hdu 3001 Travelling

Travelling

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6735    Accepted Submission(s): 2189

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 
  

 

Source

2009 Multi-University Training Contest 11 - Host by HRBEU

 

Recommend

gaojie

 

Statistic |  Submit |  Discuss |  Note 题意：给你两个数n，m，n代表有多少个城市，m代表有多少条有权值的路，一个人想逛完全部的地方，但是同一个地点不能停留超过两次，问所需的最少费用是多少。

思路：明显是状压dp，但是之前我只会二进制的所以楞是没想出来，上网看了题解才知道弄成三进制的就好了，和一般的题目一样第一维表示状态（即每个地点走过几次），第二维表示最后停留的地点，然后就是普通的状压dp。。关键是如何化成三进制，化成三进制的部分挺难说明的，看代码吧。

#include<iostream>
#include<cmath>
#include<queue>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<string>
#define maxn 100005
#define inf 0x3f3f3f3f
typedef long long LL;
using namespace std;
int digit[60000][11],dp[60000][11],dis[11][11];//digit[i][j]表示i在三进制下的第j位数是多少
int main(){
    int n,m;
    int three[11]={1};//three求出3的n次方
    for(int i=1;i<=10;i++){
        three[i]=three[i-1]*3;
    }
    for(int i=1;i<three[10];i++){
        int temp=i;
        for(int j=1;j<11;j++){
            digit[i][j]=temp%3;
            temp/=3;
            if(!temp)
              break;
        }
    }
    while(~scanf("%d%d",&n,&m)){
        memset(dis,-1,sizeof(dis));
        memset(dp,-1,sizeof(dp));
        for(int i=0;i<m;i++){
            int u,v,value;
            scanf("%d%d%d",&u,&v,&value);
            if(dis[u][v]==-1){
                dis[u][v]=dis[v][u]=value;
            }
            else{
                dis[u][v]=dis[v][u]=min(dis[u][v],value);
            }
        }
        for(int i=1;i<=n;i++){
            dp[three[i-1]][i]=0;
        }
        int ans=inf;
        for(int i=1;i<three[n];i++){
            bool jud=true;
            for(int j=1;j<=n;j++){
                if(digit[i][j]){
                    if(dp[i][j]!=-1){
                        for(int k=1;k<=n;k++){
                            if(j==k||dis[j][k]==-1)
                             continue;
                            if(digit[i][k]<2){
                                if(dp[i+three[k-1]][k]==-1)
                                    dp[i+three[k-1]][k]=dp[i][j]+dis[j][k];
                                else
                                    dp[i+three[k-1]][k]=min(dp[i+three[k-1]][k],dp[i][j]+dis[j][k]);                                      
                            }
                        }
                    }
                   
                }
                else
                  jud=false;
            }
            if(jud){
                for(int j=1;j<=n;j++){
                    if(dp[i][j]==-1)
                      continue;
                    ans=min(dp[i][j],ans);
                }
            }
        }
        if(ans==inf)
          printf("-1\n");
        else
          printf("%d\n",ans);
    }
}//g++ -o a.exe a.cpp