POJ2533 Longest Ordered Subsequence【最长上升子序列+DP】

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最新推荐文章于 2024-10-01 18:52:56 发布

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分类专栏： HDU练习题文章标签：最长上升子序列动态规划

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本文介绍了解决最长上升子序列问题的一种动态规划方法。通过定义f[i]为以a[i]为末尾的最长上升子序列的长度，并通过两层循环找出所有可能的上升子序列，最终求得最长上升子序列的长度。

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问题链接：POJ2533 Longest Ordered Subsequence

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

Source

Northeastern Europe 2002, Far-Eastern Subregion

问题分析：这是一个最长上升子序列问题，使用DP算法实现。

定义f[i]=以a[i]为末尾的最长上升子序列的长度。

那么，以a[i]为末尾的最长上升子序列有以下两种情形：

1.只包含a[i]的子序列

2.满足j<i并且a[j]<a[i]的以a[j]为结尾的上升子序列末尾，追加上a[i]后得到的子序列，通过循环找到f[j]+1的最大值。

即：f[i]=max{1,f[j]+1}

#include <iostream>
using namespace std;
///////////ac
int a[1005],f[1005];
int res(int n)
{
	int s=0;
	for(int i=0;i<n;i++)
	{
		f[i]=1;//为f[i]设置初值，最初只包含a[i] 
		for(int j=0;j<i;j++)
		{
			if(a[j]<a[i])
			{
				f[i]=max(f[i],f[j]+1);
			}
		}
		s=max(s,f[i]);//记录f[i]的最大值 
	}
	return s;
}
int main()
{
	int n;
	while(cin>>n)
	{
		for(int i=0;i<n;i++)
		{
			cin>>a[i];
		}
		cout<<res(n)<<endl;
	}
	return 0;
}