本文主要是介绍POJ 2533 Longest Ordered Subsequence（最长非递减子序列，LIS），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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Longest Ordered Subsequence
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 50206 | Accepted: 22293 |
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.
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
4
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;
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