In this paper we address the constrained longest common subsequence problem. Given two sequences $X$, $Y$ and a constrained sequence $P$, a sequence $Z$ is a constrained longest common subsequence for $X$ and $Y$ with respect to $P$ if $Z$ is the longest subsequence of $X$ and $Y$ such that $P$ is a subsequence of $Z$. Recently, Tsai \cite{Tsai} proposed an $O(n^2 \cdot m^2 \cdot r)$ time algorithm to solve this problem using dynamic programming technique, where $n$, $m$ and $r$ are the lengths of $X$, $Y$ and $P$, respectively. In this paper, we present a simple algorithm to solve the constrained longest common subsequence problem in $O(n \cdot m \cdot r)$ time and show that the constrained longest common subsequence problem is equivalent to a special case of the constrained multiple sequence alignment problem which can also be solved.

中文翻译：

---

约束序列问题的简单算法

在本文中，我们解决了受约束的最长公共子序列问题。给定两个序列$ X $，$ Y $和约束序列$ P $，如果$ Z $最长，则序列$ Z $是$ X $和$ Y $相对于$ P $的约束最长公共子序列$ X $和$ Y $的子序列，使得$ P $是$ Z $的子序列。最近，Tsai \ cite {Tsai}提出了一种$ O（n ^ 2 \ cdot m ^ 2 \ cdot r）$时间算法，使用动态编程技术来解决此问题，其中$ n $，$ m $和$ r $是$ X $，$ Y $和$ P $的长度。在本文中，我们提出了一种简单的算法来求解$ O（n \ cdot m \ cdot r）$时间内的约束最长公共子序列问题，并证明约束最长公共子序列问题等效于约束多重数的特例序列比对问题也可以解决。