SPOJ 1811 Longest Common Substring（后缀自动机模板题）

   

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

求两个字符串的最大公共子串的长度

代码如下：

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>

using namespace std;
const int N=250005;

struct State
{
    State *link,*go[26];
    int step;
    void clear()
    {
        link=0;
        step=0;
        memset(go,0,sizeof(go));
    }
}*root,*last;

State statePool[N*2],*cur;

void init()
{
    cur=statePool;
    root=last=cur++;
    root->clear();
}

void Insert(int w)
{
    State *p=last;
    State *np=cur++;
    np->clear();
    np->step=p->step+1;
    while(p&&!p->go[w])
        p->go[w]=np,p=p->link;
    if(p==0)
        np->link=root;
    else
    {
        State *q=p->go[w];
        if(p->step+1==q->step)
            np->link=q;
        else
        {
            State *nq=cur++;
            nq->clear();
            memcpy(nq->go,q->go,sizeof(q->go));
            nq->step=p->step+1;
            nq->link=q->link;
            q->link=nq;
            np->link=nq;
            while(p&&p->go[w]==q)
                p->go[w]=nq, p=p->link;
        }
    }
    last=np;
}

char A[N],B[N];

int main()
{
    int maxx,len;
    while(scanf("%s%s",A,B)!=EOF)
    {
        len=0;
        maxx=0;
        int n,m;
        n=strlen(A);
        m=strlen(B);
        init();
        State *p=root;
        for(int i=0;i<n;i++)
         Insert(A[i]-'a');
       for(int i=0;i<m;i++)
       {
           if(p->go[B[i]-'a'])
           {
               len++;
               p=p->go[B[i]-'a'];
           }
           else
           {
             while(p&&!p->go[B[i]-'a'])p=p->link;
             if(!p)
             {
              p=root;
              len=0;
             }
             else
             {
                 len=p->step+1;
                 p=p->go[B[i]-'a'];
             }
           }
           maxx=max(maxx,len);
        }
        cout<<maxx<<endl;
    }
    return 0;
}