)
                    continue;
                for(int l = 0; l < n; l++)
                {
                    if(j == l || num[i][l] >= 2 || load[l][j] == -1)
                        continue;
                    next = i + bit[l];
                    dp[next][l] = min(dp[next][l], dp[i][j] + load[j][l]);
                }//由于跟访问路径有关，所以for l 0 -> n，用来历遍所有起点
            }
            if(flag == 1)
            {
                for(int j = 0; j < n; j++)
                    minn = min(minn , dp[i][j]);
            }
        }
        if(minn == INF)
            minn = -1;
        printf("%d\n", minn);
    }
    return 0;
}