题意 A string is finite sequence of characters over a non-empty finite set Σ. In this problem, Σ is the set of lowercase letters. Substring, also called factor, is a consecutive sequence of characters occurrences at least once in a string. Now your tas…

Portal,Portal to 洛谷 Description 给出\(n(n\leq10)\)个仅包含小写字母的字符串\(s_1..s_n(|s_i|\leq10^5)\),求这些字符串的最长公共子串长度. Solution 对第一个串建立SAM,然后依次跑剩下的串. 记录\(maxL[i]\)表示对于目前做过的所有串,状态\(i\)最长能匹配的长度.跑一个串\(x\)时,记录\(tmp[i]\)表示对于串\(x\),状态\(i\)最长能匹配的长度,跑完之后将\(maxL\)与\(tmp\)取…

题目链接:http://www.spoj.com/problems/LCS2/ 其实两个串的LCS会了,多个串的LCS也就差不多了. 我们先用一个串建立后缀自动机,然后其它的串在上面跑.跑的时候算出每一个位置能往左扩展的最大长度也就是LCS. 于是对于每一个状态维护mx数组,表示当前串与SAM在此状态的LCS值.对于一个状态取所有mx中的最小值,然后答案就是所有状态最小值中的最大值,证明显然. 两个串的时候不用拿一个状态更新其祖先状态,但是这里需要.SAM是一个DAG图,我们通过l数组来基数排序…

地址:http://www.spoj.com/problems/LCS2/ 题面: LCS2 - Longest Common Substring II no tags  A string is finite sequence of characters over a non-empty finite set Σ. In this problem, Σ is the set of lowercase letters. Substring, also called factor, is a con…

LCS2 - Longest Common Substring II 多个字符串找最长公共子串 以其中一个串建\(SAM\),然后用其他串一个个去匹配,每次的匹配方式和两个串找\(LCS\)一样,就是要记录\(SAM\)的每个状态和当前匹配串匹配的最大值\(maxx\),这个在匹配完一个串之后需要通过\(parent\)树上传最大匹配值,同时要更新一个最小值\(minn\),来表示每个节点和当前已经匹配过的所有串能匹配上的最大值,这个需要每次匹配一个串之后和当前节点的\(maxx\)取\(min…

LCS2 - Longest Common Substring II A string is finite sequence of characters over a non-empty finite set Σ. In this problem, Σ is the set of lowercase letters. Substring, also called factor, is a consecutive sequence of characters occurrences at leas…

Longest Common Substring II Time Limit: 2000ms Memory Limit: 262144KB A string is finite sequence of characters over a non-empty finite set Σ. In this problem, Σ is the set of lowercase letters. Substring, also called factor, is a consecutive sequenc…

[SPOJ]Longest Common Substring II (后缀自动机) 题面 Vjudge 题意:求若干个串的最长公共子串 题解 对于某一个串构建\(SAM\) 每个串依次进行匹配 同时记录\(f[i]\)表示走到了\(i\)节点 能够匹配上的最长公共子串的长度 当然,每个串的\(f[i]\)可以更新\(f[i.parent]\) 所以需要拓扑排序 对于每个串求出每个节点的最长匹配 然后对他们取\(min\),表示某个节点大家都能匹配的最长长度 最后对于所有点的值都取个\(max\)…

[SP1812]LCS2 - Longest Common Substring II 题面 洛谷 题解 你首先得会做这题. 然后就其实就很简单了, 你在每一个状态\(i\)打一个标记\(f[i]\)表示状态\(i\)能匹配到最长的子串长度, 显然\(f[i]\)可以上传给\(f[i.fa]\). 然后去每个串和第\(1\)个串\(f\)的最小值的最大值即可. 代码 #include <iostream> #include <cstdio> #include <cstdlib&…

LCS2 - Longest Common Substring II no tags  A string is finite sequence of characters over a non-empty finite set Σ. In this problem, Σ is the set of lowercase letters. Substring, also called factor, is a consecutive sequence of characters occurrence…