Longest common subsequence problem

最新推荐文章于 2024-07-08 15:40:47 发布

weixin_30585437

最新推荐文章于 2024-07-08 15:40:47 发布

阅读量141

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/cutepig/archive/2010/05/21/1741203.html

本文介绍了最长公共子序列问题的基本概念及解决方法。通过定义两个序列X和Y的前缀，利用LCS函数来找出两个序列中最长的共同子序列。当序列元素相等时，会在LCS(Xi-1,Yj-1)的基础上加入该元素；如果不相等，则保留LCS(Xi,Yj-1)和LCS(Xi-1,Yj)中较长的那个序列。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

http://en.wikipedia.org/wiki/Longest_common_subsequence_problem

LCS function defined

Let two sequences be defined as follows: X = (x₁, x₂...x_m) and Y = (y₁, y₂...y_n). The prefixes of X are X_{1, 2,...m}; the prefixes of Y are Y_{1, 2,...n}. Let LCS(X_i, Y_j) represent the set of longest common subsequence of prefixes X_i and Y_j. This set of sequences is given by the following.

$LCS\left(X_{i},Y_{j}\right) = \begin{cases} \empty & \mbox{ if }\ i = 0 \mbox{ or } j = 0 \\ \textrm{ ( } LCS\left(X_{i-1},Y_{j-1}\right), x_i) & \mbox{ if } x_i = y_j \\ \mbox{longest}\left(LCS\left(X_{i},Y_{j-1}\right),LCS\left(X_{i-1},Y_{j}\right)\right) & \mbox{ if } x_i \ne y_j \\ \end{cases}$

To find the longest subsequences common to X_i and Y_j, compare the elements x_i and y_i. If they are equal, then the sequence LCS(X_i-1, Y_j-1) is extended by that element, x_i. If they are not equal, then the longer of the two sequences, LCS(X_i, Y_j-1), and LCS(X_i-1, Y_j), is retained. (If they are both the same length, but not identical, then both are retained.) Notice that the subscripts are reduced by 1 in these formulas. That can result in a subscript of 0. Since the sequence elements are defined to start at 1, it was necessary to add the requirement that the LCS is empty when a subscript is zero.