【DP】 hdu3001 Travelling

Travelling

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 
  

题解：类似旅行商问题，特殊的地方就是每个点最多可以经过两次，求经过每个点的路径的最小代价。改用三进制记录，每位代表经过那个点的次数。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define inf ((1<<30)-1)
int map[12][12];
int dp[59049][12];//i状态当前在j点
int dig[59049][12];//i状态j点去过几次
int thr[12]= {0,1,3,9,27,81,243,729,2187,6561,19683,59049};
int main()
{
    int n,m;
    int a,b,c;
    memset(dig,0,sizeof(dig));
    for(int i=0; i<59049; ++i)
    {
        int t=i;
        for(int j=1; j<=10&&t; ++j)
        {
            dig[i][j]=t%3;
            t/=3;
        }
    }
    for(; ~scanf("%d%d",&n,&m);)
    {
        memset(map,-1,sizeof(map));
        memset(dp,-1,sizeof(dp));
        for(; m--;)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(map[a][b]==-1)
                map[a][b]=map[b][a]=c;
            else
                map[a][b]=map[b][a]=min(map[a][b],c);//消除重边
        }
        for(int i=1; i<=n; ++i) dp[thr[i]][i]=0;
        int ans=inf;
        for(int i=0; i<thr[n+1]; ++i)
        {
            bool flag=true;
            for(int j=1; j<=n; ++j)
            {
                if(!dig[i][j]) flag=false;
                if(dp[i][j]==-1) continue;
                for(int k=1; k<=n; ++k)
                {
                    if((k==j)||(map[j][k]==-1)||(dig[i][k]>=2))
                        continue;
                    if((dp[i+thr[k]][k]==-1)||(dp[i+thr[k]][k]>dp[i][j]+map[j][k]))
                        dp[i+thr[k]][k]=dp[i][j]+map[j][k];
                }
            }
            if(flag)
            {
                for(int j=1; j<=n; ++j)
                    if(dp[i][j]!=-1)
                        ans=min(ans,dp[i][j]);
            }
        }
        if(ans==inf) puts("-1");
        else         printf("%d\n",ans);
    }
    return 0;
}

来源： http://blog.youkuaiyun.com/ACM_Ted