hdu3001 类TSP(每个点至多2次)

Travelling

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4596    Accepted Submission(s): 1517

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 
   
   
   
   

 

这道题和poj3311很像，传送门：poj3311 

但一个明显的区别是这题每个点至少访问一次，至多访问两次，因此很显然就不能floyd处理邻接矩阵了，原因可以看poj3311的题解。这题还有一个坑点是起点未知。

因此这题的dp方式很奇葩，可能有部分人暂时无法接受。dp[s][i]表示状态为s，到达点i的最小代价。这里s应该视作3进制数。由于有多个初值，因此从点i出发的dp[tri[i]][i]=0,tri数组含义见代码；

这里有一个问题初始化多个起点没问题吗？因为事实只可能有一个起点啊。

深入想想是可以的，因为本题采用的是刷表的思想(小白上有提及)，而且每次dp都是一个一个点加上去的，因此有很明显的分层现象，而本层是通过上一层比较大小得出的，虽然会有很多起点，但是经过比较，最终只会是最小的赋值给本层，有点类似打擂台的思想，可以由数学归纳法看出每个dp都只可能对应一个起点，而且不会发生都多个起点值融合成一个值，因为每个dp只可能由一个值转移过来，都是独立的，不会冲突。

这题的dp方法，真心觉得需要学习一下，这是以往的题目没出现过的。原来dp也可以这样啊，太神奇了！

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#define Maxn 59049
using namespace std;

const int inf=0x3f3f3f3f;
int tri[]={0,1,3,9,27,81,243,729,2187,6561,19683,59049};
void table(){ //打表-3的倍数
    int t=1;
    for(int i=1;i<=11;i++){
        printf("%d,",t);
        t*=3;
    }
}
int bit[Maxn][11]; //bit[i][j]表示i的三进制表示第j位数
int adj[11][11],dp[Maxn][11];
void pre(){
    for(int i=0;i<Maxn;i++){
        int t=i;
        for(int j=1;j<=10;j++){
            bit[i][j]=t%3;
            t/=3;
        }
    }
}
int main()
{
    int n,m,a,b,c;
    //table();
    pre(); //预处理bit数组
    while(cin>>n>>m){
        memset(adj,0x3f,sizeof adj);
        for(int i=0;i<m;i++){
            cin>>a>>b>>c;
            adj[a][b]=adj[b][a]=min(adj[a][b],c);
        }
        int ans=inf;
        memset(dp,0x3f,sizeof dp); //初始化为非法状态
        for(int i=1;i<=n;i++) dp[tri[i]][i]=0; //初始化多个起点,多路推进
        for(int s=0;s<tri[n+1];s++){ //枚举每个状态
            bool flag=true; //表示所有点已被遍历
            for(int i=1;i<=n;i++){ //检查s的每一位
                if(!bit[s][i]) flag=false; //点i没有被遍历
                if(dp[s][i]==inf) continue; //非法,无法刷表
                for(int j=1;j<=n;j++){ //枚举接下来需要遍历的点
                    if(bit[s][j]>=2||adj[i][j]==inf) continue; //已访问2次
                    int ns=s+tri[j];
                    dp[ns][j]=min(dp[ns][j],dp[s][i]+adj[i][j]);
                }
            }
            if(flag){ //是解
                for(int i=1;i<=n;i++) //枚举终点
                    ans=min(ans,dp[s][i]);
            }
        }
        if(ans==inf) ans=-1;
        printf("%d\n",ans);
    }
	return 0;
}