Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 32044    Accepted Submission(s): 14425

Problem Description
Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now.The game can be played by two or more than two players. It consists of a chessboard(棋盘)and some chessmen(棋子), and all chessmen are marked by a positive integer or “start” or “end”. The player starts from start-point and must jumps into end-point finally. In the course of jumping, the player will visit the chessmen in the path, but everyone must jumps from one chessman to another absolutely bigger (you can assume start-point is a minimum and end-point is a maximum.). And all players cannot go backwards. One jumping can go from a chessman to next, also can go across many chessmen, and even you can straightly get to end-point from start-point. Of course you get zero point in this situation. A player is a winner if and only if he can get a bigger score according to his jumping solution. Note that your score comes from the sum of value on the chessmen in you jumping path. Your task is to output the maximum value according to the given chessmen list.
 
Input
Input contains multiple test cases. Each test case is described in a line as follow: N value_1 value_2 …value_N  It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int. A test case starting with 0 terminates the input and this test case is not to be processed.
 
Output
For each case, print the maximum according to rules, and one line one case.
 
Sample Input
3 1 3 2
4 1 2 3 4
4 3 3 2 1
0
 
Sample Output
4
10
3
 

题解:求单调上升的序列和的最大值;简单动态规划:

代码:

import java.util.Scanner;

public class hdu1087 {
public static void main(String[] argvs){
Scanner cin = new Scanner(System.in);
while(true){
int n = cin.nextInt();
if(n == )break;
int a[] = new int[n];
int dp[] = new int[n];
int ans = -0x3f3f3f3f;
for(int i = ; i < n; i++){
a[i] = cin.nextInt();
dp[i] = a[i];
for(int j = ; j < i; j++){
if(a[j] < a[i]){
dp[i] = Math.max(dp[j] + a[i], dp[i]);
}
}
ans = Math.max(ans, dp[i]);
}
System.out.println(ans);
}
}
}

Super Jumping! Jumping! Jumping!(dp)的更多相关文章

  1. hdu 1087 Super Jumping! Jumping! Jumping! 简单的dp

    Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 ...

  2. hdu 1087 Super Jumping! Jumping! Jumping!(动态规划DP)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1087 Super Jumping! Jumping! Jumping! Time Limit: 200 ...

  3. 【HDU - 1087 】Super Jumping! Jumping! Jumping! (简单dp)

    Super Jumping! Jumping! Jumping! 搬中文ing Descriptions: wsw成功的在zzq的帮助下获得了与小姐姐约会的机会,同时也不用担心wls会发现了,可是如何 ...

  4. 杭电1087 Super Jumping! Jumping! Jumping!(初见DP)

    Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 ...

  5. HDU - 1087 Super Jumping!Jumping!Jumping!(dp求最长上升子序列的和)

    传送门:HDU_1087 题意:现在要玩一个跳棋类游戏,有棋盘和棋子.从棋子st开始,跳到棋子en结束.跳动棋子的规则是下一个落脚的棋子的号码必须要大于当前棋子的号码.st的号是所有棋子中最小的,en ...

  6. HDU-1087-Super Jumping! Jumping! Jumping!(线性DP, 最大上升子列和)

    Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 ...

  7. Codeforces Round #598 (Div. 3) C. Platforms Jumping 贪心或dp

    C. Platforms Jumping There is a river of width n. The left bank of the river is cell 0 and the right ...

  8. CodeForces - 512B Fox And Jumping[map优化dp]

    B. Fox And Jumping time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  9. DP专题训练之HDU 1087 Super Jumping!

    Description Nowadays, a kind of chess game called "Super Jumping! Jumping! Jumping!" is ve ...

  10. HDU 1087 Super Jumping! Jumping! Jumping! (DP)

    C - Super Jumping! Jumping! Jumping! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format: ...

随机推荐

  1. JSF标签大全详解

    1. JSF入门 藉由以下的几个主题,可以大致了解JSF的轮廓与特性,我们来看看网页设计人员与应用程序设计人员各负责什么. 1.1简介JSF Web应用程序的开发与传统的单机程序开发在本质上存在着太多 ...

  2. 为MyEclipse加入自己定义凝视

    非常多时候我们默认的MyEclipse的类凝视是这种,例如以下图 能够通过改动MyEclipse的凝视规则来改变,不但能够改动类的.还能够改动字段.方法等凝视规则,操作方法例如以下 1.针对方法的凝视 ...

  3. angularjs基本执行流程

    近期温习了下angularjs执行流程,备记下.以便查看. 主要的执行流程例如以下: 1.用户请求应用起始页. 2.用户的浏览器向server发起一次HTTP连接,然后载入index.html页面,这 ...

  4. LG 2.2.1 P350安卓系统刷机,问题总结,希望对需要的朋友有助

    手机误删软件导致短信,键盘等无声音提醒 我的手机前几天被我误删了一个软件,导致电话接不了,别人打电话的时候,老提示我在通话中,但是我可以在通话中看到对方的打电话记录.短信,键盘,USB连接,等等都没有 ...

  5. [RxJS] Refactoring Composable Streams in RxJS, switchMap()

    Refactoring streams in RxJS is mostly moving pieces of smaller streams around. This lessons demonstr ...

  6. Android程序Crash时的异常上报

    转载请注明来源:http://blog.csdn.net/singwhatiwanna/article/details/17289479 前言 大家都知道,android应用不可避免的会发生crash ...

  7. post请求和get请求的区别

    1:如果表单是以post方式发送,那么表单中的数据会放在请求报文体中,发送到服务端.但是如果是以get方式提交表单,那么表单中用户输入的数据都是以URL地址的方式发送到服务端. 2:在服务端接收数据时 ...

  8. word-wrap 和 word-break

    一.word-wrap 1.浏览器支持 所有主流浏览器都支持 word-wrap属性 2.定义和用法 word-wrap 属性允许长单词或 URL 地址换行到下一行. 语法 word-wrap: no ...

  9. repeater截取字数

    <%# Eval("ArticleName").ToString().Length > 14 ? Eval("ArticleName").ToStr ...

  10. Java数据输入

    以下是数据输入实例: //以下是数据输入实例 import java.util.Scanner;//导入java.util.Scanner,Scanner首字母大写 public class Test ...