ZOJ2972 Hurdles of 110m java

2023-10-12 18:40
文章标签 java hurdles zoj2972 110m

本文主要是介绍ZOJ2972 Hurdles of 110m java,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2972

                                      Hurdles of 110m

In the year 2008, the 29th Olympic Games will be held in Beijing. This will signify the prosperity of China and Beijing Olympics is to be a festival for people all over the world as well.

Liu Xiang is one of the famous Olympic athletes in China. In 2002 Liu broke Renaldo Nehemiah's 24-year-old world junior record for the 110m hurdles. At the 2004 Athens Olympics Games, he won the gold medal in the end. Although he was not considered as a favorite for the gold, in the final, Liu's technique was nearly perfect as he barely touched the sixth hurdle and cleared all of the others cleanly. He powered to a victory of almost three meters. In doing so, he tied the 11-year-old world record of 12.91 seconds. Liu was the first Chinese man to win an Olympic gold medal in track and field. Only 21 years old at the time of his victory, Liu vowed to defend his title when the Olympics come to Beijing in 2008.

                                           

In the 110m hurdle competition, the track was divided into N parts by the hurdle. In each part, the player has to run in the same speed; otherwise he may hit the hurdle. In fact, there are 3 modes to choose in each part for an athlete -- Fast Mode, Normal Mode and Slow Mode. Fast Mode costs the player T1 time to pass the part. However, he cannot always use this mode in all parts, because he needs to consume F1 force at the same time. If he doesn't have enough force, he cannot run in the part at the Fast Mode. Normal Mode costs the player T2 time for the part. And at this mode, the player's force will remain unchanged. Slow Mode costs the player T3 time to pass the part. Meanwhile, the player will earn F2 force as compensation. The maximal force of a player is M. If he already has M force, he cannot earn any more force. At the beginning of the competition, the player has the maximal force.

The input of this problem is detail data for Liu Xiang. Your task is to help him to choose proper mode in each part to finish the competition in the shortest time.

Input

Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <= 50) which is the number of test cases. And it will be followed by T consecutive test cases.

Each test case begins with two positive integers N and M. And following N lines denote the data for the N parts. Each line has five positive integers T1 T2 T3 F1 F2. All the integers in this problem are less than or equal to 110.

Output

Results should be directed to standard output. The output of each test case should be a single integer in one line, which is the shortest time that Liu Xiang can finish the competition.

Sample Input

2

1 10

1 2 3 10 10

4 10

1 2 3 10 10

1 10 10 10 10

1 1 2 10 10

1 10 10 10 10

Sample Output

1

6

Hint

For the second sample test case, Liu Xiang should run with the sequence of Normal Mode, Fast Mode, Slow Mode and Fast Mode.

题意:

跨栏跑分为几段,每段有三种跑步方式,不同方式通过每段路程时间不同,并消耗或补充体力,要求输出最短时间。

思路:

动态规划的题目。

代码:

import java.util.Arrays;
import java.util.Scanner;
//动态规划
public class Main {public static void main(String[] args) {Scanner sc=new Scanner(System.in);while(sc.hasNext()){int t=sc.nextInt();while(t-->0){int n=sc.nextInt();int m=sc.nextInt();int temp=0;int dp[][]=new int[n+1][m+1];for(int i=0;i<n+1;i++){for(int j=0;j<m+1;j++){if(i==0){dp[i][j]=0;}else{dp[i][j]=10000;}}}for(int i=1;i<=n;i++){int t1=sc.nextInt();int t2=sc.nextInt();int t3=sc.nextInt();int f1=sc.nextInt();int f2=sc.nextInt();for(int j=0;j<=m;j++){dp[i][j]=Math.min(dp[i-1][j]+t2, dp[i][j]);if(j>=f1){dp[i][j-f1]=Math.min(dp[i-1][j]+t1, dp[i][j-f1]);}temp=Math.min(j+f2,m);dp[i][temp] = Math.min(dp[i - 1][j] + t3, dp[i][temp]);}}int a=100000;for(int j=0;j<=m;j++){a=Math.min(a, dp[n][j]);}System.out.println(a);}}}
}

这篇关于ZOJ2972 Hurdles of 110m java的文章就介绍到这儿,希望我们推荐的文章对编程师们有所帮助!



