Longest Ordered Subsequence ||POJ2533

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本文介绍了一种使用动态规划解决最长递增子序列问题的方法，并提供了一个完整的C++实现示例。通过分析给定序列，算法能够找出序列中最长的递增子序列的长度。

题目链接：http://poj.org/problem?id=2533
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

题解：注意样例：输出是4，那么符合要求的序列有1 3 5 9；1 3 5 8；1 3 4 8，我的算法是选出1 3 5 9，因为9 比后两个都大，不符合算法要求
动态规划，逐步理解

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
int a[1010],dp[1010],n;
int LIS()
{
    int i,j,ans,m;
    ans=1;
    dp[1]=1;
    for(i=2;i<=n;i++)
    {
        m=0;
        for(j=1;j<i;j++)
        {
            if(dp[j]>m&&a[j]<a[i])
            {
                m=dp[j];
            }
        }
        dp[i]=m+1;
        if(dp[i]>ans)
          ans=dp[i];
    }
return ans;
} 
int main()
{
    int i;
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        printf("%d\n",LIS());
    }
return 0;
}