SPOJ 题目1811 LCS - Longest Common Substring（后缀自动机求最长公共子串）

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
sam确实快，sa超时了=_=,给a串建个sam，b串跑一遍就行了
ac代码

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
#define N 510005  
struct sam
{
	sam *pre,*son[26];
	int len,g;
}que[N],*root,*tail,*b[N];
int tot;
void add(int c,int l)
{
	sam *p=tail,*np=&que[tot++];
	np->len=l;
	tail=np;
	while(p&&p->son[c]==NULL)
	{
		p->son[c]=np;
		p=p->pre;
	}
	if(p==NULL)
		np->pre=root;
	else
	{
		sam *q=p->son[c];
		if(p->len+1==q->len)
			np->pre=q;
		else
		{
			sam *nq=&que[tot++];
			*nq=*q;
			nq->len=p->len+1;
			np->pre=q->pre=nq;
			while(p&&p->son[c]==q)
			{
				p->son[c]=nq;
				p=p->pre;
			} 
		}
	}
}
char str1[N>>1],str2[N>>1];
int dp[N>>1];
int main()
{
	while(scanf("%s%s",str1,str2)!=EOF)
	{
		tot=0;
		root=tail=&que[tot++];
		int l=0;
		int len=strlen(str1),i;
		for(i=0;i<len;i++)
		{
			add(str1[i]-'a',i+1);
		}
		len=strlen(str2);
		sam *p=root;
		int ans=0;
		for(i=0;i<len;i++)
		{
			int now=str2[i]-'a';
			if(p->son[now])
			{
				p=p->son[now];
				l++;
			}
			else
			{
				while(p&&p->son[now]==NULL)
					p=p->pre;
				if(p==NULL)
				{
					p=root;
					l=0;
				}
				else
				{
					l=p->len+1;
					p=p->son[now];
				}
			}
			if(l>ans)
				ans=l;
		}
		printf("%d\n",ans);
	}
}