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

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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:alsdfkjfjkdsalfdjskalajfkdslaOutput: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);}}

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