hdu3001 travelling

Travelling

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2524    Accepted Submission(s): 738

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 11 2 1003 21 2 402 3 503 31 2 31 3 42 3 10

 

Sample Output

100907 

 

Source

2009 Multi-University Training Contest 11 - Host by HRBEU

 

Recommend

gaojie

 

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这是队友写的我还没搞懂所以先转过来以后好看看Memory: 7228 KB Time: 359 MSLanguage: G++Result: Accepted

#include <iostream>#include <stdio.h>using namespace std;int threenum[11]={1,3,9,27,81,243,729,2187,6561,19683,59049};#define inf 0x4f4f4f4fint dp[600000][15],n,m;int threeevery[600000][15],dis[15][15];int fmin(int x,int y){if(x<y)return x;return y;}void makethree()//先预处理，算出所有的值{    int i,temp,j;    for(i=0;i<threenum[10];i++)//3的１１次之内的所有状态    {        temp=i;       for(j=0;j<10;j++)       {           threeevery[i][j]=temp%3;//写出i的三进制           temp=temp/3;       }    }}void init(){    int i,j,s,e,val;    for(i=0;i<threenum[n];i++)    {        for(j=0;j<n;j++)        {            dp[i][j]=inf;        }    }    for(i=0;i<n;i++)        for(j=0;j<n;j++)            {                dis[i][j]=inf;            }      for(i=0;i<n;i++)    {        dp[threenum[i]][i]=0;//到自已的点初始化为０    }    for(i=0;i<m;i++)    {            scanf("%d%d%d",&s,&e,&val);            s--;e--;//全部从0开始           // if(dis[s][e]==-1)            //{            //    dis[s][e]=dis[e][s]=val;            //}            //else             if(val<dis[s][e])                dis[s][e]=dis[e][s]=val;//排除重边取最小值    }}int makedp(){    int i,j,k,minx;    bool flag;    minx=inf;    for(i=0;i<threenum[n];i++)//对于每一位都要计算    {        flag=true;//标记是否每一个点都已经经过        for(j=0;j<n;j++)        {            if(threeevery[i][j]==0)//是这个点没有走                flag=false;            if(dp[i][j]==inf)//没有走过            {                continue;            }            for(k=0;k<n;k++)//加入k这个点            {                if((j==k)||threeevery[i][k]>=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<n;j++)            {                minx=fmin(dp[i][j],minx);            }        }    }    if(minx==inf)    {       minx=-1;    }    printf("%d\n",minx);    return 1;}int main(){    makethree();    while(scanf("%d%d",&n,&m)!=EOF)    {        init();        makedp();    }    return 0;}