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

Notice:

new testcases added

思路

代码

#include <cstdio>
#include <cstring>const int maxn=250000;struct node
{node* tr[26];node* par;int maxl;node(int l=0){memset(tr,0,sizeof tr);par=NULL;maxl=l;}
};struct suffix_automaton
{node* qs;node* qlast;node* qn[maxn+10];int l[maxn+10];inline int clear(){delete qs;delete qlast;qlast=qs=new node();return 0;}inline int addchr(int ch){node* p=qlast;node* np=new node(qlast->maxl+1);qlast=np;while((p!=NULL)&&(p->tr[ch]==NULL)){p->tr[ch]=np;p=p->par;}if(p==NULL){np->par=qs;return 0;}node* q=p->tr[ch];if(q->maxl!=p->maxl+1){node* nq=new node(p->maxl+1);memcpy(nq->tr,q->tr,sizeof q->tr);nq->par=q->par;q->par=np->par=nq;while((p!=NULL)&&(p->tr[ch]==q)){p->tr[ch]=nq;p=p->par;}}else{np->par=q;}return 0;}inline int run(char* s,int len){qn[0]=qs;l[0]=0;for(register int i=1; i<=len; ++i){node* nq=qn[i-1];if(nq->tr[s[i]-'a']!=NULL){l[i]=l[i-1]+1;qn[i]=nq->tr[s[i]-'a'];continue;}while((nq!=NULL)&&(nq->tr[s[i]-'a']==NULL)){nq=nq->par;}if(nq==NULL){qn[i]=qs;l[i]=0;continue;}qn[i]=nq->tr[s[i]-'a'];l[i]=nq->maxl+1;}int ans=0;for(register int i=1; i<=len; ++i){if(l[i]>ans){ans=l[i];}}return ans;}
};suffix_automaton sam;
int la,lb,l[maxn+10];
char a[maxn+10],b[maxn+10];int main()
{scanf("%s%s",a+1,b+1);la=strlen(a+1);lb=strlen(b+1);sam.clear();for(register int i=1; i<=lb; ++i){sam.addchr(b[i]-'a');}printf("%d\n",sam.run(a,la));return 0;
}