后缀自动机(SAM) ：SPOJ LCS - Longest Common Substring

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

　　这题就是求最长公共子串，于是我直接套用了SAM。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;
 const int MAXN=;
 struct SAM{
     int fa[MAXN],ch[MAXN][],len[MAXN],cnt,last;
     void Init()
     {
         last=cnt=;
     }
     void Insert(int c)
     {
         int p=last,np=last=++cnt;
         len[np]=len[p]+;
         while(!ch[p][c]&&p)
             ch[p][c]=np,p=fa[p];
         if(!p)fa[np]=;
         else{
             int q=ch[p][c];
             if(len[q]==len[p]+)
                 fa[np]=q;
             else{
                 int nq=++cnt;len[nq]=len[p]+;
                 memcpy(ch[nq],ch[q],sizeof(ch[p]));
                 fa[nq]=fa[q];fa[q]=fa[np]=nq;
                 while(ch[p][c]==q)
                     ch[p][c]=nq,p=fa[p];
             }
         }
     }
 }sam;
 char str1[MAXN],str2[MAXN];
 int main()
 {
     int len1=,len2=,ans=;
     scanf("%s%s",str1,str2);
     sam.Init();
     while(str1[len1])
         sam.Insert(str1[len1++]-'`');
     int node=,tmp=;
     for(;str2[len2];len2++){
         int c=str2[len2]-'`';
         if(sam.ch[node][c])
             node=sam.ch[node][c],tmp++;
         else{
             while(node&&!sam.ch[node][c])
                 node=sam.fa[node];
             if(node==)
                 node=,tmp=;
             else
                 tmp=sam.len[node]+,node=sam.ch[node][c];
         }
         ans=max(ans,tmp);
     }
     printf("%d\n",ans);
     return ;
 }