【Poj 2533】 Longest Ordered Subsequence 最长上升子序列

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本文介绍了一种解决最长递增子序列问题的有效算法。通过动态规划的方法，该算法能够在给定的时间和空间限制内找到序列中最长递增子序列的长度。文章提供了完整的代码实现，并附有详细的解释。

Longest Ordered Subsequence
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 43410 Accepted: 19139

Description
A numeric sequence of ai is ordered if a1 < a2 < … < aN. Let the subsequence of the given numeric sequence (a1, a2, …, aN) be any sequence (ai1, ai2, …, 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

题目链接：http://poj.org/problem?id=2533

题意：

Longest Ordered Subsequence

思路：Longest Ordered Subsequence

代码：

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int n;
int dp[1005];
int a[1005];


int maxx;
int main()
{
    scanf("%d",&n);

    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        dp[i]=1;
    }

    maxx=0;

    for(int i=1;i<=n;i++)
    for(int j=1;j<i;j++)
    if(a[i]>a[j])
    {
        dp[i]=max(dp[i],dp[j]+1);
    }
    for(int i=1;i<=n;i++)
    maxx=max(maxx,dp[i]);
    printf("%d\n",maxx);



}