P3133 [USACO16JAN] Radio Contact G 无线电通话
P3133 [USACO16JAN] Radio Contact G 无线电通话
[USACO16JAN] Radio Contact G
题目描述
Farmer John has lost his favorite cow bell, and Bessie the cow has agreed to help him find it! They both fan out and search the farm along different paths, but stay in contact via radio so they can keep in touch with each-other. Unfortunately, the batteries in their radios are running low, so they want to plan their movements so as to conserve power, by trying to stay always within a short distance apart.
Farmer John starts at location (\(f_x, f_y\)) and plans to follow a path consisting of \(N\) steps, each of which is either 'N' (north), 'E' (east), 'S' (south), or 'W' west. Bessie starts at location (\(b_x, b_y\)) and follows a similar path consisting of \(M\) steps. Both paths may share points in common. At each time step, Farmer John can either stay put at his current location, or take one step forward along his path, in whichever direction happens to be next (assuming he has not yet reached the final location in his path). Bessie can make a similar choice. At each time step (excluding the first step where they start at their initial locations), their radios consume energy equal to the square of the distance between them.
Please help FJ and Bessie plan a joint movement strategy that will minimize the total amount of energy consumed up to and including the final step where both of them first reach the final locations on their respective paths.
FJ失去了他最喜欢的牛铃,而Bessie已经同意帮助他找到它!他们用不同的路径搜索农场,通过无线电保持联系。不幸的是,无线电中的电池电量不足,所以他们设法尽可能保持两者位置的距离最小,以节省电量。
FJ从位置(fx,fy)开始,并计划遵循由N步骤组成的路径,每个步骤都是“N”(北),“E”(东),“S”(南),或“W”(西)。Bessie从位置(bx,by)开始,并遵循由M步骤组成的类似路径。两个路径可以经过相同的点。在每个时间段,FJ可以保持在他现在的位置,或沿着他的道路前进一步,无论哪个方向恰好在下一个(假设他还没有到达他的路径的最后位置)。Bessie可以做出类似的选择。在每个时间步(不包括从初始位置开始的第一步),他们的无线电消耗的能量等于它们之间距离的平方。
请帮助FJ和Bessie计划行动策略,最大限度地减少消耗的能量总量。总量包括最终步骤,这时两者首先到达各自路径上的最终位置。
输入格式
The first line of input contains \(N\) and \(M\) (\(1 \leq N, M \leq 1000\)). The
second line contains integers \(f_x\) and \(f_y\), and the third line contains \(b_x\)
and \(b_y\) (\(0 \leq f_x, f_y, b_x, b_y \leq 1000\)). The next line contains a
string of length \(N\) describing FJ's path, and the final line contains a string
of length \(M\) describing Bessie's path.
It is guranteed that Farmer John and Bessie's coordinates are always in the
range (\(0 \leq x,y \leq 1000\)) throughout their journey. Note that East points in the positive x direction and North points in the positive y direction.
第一行输入N和M(1≤N,M≤1000)。
第二行输入整数fx和fy,第三行输入bx和by(0≤fx,fy,bx,≤1000)。下一行包含一个长度为N的字符串描述FJ的路径,最后一行包含一个字符串的长度M描述Bessie的路径。
数据满足(0≤x,y≤1000)。注意,东方向为正X方向,北方向为正Y方向。
输出格式
Output a single integer specifying the minimum energy FJ and Bessie can use
during their travels.
输出一个整数,表示最小能量。
样例 #1
样例输入 #1
2 7
3 0
5 0
NN
NWWWWWN
样例输出 #1
28
提示
感谢@ prcups 改进翻译
思路
先分别求出农夫和奶牛没走一步到达的点
设 \(dp_{i , j}\) 当农夫走了 \(i\) 步,奶牛走了 \(j\) 步后的最小花费
那么
\]
注意判断越界条件,\(dp_{0 , 0} = 0\)
后记
考试时 \(dis\) 打错了,差点就寄了。
P3133 [USACO16JAN] Radio Contact G 无线电通话的更多相关文章
- 洛谷 P3133 [USACO16JAN]Radio Contact G
题目传送门 解题思路: f[i][j]表示FJ走了i步,Bessie走了j步的最小消耗值.方程比较好推. 横纵坐标要搞清楚,因为这东西WA了半小时. AC代码: #include<iostrea ...
- 洛谷 P3133 [USACO16JAN]无线电联系Radio Contact
P3133 [USACO16JAN]无线电联系Radio Contact 题目描述 Farmer John has lost his favorite cow bell, and Bessie the ...
