题目链接:http://begin.lydsy.com/JudgeOnline/problem.php?id=2796 把一个字符串做出后缀自动机,另一个字符串与之匹配. #include<cstdio> #include<cstring> #include<iostream> #define inf 1<<30 #define maxn 250005 using namespace std; int tot,last,root,ans,sum,n,m; c…

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #define maxn 500005 #define maxl 250005 using namespace std; ],fa[maxn],dist[maxn]; char s1[maxl],s2[maxl]; struct Tsegment{ ;} in…

题目链接:http://www.spoj.com/problems/LCS/ 题意如题目,求两个串的最大公共子串LCS. 首先对其中一个字符串A建立SAM,然后用另一个字符串B在上面跑. 用一个变量Lcs来记录当前答案,如果能转移到下一个状态则Lcs++. 若不能转移说明需要重新选择状态,则不断地跳到当前状态的fa状态,如果发现能够转移就停下来,Lcs变为当前状态的len+1并转移到下一个可行状态. 这里很像AC自动机的fail指针的跳跃.因为fa状态是当前状态的后缀,已经被匹配过,而fa状态有…

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 task i…

A string is finite sequence of characters over a non-empty finite set \(\sum\). In this problem, \(\sum\) is the set of lowercase letters. Substring, also called factor, is a consecutive sequence of characters occurrences at least once in a string. N…

地址:http://www.spoj.com/problems/LCS/ 题面: LCS - Longest Common Substring 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 consecut…

Write a function to find the longest common prefix string amongst an array of strings. 这道题让我们求一系列字符串的共同前缀,没有什么特别的技巧,无脑查找即可,我们定义两个变量i和j,其中i是遍历搜索字符串中的字符,j是遍历字符串集中的每个字符串.这里将单词上下排好,则相当于一个各行长度有可能不相等的二维数组,我们遍历顺序和一般的横向逐行遍历不同,而是采用纵向逐列遍历,在遍历的过程中,如果某一行没有了,说明其为…

1. 问题描述 子串应该比较好理解,至于什么是子序列,这里给出一个例子:有两个母串 cnblogs belong 比如序列bo, bg, lg在母串cnblogs与belong中都出现过并且出现顺序与母串保持一致,我们将其称为公共子序列.最长公共子序列(Longest Common Subsequence, LCS),顾名思义,是指在所有的子序列中最长的那一个.子串是要求更严格的一种子序列,要求在母串中连续地出现.在上述例子的中,最长公共子序列为blog(cnblogs, belong),最长公…

题目简述: Write a function to find the longest common prefix string amongst an array of strings. 解题思路: class Solution: # @return a string def longestCommonPrefix(self, strs): if strs == []: return '' minl = 99999 for i in strs: if len(i) < minl: minl = l…

  public class Solution { /** * @param strs: A list of strings * @return: The longest common prefix */ public String longestCommonPrefix(String[] strs) { // write your code here if(strs.length == 0){ return ""; } String prefix = strs[0]; int i =…