Travelling(三进制状压dp)

本文介绍了一个旅行者计划访问多个城市的优化问题,他希望每座城市最多访问两次,并最小化总费用。给出的算法使用三进制状态压缩来解决,其中0、1、2分别表示未访问、访问一次和访问两次。通过预处理每种状态的三进制表示,计算以特定城市为最后一个访问点的最短路径。

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描述

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.

输入

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.

输出

Output the minimum fee that he should pay,or -1 if he can’t find such a route.

样例

  • 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

  • Output
    100
    90
    7

题解

  • 题意:从任意起点开始走完所有的点,每个点最多走两次,求最短路径
  • 与经典TSP问题不同的是,这里每个点可以走两次,所以采用三进制状压,012分别表示没走过,走过一次,走过两次。
  • 与二进制状压不同的是,我们不能通过位运算很方便地知道每一位的数值或者状态转移,所以我们要预处理出来每种状态的三进制下每一位的数值。
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