POJ 2533 Longest Ordered Subsequence

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本文介绍了一种使用动态规划解决最长递增子序列问题的方法，并提供了完整的代码实现。通过对给定数值序列进行分析，该算法能够找到序列中最长递增子序列的长度。

题目

总时间限制: 2000ms 内存限制: 65536kB
描述
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.
输入
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
输出
Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.
样例输入
7
1 7 3 5 9 4 8
样例输出
4
来源
Northeastern Europe 2002, Far-Eastern Subregion

思路

动态规划。 $maxlen(k)$ 表示以 $a_k$ 为终点的最长上升子序列长度.

m a x l e n (1) = 1 m a x l e n (k) = M a x {m a x l e n (i) : 1 < i < k a n d a i < a k a n d k \neq 1} + 1

$maxlen(1)=1\\maxlen(k)=Max\{maxlen(i):1<i<k \ and\ a_i<a_k \ and\ k \neq 1\}+1$

代码

while True:
    try:
        N = int(input().strip())
        nums = input().strip().split()
        nums = [int(i) for i in nums]
        dp = []
        dp.append(1)
        for i in range(1, N):
            maxnum = 0
            for j in range(0, i):
                if nums[i] > nums[j] and maxnum < dp[j]:
                    maxnum = dp[j]
            dp.append(maxnum + 1)
        print(max(dp))
    except:
        break