题意: 给i到j花费的地图 1到n编号   一个人要从1遍历n个城市后回到1 求最小的花费(可以重复走) 分析 http://www.cnblogs.com/Empress/p/4039240.html TSP 因为可以重复走 所以先floyd一下求最短路 #include <cstdio> #include <cstdlib> #include <cstring> #include <climits> #include <cctype> #in…

题目链接:http://poj.org/problem?id=3311 Hie with the Pie Time Limit: 2000MS   Memory Limit: 65536K Total Submissions:12225   Accepted: 6441 Description The Pizazz Pizzeria prides itself in delivering pizzas to its customers as fast as possible. Unfortuna…

Hie with the Pie Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 3160   Accepted: 1613 Description The Pizazz Pizzeria prides itself in delivering pizzas to its customers as fast as possible. Unfortunately, due to cutbacks, they can affo…

题意:tsp问题,经过图中所有的点并回到原点的最短距离. 解题关键:floyd+状态压缩dp,注意floyd时k必须在最外层 转移方程:$dp[S][i] = \min (dp[S \wedge (1 <  < (i - 1))][k] + dis[k][j],dp[S][i])$ #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include&…

本题是经典的Tsp问题的变形,Tsp问题就是要求从起点出发经过每个节点一次再回到起点的距离最小值,本题的区别就是可以经过一个节点不止一次,那么先预处理出任意两点之间的最短距离就行了,因为再多走只会浪费更多的距离. dp[S][u]表示当前已访问的节点集合为S,从u出发走完剩余节点回到起点的最短距离. 边界条件:dp[(1<<n)-1][0]=0,最后的答案就是dp[0][0]: 记忆化递归代码: 1 #include<cstdio> 2 #include<cstring>…

题解网上一搜一大坨的,不用复述了吧. 只是觉得网上dp方程没多大问题,但是状态的表示含义模糊.不同于正常哈密顿路径求解,状态表示应当改一下. 首先定义一次移动为从一个点经过若干个点到达另一个点,则$f[S][i]$个人认为应当表示经过若干次移动,每次移动的终点状态记为$1$,由此构成的集合$S$,也就是说,每次移动的中间路过点都不算在内.$i$是最后一次移动的终点. 下面重点解决两个问题: 为什么不记路过点,状态表示仍然是对的? 不记路过点,走过的一条路,必然可以通过相同的一步一步走的路径把每一…

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 be…

题目链接:http://poj.org/problem?id=3311 学习博客:https://blog.csdn.net/u013480600/article/details/19692985 Hie with the Pie Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9954   Accepted: 5368 Description The Pizazz Pizzeria prides itself in de…

Hie with the Pie Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 4491   Accepted: 2376 Description The Pizazz Pizzeria prides itself in delivering pizzas to its customers as fast as possible. Unfortunately, due to cutbacks, they can affo…

Hie with the Pie poj-3311 题目大意:n+1个点,伪旅行商问题. 注释:n<=10. 想法:咳咳,第一道状压dp,下面我来介绍一下状压dp. 所谓dp,就是动态性决策规划,通过上一时刻或上几时刻的状态来更新当前状态并且无后效性.而状压dp就是将之前的状态通过二进制表现出来.几个例子.有五个格子_ _ _ _ _.上面可以放棋子或者不放.我们将放棋子的格子标注为1,不放棋子的格子标注为0.那么,我们就可以用一个二进制数来表达出人任何一个的完整状态而不是片面的,这就是状压dp…