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

Longest Common Substring II

Time Limit: 2000ms

Memory Limit: 262144KB

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 a bit harder, for some 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 at most 10 lines, each line consists of no more than 100000 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
aaaajfaaaa

Output:
2

　　这题是找一些字符串的最长公共子串。
　　这里对其中一个建SAM，然后跑其他字符串，在每一个字符串中，记录SAM节点中每个节点对应的最长长度，最后在所有字符串中SAM的最长长度取MIN,答案最后再在对每个位置的最小值取MAX。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;

 struct SAM{
     int ch[][],len[],fa[],last,cnt;
     void Init()
     {
         memset(fa,,sizeof(fa));
         last=cnt=;
     }
     void Insert(int c)
     {
         int p=last,np=last=++cnt;
         len[np]=len[p]+;
         while(p&&!ch[p][c])
             ch[p][c]=np,p=fa[p];
         if(!p)
             fa[np]=;
         else{
             int q=ch[p][c];
             if(len[p]+==len[q])
                 fa[np]=q;
             else{
                 int nq=++cnt;
                 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][c]==q)
                     ch[p][c]=nq,p=fa[p];
             }
         }
     }
 }sam;
 char s[];
 int ans[],f[];
 int main()
 {
     memset(ans,,sizeof(ans));
     sam.Init();
     scanf("%s",&s);
     for(int i=;s[i];i++)
         sam.Insert(s[i]-'a');
     while(~scanf("%s",s))
     {
         memset(f,,sizeof(f));
         int node=,len=;
         for(int i=;s[i];i++)
         {
             int c=s[i]-'a';
             while(node&&!sam.ch[node][c])
                 node=sam.fa[node];
             if(!node)
                 node=,len=;
             else
                 len=sam.len[node]+,node=sam.ch[node][c];
             if(len>f[node])f[node]=len;
         }
         for(int i=;i<=sam.cnt;i++)
             ans[i]=min(ans[i],f[i]);
     }
     int ret=;
     for(int i=;i<=sam.cnt;i++)
         ret=max(ret,ans[i]);
     printf("%d\n",ret);
     return ;
 }