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POJ 2533 Longest Ordered Subsequence C++版

POJ题解 Slyar 180浏览 0评论

文章作者:姜南(Slyar) 文章来源:Slyar Home (www.slyar.com) 转载请注明,谢谢合作。

纠结DP,从最长上升子序列开始。

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

Slyar:属于简单的经典的DP,求最长上升子序列(LIS)。先研究了O(n^2)的思路。

令A[i]表示输入第i个元素,D[i]表示从A[1]到A[i]中以A[i]结尾的最长子序列长度。对于任意的0 <  j <= i-1,如果A(j) < A(i),则A(i)可以接在A(j)后面形成一个以A(i)结尾的新的最长上升子序列。对于所有的 0 <  j <= i-1,我们需要找出其中的最大值。

DP状态转移方程:

D[i] = max{1, D[j] + 1} (j = 1, 2, 3, ..., i-1 且 A[j] < A[i])

解释一下这个方程,i, j在范围内:

如果 A[j] < A[i] ,则D[i] = D[j] + 1

如果 A[j] >= A[i] ,则D[i] = 1

转载请注明:Slyar Home » POJ 2533 Longest Ordered Subsequence C++版

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网友最新评论 (2)

  1. 我错了,没想到博主已经给出解答了。
    shilong5年前 (2011-09-07)回复
  2. 博主,非O(n^2)的思路是什么啊?
    shilong5年前 (2011-09-07)回复