[POJ]2533 Longest Ordered Subsequence-优快云博客

本文链接：https://blog.youkuaiyun.com/Miner_Sty/article/details/50145503

本文通过动态规划解决寻找给定序列中最长非降序子序列的问题，详细阐述了解题思路并提供优化后的源代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

[POJ]2533 Longest Ordered Subsequence

问题

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

4

分析

题目要求找最长非降序子序列，可以采用动态规划的思想做。

f (i) = ⎧ ⎩ ⎨ 1, f (j) + 1, 1, i = 1 seq[i] > seq[j](1 < j < i) 不 存 在 seq[i] < seq[j]

$f(i) = \begin{cases} 1, &\text{i = 1}\\ f(j)+1, &\text{seq[i] > seq[j](1 < j < i)}\\ 1, &\text{不存在seq[i] < seq[j]} \end{cases}$

源代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int N, ans = 1;
    cin >> N;
    vector<int> seq(N, 0), dp(N, 0);
    for (int i = 0; i < N; ++i)
        cin >> seq[i];
    dp[0] = 1;
    for (int i = 1; i < N; ++i) {
        int tmp = 0;
        for (int j = 0; j < i; ++j)
            if (seq[i] > seq[j])
                tmp = max(tmp, dp[j]);
        dp[i] = tmp + 1;
        ans = max(dp[i], ans);
    }
    cout << ans << endl;
    return 0;
}