地址：http://www.spoj.com/problems/LCS/
题面：
LCS - Longest Common Substring
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
思路：
对于串a建立后缀自动机， 然后让串b在sam上能走，记录到达每个状态时的长度，取个max就行。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 struct SAM 6 { 7 static const int MAXN = 3e5 * 2;//大小为字符串长度两倍 8 static const int LetterSize = 26; 9 int tot, last, ch[MAXN][LetterSize], fa[MAXN], len[MAXN]; 10 11 void init( void) 12 { 13 last = tot = 1; 14 len[1] = 0; 15 memset(ch,0,sizeof ch); 16 memset(fa,0,sizeof fa); 17 } 18 19 void add( int x) 20 { 21 int p = last, np = last = ++tot; 22 len[np] = len[p] + 1; 23 while( p && !ch[p][x]) ch[p][x] = np, p = fa[p]; 24 if( p == 0) 25 fa[np] = 1; 26 else 27 { 28 int q = ch[p][x]; 29 if( len[q] == len[p] + 1) 30 fa[np] = q; 31 else 32 { 33 int nq = ++tot; 34 memcpy( ch[nq], ch[q], sizeof ch[q]); 35 len[nq] = len[p] + 1; 36 fa[nq] = fa[q], fa[q] = fa[np] = nq; 37 while( p && ch[p][x] == q) 38 ch[p][x] = nq, p = fa[p]; 39 } 40 } 41 } 42 }; 43 char sa[300000],sb[300000]; 44 SAM sam; 45 int main(void) 46 { 47 scanf("%s%s",sa,sb); 48 int len=strlen(sa),ans=0; 49 sam.init(); 50 for(int i=0;i<len;i++) 51 sam.add(sa[i]-'a'); 52 len=strlen(sb); 53 for(int i=0,p=1,cnt=0;i<len;i++) 54 { 55 int k=sb[i]-'a'; 56 if(sam.ch[p][k]) 57 p=sam.ch[p][k],cnt++; 58 else 59 { 60 while(p&&!sam.ch[p][k]) p=sam.fa[p]; 61 if(p==0) 62 p=1,cnt=0; 63 else 64 cnt=sam.len[p]+1,p=sam.ch[p][k]; 65 } 66 ans=max(ans,cnt); 67 } 68 printf("%d ",ans); 69 return 0; 70 }