Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 49436    Accepted Submission(s): 22871

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

题意

求上升子序列的和的最大值,注意对dp数组的初始化

AC代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
const double E=exp(1);
const int maxn=1e6+10;
using namespace std;
int a[maxn];
int b[maxn];
int dp[maxn];
int main(int argc, char const *argv[])
{
int n;
while(~scanf("%d",&n)&&n)
{
ms(a);
ms(dp);
for(int i=0;i<n;i++)
{
cin>>a[i];
// 对dp进行初始化
dp[i]=a[i];
}
int ans=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<i;j++)
{
// 如果是递增的
if(a[i]>a[j])
// 加起来和原本的作比较
dp[i]=max(dp[j]+a[i],dp[i]);
}
// 如果一开始不对dp初始化,要加上下面这一句
// dp[i]=max(dp[i],a[i]);
ans=max(ans,dp[i]);
}
cout<<ans<<endl;
}
return 0;
}

HDU 1087:Super Jumping! Jumping! Jumping!(LIS)的更多相关文章

  1. HDU 1024:Max Sum Plus Plus(DP)

    http://acm.hdu.edu.cn/showproblem.php?pid=1024 Max Sum Plus Plus Problem Description Now I think you ...

  2. 动态规划:最长上升子序列(LIS)

    转载请注明原文地址:http://www.cnblogs.com/GodA/p/5180560.html 学习动态规划问题(DP问题)中,其中有一个知识点叫最长上升子序列(longest  incre ...

  3. Android(java)学习笔记232:Android进程间通讯(IPC)之AIDL

    一.IPC inter process communication  进程间通讯 二.AIDL android  interface  defination  language  安卓接口定义语言 满 ...

  4. 黑马程序员:Java基础总结----泛型(高级)

    黑马程序员:Java基础总结 泛型(高级)   ASP.Net+Android+IO开发 . .Net培训 .期待与您交流! 泛型(高级) 泛型是提供给javac编译器使用的,可以限定集合中的输入类型 ...

  5. Android(java)学习笔记175:Android进程间通讯(IPC)之AIDL

    一.IPC inter process communication  进程间通讯 二.AIDL android  interface  defination  language  安卓接口定义语言 满 ...

  6. Android BottomSheet:以选取图片为例(2)

     Android BottomSheet:以选取图片为例(2) 附录文章5简单介绍了常见的分享面板在BottomSheet中的具体应用.本文再以常见的选取图片为例写一个例子. 布局文件: < ...

  7. Android零基础入门第80节:Intent 属性详解(下)

    上一期学习了Intent的前三个属性,本期接着学习其余四个属性,以及Android系统常用内置组件的启动. 四.Data和Type属性 Data属性通常用于向Action属性提供操作的数据.Data属 ...

  8. Spring Boot 2.X(九):Spring MVC - 拦截器(Interceptor)

    拦截器 1.简介 Spring MVC 中的拦截器(Interceptor)类似于 Servlet 开发中的过滤器 Filter,它主要用于拦截用户请求并作相应的处理,它也是 AOP 编程思想的体现, ...

  9. [易学易懂系列|rustlang语言|零基础|快速入门|(25)|实战2:命令行工具minigrep(2)]

    [易学易懂系列|rustlang语言|零基础|快速入门|(25)|实战2:命令行工具minigrep(2)] 项目实战 实战2:命令行工具minigrep 我们继续开发我们的minigrep. 我们现 ...

随机推荐

  1. Chrome使用的plugin

    Chrome使用的plugin   翻译 google翻译 youlict划词翻译     书签   查询书签 neater bookmarks 博客园收藏网页 书签保存 bookmark sysnc ...

  2. ID基本操作(新建文档,页面编码)5.8

    “文件”“新建”“文档”选择页数,页面大小.页面方向,“边距和分栏”设置上下左右的边距,栏数,如三栏 还可以改变分栏距离·改变排版方向,如图,垂直 单击“页面”可以查看我们的页面情况 超过两页会可以看 ...

  3. 模块化&os&sys

    syspath python 使用import模块调用的优先级是根据sys.path路径来的,此变量中位置在列表中的先后顺序来调用,如果先找到对应的模块,则先调用此模块. import sys pri ...

  4. QuickHit 项目

    package cn.javaoppday01; import java.util.Random; public class Game { public Player player; public G ...

  5. SpringMVC防止表单重复提交

    最近公司上线,有同志进行攻击,表当防重复提交也没有弄,交给我 ,本人以前也没弄过,知道大概的思路,但是那样实在是太麻烦了,虽然后面试过使用过滤器加拦截器实现,不过还是有点小麻烦. 后来在网上搜索后发现 ...

  6. 用Python的导入csv、文本文件、Excel文件的数据

    使用read_csv函数导入CSV文件 read.csv函数语法 read_csv(file,encoding) 例子: Age,Name 22,wangwei 23,lixin 24,liqing ...

  7. form 表单模板

    <div class="modal-dialog modal-lg"> //大布局modal-lg <div class="modal-content& ...

  8. perl代码调试

    perl调试教程 一.DESCRIPTIONA (very) lightweight introduction in the use of the perl debugger, and a point ...

  9. k8s weave network IP回收冲突

    问题:将备用集群的一个机器加入到新的集群中的时候,出现该机器上的pod都不能被访问. 查明原因是weave 没有删除干净 https://github.com/weaveworks/weave/iss ...

  10. day 36 关于io模型的问题 阻塞 和多路复用

    # from gevent import spawn,monkey;monkey.patch_all()# from socket import *# def server(ip,port):# se ...