HDOJ/HDU 1087 Super Jumping! Jumping! Jumping!(经典DP~)
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
就是找最大的递增子序列!!!
用动态规划做~
从前往后依次计算出当前递增子序列的值~dp[i]
最后找出最大的dp[i]就是答案~~
import java.util.Scanner;
/**
* @author 陈浩翔
*
* 2016-5-26
*/
public class Main{
public static void main(String[] args) {
Scanner sc =new Scanner(System.in);
while(sc.hasNext()){
int n=sc.nextInt();
if(n==0){
break;
}
int a[]=new int[n];
int dp[]=new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
dp[0]=a[0];
for(int i=1;i<n;i++){
int max=0;
for(int j=0;j<i;j++){
if(a[j]<a[i]&&dp[j]>max){
max=dp[j];
}
}
dp[i]=a[i]+max;
}
int max=dp[0];
for(int i=1;i<n;i++){
if(dp[i]>max){
max=dp[i];
}
}
System.out.println(max);
}
}
}
HDOJ/HDU 1087 Super Jumping! Jumping! Jumping!(经典DP~)的更多相关文章
- HDU 1087 Super Jumping! Jumping! Jumping
HDU 1087 题目大意:给定一个序列,只能走比当前位置大的位置,不可回头,求能得到的和的最大值.(其实就是求最大上升(可不连续)子序列和) 解题思路:可以定义状态dp[i]表示以a[i]为结尾的上 ...
- HDU 1087 Super Jumping! Jumping! Jumping!(求LSI序列元素的和,改一下LIS转移方程)
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1087 Super Jumping! Jumping! Jumping! Time Limit: 20 ...
- 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 ...
- HDU 1087 Super Jumping! Jumping! Jumping! 最长递增子序列(求可能的递增序列的和的最大值) *
Super Jumping! Jumping! Jumping! Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64 ...
- hdu 1087 Super Jumping! Jumping! Jumping!(dp 最长上升子序列和)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1087 ------------------------------------------------ ...
- DP专题训练之HDU 1087 Super Jumping!
Description Nowadays, a kind of chess game called "Super Jumping! Jumping! Jumping!" is ve ...
- 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 ...
- HDU 1087 Super Jumping! Jumping! Jumping! 最大递增子序列
Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 ...
- 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
- Codevs 1183 泥泞的道路
1183 泥泞的道路 时间限制: 1 s 空间限制: 128000 KB 题目等级 : 钻石 Diamond 传送门 题目描述 Description CS有n个小区,并且任意小区之间都有两条单向道路 ...
- 选择第n小的元素之python实现源码
def partition(A, p, r): j = p+1 for i in range(p+1, r+1): if(A[i] < A[p]): tmp = A[i] A[i] = A[j] ...
- 九度OJ 1087 约数的个数
题目地址:http://ac.jobdu.com/problem.php?pid=1087 题目描述: 输入n个整数,依次输出每个数的约数的个数 输入: 输入的第一行为N,即数组的个数(N<=1 ...
- $.ajax参数备注-转转转
jquery中的ajax方法参数总是记不住,这里记录一下. $,ajax()方法参数详解 1.url: 要求为String类型的参数,(默认为当前页地址)发送请求的地址. 2.type: 要求为St ...
- Swift(三.函数)
一.swift中的函数分为以下几类吧 1>无参无返 2>无参有返 3>有参无返 4>有参有返 5>有参多返 二.看下面几个例子吧 1>无参无返 func a ...
- Core Animation
position和anchorPoint的区别 -整理自苹果官方文档- Layers使用两种坐标系: 1. point-based :1)当需要定义layer在屏幕中或是距另一个layer的位置时 ...
- 对ARM9哈佛结构的认识
书本上都说ARM是哈佛结构,但是我总感觉好像看不出来.后来针对S3C2440的ARM9核进行分析,我有了自己的见解. 我的结论是“ARM9被称为是哈佛结构是从它拥有指令cache和数据cache”来说 ...
- 系统调用与API的区别
整理自系统调用与API的区别 1.为什么用户程序不能直接访问系统内核模式提供的服务? 答:在linux中,将程序的运行空间分为内核与用户空间(内核态和用户态),在逻辑上它们之间是相互隔离的,因此用户程 ...
- SpringSecurity简单应用(二)
这里我首先对我上一篇博文的第三个实例做一下讲解,下面是applicationContext-security.xml内容如下: <?xml version="1.0" enc ...
- USB Type-C“三剑客”: 连接器、控制器和电缆
USB Type-C™是最新的有关电缆布线的USB连接器标准.您会看到,从笔记本电脑.智能手机.闪存到视频系统,这些设备上有一个小型可逆的Type-C连接器.由于Type-C电缆既可以给主机和设备提供 ...