- 「BZOJ4510」「Usaco2016 Jan」Radio Contact 解题报告
无线电联系 Radio Contact 题目描述 Farmer John has lost his favorite cow bell, and Bessie the cow has agreed t ...
- P3133 [USACO16JAN]无线电联系Radio Contact
题目描述 Farmer John has lost his favorite cow bell, and Bessie the cow has agreed to help him find it! ...
- [USACO16JAN]Angry Cows G 解题报告
一图流 参考代码: #include<bits/stdc++.h> #define ll long long #define db double #define filein(a) fre ...
- 越狱Season 1-Episode 7: Riots, Drills and the Devil: Part 2
Season 1, Episode 7: Riots, Drills and the Devil: Part 2 -Pope: Belick, get those guys in line guy: ...
- VCF文件导入导出
参考资料 通讯录导入导出vcf格式文件方法可参考: https://qiaodahai.com/android-iphone-mobile-phones-contacts-import-and-exp ...
- [SinGuLaRiTy] 2017-07-21 综合性测试
[SinGuLaRiTy-1028] Copyright (c) SinGuLaRiTy 2017. All Rights Reserved. 对于所有题目:Time Limit: 1s | Memo ...
- bzoj AC倒序
Search GO 说明:输入题号直接进入相应题目,如需搜索含数字的题目,请在关键词前加单引号 Problem ID Title Source AC Submit Y 1000 A+B Problem ...
- BlackArch-Tools
BlackArch-Tools 简介 安装在ArchLinux之上添加存储库从blackarch存储库安装工具替代安装方法BlackArch Linux Complete Tools List 简介 ...
随机推荐
- 【Diary】JZSC 2020 旅 游 记(迫真
Day-2 期末考试的day1. 科目是数学 政治 语文.数学25.(3)没动.政治各种抄选择题选项()语文难得写完了.作文压根不知道写的什么 Day-1 期末考试的day2. 科目是英语 物理 历史 ...
- mariabackup -prepare step on increment backup failed
问题描述:使用mariabackup对maridb10.6.4进行物理备份,进行增量恢复的时候报错.截止到目前,还是mariadb的一个bug,还没有修复.在增备的过程中如果出现新库的建立,数据库就会 ...
- 【Vue项目】尚品汇(四)Search组件开发
Search模块开发 分析:1)编写静态页面 2)编写api 3)编写vuex三大件 4)组件获取仓库数据,并进行动态展示 1 SearchSelector 1 编写api export const ...
- 【谷粒商城】(二)SpringCloudAlibaba分布式组件
微服务 分布式组件 注册中心:每一个微服务上线都应该注册到注册中心.这样做的好处在于方便微服务之间的相互调用,比如订单服务想要调用商品服务,就可以通过注册中心查看有哪几台主机的商品服务进行了注册,然后 ...
- vue侦听器(引入vue.js写法)
首先在html中引入vue.js,具体怎么下载可以参考https://blog.csdn.net/lvoelife/article/details/129254906,下载后在html中引入: 侦听d ...
- 搞懂Python正则表达式,这一篇就够了
本文代码基于Python3.11解释器,除了第一次示例,代码将省略 import re 这个语句 所有示例代码均可以在我的github仓库中的 code.py文件内查看 [我的仓库](PythonLe ...
- Portainer安装
个人博客地址: https://note.raokun.top 拥抱ChatGPT,国内访问网站:https://www.playchat.top Portainer是一个可视化的容器镜像的图形管理工 ...
- 如何生成文本: 通过 Transformers 用不同的解码方法生成文本
简介 近年来,随着以 OpenAI GPT2 模型 为代表的基于数百万网页数据训练的大型 Transformer 语言模型的兴起,开放域语言生成领域吸引了越来越多的关注.开放域中的条件语言生成效果令人 ...
- MySQL高频面试题
什么是MySQL MySQL是一个关系型数据库,它采用表的形式来存储数据.你可以理解成是Excel表格,既然是表的形式存储数据,就有表结构(行和列).行代表每一行数据,列代表该行中的每个值.列上的值是 ...
- OpenResty学习笔记03:再探WAF
一. 再谈WAF 我们上一篇安装的WAF来自另一位技术大神 赵舜东,花名 赵班长,一直从事自动化运维方面的架构设计工作.阿里云MVP.华为云MVP.中国SaltStack用户组发起人 .新运维社区发起 ...