spoj1811 LCS - Longest Common Substring

地址：http://www.spoj.com/problems/LCS/

题面：

LCS - Longest Common Substring

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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。

 #include <bits/stdc++.h>

 using namespace std;

 struct SAM
 {
     static const int MAXN = 3e5 * ;//大小为字符串长度两倍
     static const int LetterSize = ;
     int tot, last, ch[MAXN][LetterSize], fa[MAXN], len[MAXN];

     void init( void)
     {
         last = tot = ;
         len[] = ;
         memset(ch,,sizeof ch);
         memset(fa,,sizeof fa);
     }

     void add( int x)
     {
         int p = last, np = last = ++tot;
         len[np] = len[p] + ;
         while( p && !ch[p][x]) ch[p][x] = np, p = fa[p];
         if( p == )
             fa[np] = ;
         else
         {
             int q = ch[p][x];
             if( len[q] == len[p] + )
                 fa[np] = q;
             else
             {
                 int nq = ++tot;
                 memcpy( ch[nq], ch[q], sizeof ch[q]);
                 len[nq] = len[p] + ;
                 fa[nq] = fa[q], fa[q] = fa[np] = nq;
                 while( p && ch[p][x] == q)
                     ch[p][x] = nq, p = fa[p];
             }
         }
     }
 };
 char sa[],sb[];
 SAM sam;
 int main(void)
 {
     scanf("%s%s",sa,sb);
     int len=strlen(sa),ans=;
     sam.init();
     for(int i=;i<len;i++)
         sam.add(sa[i]-'a');
     len=strlen(sb);
     for(int i=,p=,cnt=;i<len;i++)
     {
         int k=sb[i]-'a';
         if(sam.ch[p][k])
             p=sam.ch[p][k],cnt++;
         else
         {
             while(p&&!sam.ch[p][k])  p=sam.fa[p];
             if(p==)
                 p=,cnt=;
             else
                 cnt=sam.len[p]+,p=sam.ch[p][k];
         }
         ans=max(ans,cnt);
     }
     printf("%d\n",ans);
     return ;
 }