spoj1812 LCS2 - Longest Common Substring II

地址：http://www.spoj.com/problems/LCS2/

题面：

LCS2 - Longest Common Substring II

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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 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

Notice: new testcases added

思路：

　　对第一个穿构建sam，然后让后面的串在sam上走，同时记录每个串跑完后各个状态的所能匹配的最长长度，然后对所有串取个最小值，之后再取个最大值。

　　注意：如果状态p能够到达，那么状态fa[p]也必然能够到达，但可能在匹配时不会走到fa[p]状态，所以当p可达时，要更新fa[p]。

 #include <bits/stdc++.h>

 using namespace std;

 struct SAM
 {
     static const int MAXN = <<;//大小为字符串长度两倍
     static const int LetterSize = ;

     int tot, last, ch[MAXN][LetterSize], fa[MAXN], len[MAXN];
     int sum[MAXN], tp[MAXN]; //用于拓扑排序，tp为排序后的数组

     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];
             }
         }
     }

     void toposort( void)
     {
         for(int i = ; i <= len[last]; i++)   sum[i] = ;
         for(int i = ; i <= tot; i++)   sum[len[i]]++;
         for(int i = ; i <= len[last]; i++)   sum[i] += sum[i-];
         for(int i = ; i <= tot; i++)   tp[sum[len[i]]--] = i;
     }
 } sam;
 char ss[];
 int ans,mx[<<],mi[<<];

 int main(void)
 {
     //freopen("in.acm","r",stdin);
     sam.init();
     scanf("%s",ss);
     for(int i=,len=strlen(ss);i<len;i++)  sam.add(ss[i]-'a');
     for(int i=;i<=sam.tot;i++) mi[i]=1e6;
     sam.toposort();
     while(~scanf("%s",ss))
     {
         for(int i=,len=strlen(ss),p=,cnt=;i<len;i++)
         {
             int c=ss[i]-'a';
             if(sam.ch[p][c])
                 p=sam.ch[p][c],cnt++;
             else
             {
                 while(p&&!sam.ch[p][c]) p=sam.fa[p];
                 if(!p)
                     p=,cnt=;
                 else
                     cnt=sam.len[p]+,p=sam.ch[p][c];
             }
             mx[p]=max(mx[p],cnt);
         }
         for(int i=sam.tot;i;i--)
         {
             int p=sam.tp[i];
             mi[p]=min(mx[p],mi[p]);
             if(mx[p]&&sam.fa[p]) mx[sam.fa[p]]=sam.len[sam.fa[p]];
             mx[p]=;
         }
     }
     for(int i=;i<=sam.tot;i++) ans=max(ans,mi[i]);
     printf("%d\n",ans);
     return ;
 }