POJ 2533 Longest Ordered Subsequence

最新推荐文章于 2018-07-09 23:12:26 发布

原创最新推荐文章于 2018-07-09 23:12:26 发布 · 203 阅读

1 ·

CC 4.0 BY-SA版权

DP 专栏收录该内容

34 篇文章

订阅专栏

本文介绍了一种求解最长递增子序列问题的有效算法，并提供了详细的实现代码。该算法通过动态规划的方法，逐步构建解决方案，最终确定序列中最长递增子序列的长度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

今天手感不好，来几道裸题。

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 51872		Accepted: 23089

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

题目道出天机：最长递增子序列。

f[i]表示到i为止且选i的情况下，最长的子序列。附代码

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<cstdlib>
#include<ctime>

using namespace std;

const int MAXN=1010;
int f[MAXN],a[MAXN];

int main()
{
	int n;
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		f[i]=1;
	}
	for (int i=2;i<=n;i++)
	{
		for (int j=1;j<i;j++)
		{
			if (a[i]>a[j])
			{
				f[i]=max(f[i],f[j]+1);
			}
		}
	}
	int ans=0;
	for (int i=1;i<=n;i++)
	{
		ans=max(ans,f[i]);
	}
	cout<<ans;
    return 0;
}