hdu3001 Travelling

Travelling
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2520 Accepted Submission(s): 735

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
用三进制来进行壮态压缩，这样就可以大大减少状态数，在预处理一下，就可以了，其次，每一个点都要走一次！

#include   #include   using namespace std;  int threenum[11]={1,3,9,27,81,243,729,2187,6561,19683,59049};  #define inf 0x4f4f4f4f  int dp[600000][15],n,m;  int threeevery[600000][15],dis[15][15];  int fmin(int x,int y){if(x
#include 
using namespace std;
int threenum[11]={1,3,9,27,81,243,729,2187,6561,19683,59049};
#define inf 0x4f4f4f4f
int dp[600000][15],n,m;
int threeevery[600000][15],dis[15][15];
int fmin(int x,int y){if(xhtml] view plaincopyprint              if(val=2||(dis[j][k]==inf))//自已到自已的地方，经过超过两次，不相连通去掉                  continue;                  dp[i+threenum[k]][k]=fmin(dp[i+threenum[k]][k],dp[i][j]+dis[j][k]);//从j到k              }            }          if(flag)//满足条件，对最小值进行更新          {              for(j=0;j=2||(dis[j][k]==inf))//自已到自已的地方，经过超过两次，不相连通去掉
                continue;
                dp[i+threenum[k]][k]=fmin(dp[i+threenum[k]][k],dp[i][j]+dis[j][k]);//从j到k
            }

        }
        if(flag)//满足条件，对最小值进行更新
        {
            for(j=0;j

hdu3001 ﻿﻿﻿﻿Travelling