Longest Ordered Subsequence
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 41984   Accepted: 18481

Description

A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1a2, ..., aN) be any sequence (ai1ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK<= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).

Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.

Input

The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000

Output

Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.

Sample Input

7
1 7 3 5 9 4 8

Sample Output

4

Source

Northeastern Europe 2002, Far-Eastern Subregion

[Submit]   [Go Back]   [Status]   [Discuss]

 #include<cstdio>
#include<iostream>
using namespace std; int main()
{
int n, a[], cn = , Maxlenth[], M = ;
cin >> n;
for(int i = ; i < n; i++) {
cin >> a[i];
Maxlenth[i] = ;
}
for(int i = n - ; i >= ; i--) {
for(int j = i + ; j < n; j++) {
if(a[i] < a[j] && Maxlenth[i] <= Maxlenth[j]) Maxlenth[i] = + Maxlenth[j];
if(Maxlenth[i] > M) M = Maxlenth[i];
}
}
cout << M << endl;
return ;
}

POJ2533 Longest ordered subsequence的更多相关文章

  1. POJ2533——Longest Ordered Subsequence(简单的DP)

    Longest Ordered Subsequence DescriptionA numeric sequence of ai is ordered if a1 < a2 < ... &l ...

  2. POJ2533 Longest Ordered Subsequence 【最长递增子序列】

    Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 32192   Acc ...

  3. (线性DP LIS)POJ2533 Longest Ordered Subsequence

    Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 66763   Acc ...

  4. poj-2533 longest ordered subsequence(动态规划)

    Time limit2000 ms Memory limit65536 kB A numeric sequence of ai is ordered if a1 < a2 < ... &l ...

  5. POJ2533 Longest Ordered Subsequence —— DP 最长上升子序列(LIS)

    题目链接:http://poj.org/problem?id=2533 Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 6 ...

  6. [POJ2533]Longest Ordered Subsequence<dp>

    题目链接:http://poj.org/problem?id=2533 描述: A numeric sequence of ai is ordered if a1 < a2 < ... & ...

  7. POJ-2533.Longest Ordered Subsequence (LIS模版题)

    本题大意:和LIS一样 本题思路:用dp[ i ]保存前 i 个数中的最长递增序列的长度,则可以得出状态转移方程dp[ i ] = max(dp[ j ] + 1)(j < i) 参考代码: # ...

  8. POJ2533 Longest Ordered Subsequence (线性DP)

    设dp[i]表示以i结尾的最长上升子序列的长度. dp[i]=max(dp[i],dp[j]+1). 1 #include <map> 2 #include <set> 3 # ...

  9. POJ2533:Longest Ordered Subsequence(LIS)

    Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence ...

随机推荐

  1. Rational rose下载,安装,破解

    rationalrose是一个镜像文件,后缀名是bin 之前尝试过用虚拟光驱来打开,不知道为什么,在win10的环境下,虚拟光驱硬是不能加载bin文件,后来拷到虚拟机上,打开了bin镜像文件,得到了一 ...

  2. HTML案例练习一

    发现其实JS也是挺容易的,也挺好玩的,写的一个控制图片移动的小案例,对DOM机制还是不怎么熟. <html> <head> <style type = "tex ...

  3. 关于图表的JS插件

    http://echarts.baidu.com/ http://echarts.baidu.com/tutorial.html#5%20%E5%88%86%E9%92%9F%E4%B8%8A%E6% ...

  4. HTML5 Canvas Text文本居中实例

    1.代码: <canvas width="700" height="300" id="canvasOne" class="c ...

  5. crud的意识

    CRUD说的就是增查改删C:Create 增加对应CREATE TBL ...: ADD TBL IN (...) VALUES (...)R:Retrieve查询SELECT * from TBLU ...

  6. Oracle 创建分页存储过程(转帖)

    原贴地址:http://19880614.blog.51cto.com/4202939/1316560 ps:源代码还有很多错误,我修改了 ------------------------------ ...

  7. 四、C#方法和参数

    方法是一种组合一系列语句以执行一个特定操作或计算一个特殊结果的方式. 它能够为构成程序的语句提供更好的结构和组织.   在面向对象的语言中,方法总是和类关联在一起的,我们用类将相关的方法分为一组. 方 ...

  8. Eclipse怎么忽略掉报错的js文件

    第一步,我们要先定位错误在哪里,选择菜单里window——show view——other,选择Problems. 第二步,点击有红叉的项目,在Problems视图中,可以看到是什么错,哪个文件夹中的 ...

  9. JavaScript 学习-变量的作用域和块级作用域

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...

  10. 常用命令(ubuntu)

    1.打开终端的方法 Ubuntu 中按左侧栏的第一个“面板主页(Dash 主页)”(可以按win键调出),在里面输入terminal可以打开终端,另外打开终端的快捷键是Ctrl+Alt+T 2.修改用 ...