POJ 2533 Longest Ordered Subsequence（最长非递减子序列，LIS）

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本文介绍了一道经典的编程竞赛题目——寻找给定序列中最长的非降子序列。文章提供了一个完整的C++实现代码示例，并通过注释解释了核心思路。

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Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 50206		Accepted: 22293

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

Source

Northeastern Europe 2002, Far-Eastern Subregion

题意：

给定的 n 个数，求出从左往右非递减的子序列的最大值

思路：

模板题，就过来水水题的！

AC CODE：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
const int MM = 200100+4;
using namespace std;

int h[MM], d[MM], Ans[MM];

int main()
{
	int n;
	cin >> n;
	for(int i = 1; i <= n; i++) cin >> h[i];

	int ans = 0, vv = 0;
	for(int i = 1; i <= n; i++)
		{
			d[i] = 1;
			for(int j = 1; j <= i-1; j++)
				{
					if(h[j] < h[i] && d[i] < d[j]+1)
						{
							d[i] = d[j] + 1;
						}


				}
			if(d[i] > ans) ans = d[i];
		}

	printf("%d\n", ans);
	return 0;
}