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 is simple, for two given strings, find the length of the longest common substring of them.
Here common substring means a substring of two or more strings.
Input
The input contains exactly two lines, each line consists of no more than 250000 lowercase letters, representing a string.
Output
The length of the longest common substring. If such string doesn't exist, print "0" instead.
Example
Input: alsdfkjfjkdsal fdjskalajfkdsla Output: 3
题解：
哈哈很意外啊，在电脑城随便找台机子用洛谷在线IDE 1A这题啊
讲讲思路:
以A建立SAM 让B在SAM上匹配
可以类比于kmp思想，我们知道在Parent树上，fa是当前节点的子集，也就是说满足最大前缀，利用这个就可以做题了
步骤如下：
1.如果满足ch[p][c]存在，那么就直接走就是了，和AC自动机类似，len++
2.如果不存在，那么就沿fa上跳，找到最大dis[fa]且满足存在c这个儿子的.类似于跳fail链
3.跳到根就直接初始化，重新开始
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 using namespace std; 7 const int N=550005; 8 char S[N];int n=0,cnt=1,cur=1,last=1,ch[N][26],fa[N],dis[N]; 9 void build(int c){ 10 last=cur;cur=++cnt;dis[cur]=++n; 11 int p=last; 12 for(;p && !ch[p][c];p=fa[p])ch[p][c]=cur; 13 if(!p)fa[cur]=1; 14 else{ 15 int q=ch[p][c]; 16 if(dis[q]==dis[p]+1)fa[cur]=q; 17 else{ 18 int nt=++cnt;dis[nt]=dis[p]+1; 19 memcpy(ch[nt],ch[q],sizeof(ch[q])); 20 fa[nt]=fa[q];fa[q]=fa[cur]=nt; 21 for(;ch[p][c]==q;p=fa[p])ch[p][c]=nt; 22 } 23 } 24 } 25 int ans=0; 26 void Flr() 27 { 28 int p=1,nxt,lt; 29 scanf("%s",S); 30 int l=strlen(S),len=0,c; 31 for(int i=0;i<l;i++){ 32 c=S[i]-'a'; 33 if(ch[p][c]){ 34 len++;p=ch[p][c]; 35 } 36 else{ 37 lt=0;nxt=0; 38 while(p!=1){ 39 if(ch[p][c] && dis[p]>lt)lt=dis[p],nxt=p; 40 p=fa[p]; 41 } 42 if(nxt) 43 p=ch[nxt][c],len=dis[nxt]+1; 44 else 45 p=1,len=0; 46 } 47 if(len>ans)ans=len; 48 } 49 printf("%d ",ans); 50 } 51 void work() 52 { 53 scanf("%s",S); 54 for(int i=0,l=strlen(S);i<l;i++)build(S[i]-'a'); 55 Flr(); 56 } 57 int main() 58 { 59 work(); 60 return 0; 61 }