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Resequencing a Set of Strings Based on a Target String

机译:根据目标字符串重新排序一组字符串

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摘要

Given a set S={S (1),S (2),aEuro broken vertical bar,S (l) } of l strings, a text T, and a natural number k, find a string M, which is a concatenation of k strings (not necessarily distinct, i.e., a string in S may occur more than once in M) from S, whose longest common subsequence with T is largest, where a string in S may occur more than once in M. Such a string is called a k-inlay. The resequencing longest common subsequence problem (resequencing LCS problem for short) is to find a k-inlay for each query with parameter k after T and S are given. In this paper, we propose an algorithm for solving this problem which takes O(nml) preprocessing time and O(I (k) k) query time for each query with parameter k, where n is the length of T, m is the maximal length of strings in S, and I (k) is the length of the longest common subsequence between a k-inlay and T.
机译:给定一个由l个字符串,一个文本T和一个自然数k组成的集合S = {S(1),S(2),一个欧洲的垂直竖线,S(l)},找到一个字符串M,它是与S的k个字符串(不一定是不同的,即S中的字符串在M中可能出现不止一次),后者与T的最长共同子序列最大,其中S中的字符串在M中可能出现不止一次。称为k镶嵌。重排序最长的公共子序列问题(简称重排序LCS问题)是在给定T和S之后为每个带有参数k的查询找到k嵌入。在本文中,我们提出了一种用于解决此问题的算法,该算法针对参数为k的每个查询都占用O(nml)预处理时间和O(I(k)k)查询时间,其中n是T的长度,m是最大值S中的字符串长度,而I(k)是k嵌体和T之间的最长公共子序列的长度。

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