多串LCS很适合SA但是我要学SAM 对第一个串求SAM,然后把剩下的串在SAM上跑,也就是维护p和len,到一个点,如果有ch[p][c],就p=ch[p][c],len++,否则向fa找最下的有c[p][c]的p,然后len=dis[p]+1,p=ch[p][c],否则就p=root,len=0(这个len每到一个节点就更新这个节点的f) 然后注意到在parent树上,因为每个节点代表的right集合是儿子的并集,所以f[u]是可以更新f[fa[u]]的,所以从底向上更新一遍(注意先更新!!…
167. Two Sum II - Input array is sorted[easy] Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such…
题意 求出多个串的最长公共子串. 分析 刚学SAM想做这个题的话最好先去做一下那道codevs3160.求两个串的LCS应该怎么求?把一个串s1建自动机,然后跑另一个串s2,然后找出s2每个前缀的最长公共后缀.那么多个的时候,我们也用这种类似的方法,但是我们求最长公共后缀的时候要求第一个串的.我们把其中一个串建SAM,然后把其他的串都在上面跑,维护两个值,Max[u]和Min[u].自动机中每个状态u的Right存的是结尾集合.那么对于一个字符串,我们可以求出他和自动机中每个状态的最长公共后缀.…
先求出SAM,然后考虑定义,点u是一个right集合,代表了长为dis[son]+1~dis[u]的串,然后根据有向边转移是添加一个字符,所以可以根据这个预处理出si[u],表示串u后加字符能有几个本质不同子串 然后回答的时候在树上跑一下即可 #include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=300005; int n,m,fa[N],ch[N][…
You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters. For example, giv…
This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence. InputEach sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.Outpu…
Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such that they add up to the target, where index1 m…
先求个SAM,然后再每个后缀的对应点上标记si[nw]=1,造好SAM之后用吧parent树建出来把si传上去,然后用si[u]更新f[max(u)],最后用j>i的[j]更新f[i] 因为每个点u对应长为min(u)~max(u)的串,我们就把它记在max(u)上,最后再统一向前更新,然后更新后的si就表示right大小,也就是这个串对应的后缀个数 #include<iostream> #include<cstdio> #include<cstring> usi…
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…
spoj 1812 LCS2 - Longest Common Substring II 题意: 给出最多n个字符串A[1], ..., A[n], 求这n个字符串的最长公共子串. 限制: 1 <= n <= 10 |A[i]| <= 1e5 思路: 和spoj 1811 LCS几乎相同的做法 把当中一个A建后缀自己主动机 考虑一个状态s, 假设A之外的其它串对它的匹配长度各自是a[1], a[2], ..., a[n - 1], 那么min(a[1], a[2], ..., a[n -…