Hie with the Pie POJ - 3311 （状压dp+floyd）

最新推荐文章于 2019-08-17 16:15:20 发布

elbadaernu

最新推荐文章于 2019-08-17 16:15:20 发布

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分类专栏：状压相关

本文链接：https://blog.youkuaiyun.com/elbadaernu/article/details/78129442

该博客介绍了如何处理一个关于外卖配送的问题，即在一个包含起点和多个目的地的网络中，找到从起点出发，送达所有目的地后再返回起点的最短时间。博主首先使用Floyd算法计算任意两点间的最短距离，接着采用状态压缩动态规划方法（状压dp）求解最优路径，最后给出了实现代码。

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Description

The Pizazz Pizzeria prides itself in delivering pizzas to its customers as fast as possible. Unfortunately, due to cutbacks, they can afford to hire only one driver to do the deliveries. He will wait for 1 or more (up to 10) orders to be processed before he starts any deliveries. Needless to say, he would like to take the shortest route in delivering these goodies and returning to the pizzeria, even if it means passing the same location(s) or the pizzeria more than once on the way. He has commissioned you to write a program to help him.

Input

Input will consist of multiple test cases. The first line will contain a single integer n indicating the number of orders to deliver, where 1 ≤ n ≤ 10. After this will be n + 1 lines each containing n + 1 integers indicating the times to travel between the pizzeria (numbered 0) and the n locations (numbers 1 to n). The jth value on the ith line indicates the time to go directly from location i to location j w