http://www.chinasem.cn/article/197712

相关文章

Java 正则表达式URL 匹配与源码全解析

《Java正则表达式URL匹配与源码全解析》在Web应用开发中,我们经常需要对URL进行格式验证,今天我们结合Java的Pattern和Matcher类,深入理解正则表达式在实际应用中... 目录1.正则表达式分解:2. 添加域名匹配 (2)3. 添加路径和查询参数匹配 (3) 4. 最终优化版本5.设计思

Java使用ANTLR4对Lua脚本语法校验详解

《Java使用ANTLR4对Lua脚本语法校验详解》ANTLR是一个强大的解析器生成器,用于读取、处理、执行或翻译结构化文本或二进制文件,下面就跟随小编一起看看Java如何使用ANTLR4对Lua脚本... 目录什么是ANTLR?第一个例子ANTLR4 的工作流程Lua脚本语法校验准备一个Lua Gramm

Java字符串操作技巧之语法、示例与应用场景分析

《Java字符串操作技巧之语法、示例与应用场景分析》在Java算法题和日常开发中,字符串处理是必备的核心技能,本文全面梳理Java中字符串的常用操作语法,结合代码示例、应用场景和避坑指南,可快速掌握字... 目录引言1. 基础操作1.1 创建字符串1.2 获取长度1.3 访问字符2. 字符串处理2.1 子字

Java Optional的使用技巧与最佳实践

《JavaOptional的使用技巧与最佳实践》在Java中,Optional是用于优雅处理null的容器类,其核心目标是显式提醒开发者处理空值场景,避免NullPointerExce... 目录一、Optional 的核心用途二、使用技巧与最佳实践三、常见误区与反模式四、替代方案与扩展五、总结在 Java

基于Java实现回调监听工具类

《基于Java实现回调监听工具类》这篇文章主要为大家详细介绍了如何基于Java实现一个回调监听工具类,文中的示例代码讲解详细,感兴趣的小伙伴可以跟随小编一起学习一下... 目录监听接口类 Listenable实际用法打印结果首先,会用到 函数式接口 Consumer, 通过这个可以解耦回调方法,下面先写一个

使用Java将DOCX文档解析为Markdown文档的代码实现

《使用Java将DOCX文档解析为Markdown文档的代码实现》在现代文档处理中,Markdown(MD)因其简洁的语法和良好的可读性,逐渐成为开发者、技术写作者和内容创作者的首选格式,然而,许多文... 目录引言1. 工具和库介绍2. 安装依赖库3. 使用Apache POI解析DOCX文档4. 将解析

Java字符串处理全解析(String、StringBuilder与StringBuffer)

《Java字符串处理全解析(String、StringBuilder与StringBuffer)》:本文主要介绍Java字符串处理全解析(String、StringBuilder与StringBu... 目录Java字符串处理全解析:String、StringBuilder与StringBuffer一、St

springboot整合阿里云百炼DeepSeek实现sse流式打印的操作方法

《springboot整合阿里云百炼DeepSeek实现sse流式打印的操作方法》:本文主要介绍springboot整合阿里云百炼DeepSeek实现sse流式打印,本文给大家介绍的非常详细,对大... 目录1.开通阿里云百炼,获取到key2.新建SpringBoot项目3.工具类4.启动类5.测试类6.测

Spring Boot循环依赖原理、解决方案与最佳实践(全解析)

《SpringBoot循环依赖原理、解决方案与最佳实践(全解析)》循环依赖指两个或多个Bean相互直接或间接引用,形成闭环依赖关系,:本文主要介绍SpringBoot循环依赖原理、解决方案与最... 目录一、循环依赖的本质与危害1.1 什么是循环依赖?1.2 核心危害二、Spring的三级缓存机制2.1 三

在Spring Boot中浅尝内存泄漏的实战记录

《在SpringBoot中浅尝内存泄漏的实战记录》本文给大家分享在SpringBoot中浅尝内存泄漏的实战记录,结合实例代码给大家介绍的非常详细,感兴趣的朋友一起看看吧... 目录使用静态集合持有对象引用,阻止GC回收关键点:可执行代码:验证:1,运行程序(启动时添加JVM参数限制堆大小):2,访问 htt