poj 2533 Longest Ordered Subsequence-优快云博客

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O(n^2)的时间复杂度内找到给定序列中最长递增子序列的长度。文章还提供了一个具体的示例，帮助理解如何实现这一算法。

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Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 33662		Accepted: 14734

Description

A numeric sequence of a_i is ordered if a₁ < a₂ < ... < a_N. Let the subsequence of the given numeric sequence (a₁, a₂, ..., a_N) be any sequence (a_i₁, a_i₂, ..., a_{i_K}), where 1 <= i₁ < i₂ < ... < i_K <= 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

解题思路：我们可以先确定一个终点，表示子序列最后一个元素，因此我们可以从左往右一个个确定终点，其目的在于当我们确定好前i个终点所构成子序列的最长长度时，第i+1个终点所构成的子序列的最长长度等于满足array[i+1]>array[k](0<=k<=i)中dp[k]的最大值加1；该算法的复杂度为O(n^2)，只适用于数组小于1000的情况；

#include <iostream>
using namespace std;
#define MAX 50005
int main()
{
	int n,a[MAX],dp[MAX];
	while (cin>>n)
	{
		for (int i=0;i<n;i++)
			cin>>a[i];
		dp[0]=1;
		int answer=dp[0];
		for (int i=1;i<n;i++)
		{
			int temp=1,max=0;
			for (int j=0;j<i;j++)
			{
				if (a[i]>a[j])
					temp=dp[j]+1;
				if (temp>max)
					max=temp;
			}
			dp[i]=max;
			if (dp[i]>answer)
				answer=dp[i];
		}
		cout<<answer<<endl;
	}
	return 0;
}