For a string S, a palindromic substring S[i..j] is said to be a shortest unique palindromic substring (SUPS) for an interval [s,t] in S, if S[i..j] occurs exactly once in S, the interval [i,j] contains [s,t], and every palindromic substring containing [s,t] which is shorter than S[i..j] occurs at least twice in S. In this paper, we study the problem of answering SUPS queries on run-length encoded strings. We show how to preprocess a given run-length encoded string RLE_S of size m in O(m) space and \(O(m \log \sigma _{\mathit {RLE}_{S}} + m \sqrt {\log m / \log \log m})\) time so that all SUPSs for any subsequent query interval can be answered in \(O(\sqrt {\log m / \log \log m} + \alpha )\) time, where α is the number of outputs, and \(\sigma _{\mathit {RLE}_{S}}\) is the number of distinct runs of RLE_S. Additionaly, we consider a variant of the SUPS problem where a query interval is also given in a run-length encoded form. For this variant of the problem, we present two alternative algorithms with faster queries. The first one answers queries in \(O(\sqrt {\log \log m /\log \log \log m} + \alpha )\) time and can be built in \(O(m \log \sigma _{\mathit {RLE}_{S}} + m \sqrt {\log m / \log \log m})\) time, and the second one answers queries in \(O(\log \log m + \alpha )\) time and can be built in \(O(m \log \sigma _{\mathit {RLE}_{S}})\) time. Both of these data structures require O(m) space.

的字符串小号，回文串小号[我.. Ĵ ]被说成是一个最短独特回文子（小号û P小号一个时间间隔）[小号，吨]在小号，如果小号[我.. Ĵ ]发生在正好一次小号，区间[我，Ĵ ]包含[小号，吨]，和含有每回文子串[小号，吨]，这是短于小号[i .. j ]在S中至少出现两次。在本文中，我们研究了在游程长度编码的字符串上回答S U P S查询的问题。我们展示了如何在O（m）空间中预处理给定的长度为m的游程编码字符串R L E _S和\（O（m \ log \ sigma _ {\ mathit {RLE} _ {S}} + m \ SQRT {\日志米/ \日志\日志米}）\）的时间，使所有小号ü P小号小号对于任何后续的查询间隔可以在回答\（O（\ SQRT {\日志米/ \日志\日志米} + \ alpha）\）时间，其中α是输出的数量，\（\ sigma _ {\ mathit {RLE} _ {S}} \）是R L E _S的不同行程的数量。另外，我们考虑SUPS问题的一种变体，其中查询间隔也以游程长度编码形式给出。对于这个问题的变体，我们提出了两种具有更快查询的替代算法。第一个以\（O（\ sqrt {\ log \ log m / \ log \ log \ log m} + \ alpha）\）的时间回答查询，并且可以以\（O（m \ log \ sigma _ { \ mathit {RLE} _ {S}} + m \ sqrt {\ log m / \ log \ log m}）\）时间，第二个以\（O（\ log \ log m + \ alpha）回答查询\）时间，并且可以内置于\（O（m \ log \ sigma _ {\ mathit {RLE} _ {S}}）\）时间。这两个数据结构都需要O（m）空间。