SPOJ LCS(Longest Common Substring-后缀自动机-结点的Parent包含关系)

 

1811. Longest Common Substring Problem code: LCS

 

 

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

 

本题要求2个字符串的LCS.

那么先将A串读入，再在A串上匹配B串显然

但是一个结点能表示的后缀有|maxS|-|minS|+1个。

那么就这样：

假设当前在结点x，下一个B串要匹配的字母为c

如果x有ch[c]，走过去len+1

否则找pre，更新len=pre的maxS

理由如下：

设原有串=ababd,babd,abd ,匹配为其中某一个

不存在abdx无法匹配.

取pre=bd,d,(由定义可知不存在‘dc存在匹配,bdc不存在’的情况，故它俩能为1个状态)

那么len=pre的maxS(step)

否则返回根,len=0

 

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<functional>

#include<iostream>

#include<cmath>

#include<cctype>

#include<ctime>

using namespace std;

#define For(i,n) for(int i=1;i<=n;i++)

#define Fork(i,k,n) for(int i=k;i<=n;i++)

#define Rep(i,n) for(int i=0;i<n;i++)

#define ForD(i,n) for(int i=n;i;i--)

#define RepD(i,n) for(int i=n;i>=0;i--)

#define Forp(x) for(int p=pre[x];p;p=next[p])

#define Lson (x<<1)

#define Rson ((x<<1)+1)

#define MEM(a) memset(a,0,sizeof(a));

#define MEMI(a) memset(a,127,sizeof(a));

#define MEMi(a) memset(a,128,sizeof(a));

#define INF (2139062143)

#define F (100000007)

#define MAXN (500000+10)

long long mul(long long a,long long b){return (a*b)%F;}

long long add(long long a,long long b){return (a+b)%F;}

long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}

typedef long long ll;

char s1[MAXN],s2[MAXN];

struct node

{

	int pre,step,ch[26];

	char c;

	node(){pre=step=c=0;memset(ch,sizeof(ch),0);}

}a[MAXN];

int last=0,total=0;

void insert(char c)

{

	int np=++total;a[np].c=c+'a',a[np].step=a[last].step+1;

	int p=last;

	for(;!a[p].ch[c];p=a[p].pre) a[p].ch[c]=np;

	if (a[p].ch[c]==np) a[np].pre=p;

	else

	{

		int q=a[p].ch[c];

		if (a[q].step>a[p].step+1)

		{

			int nq=++total;a[nq]=a[q];a[nq].step=a[p].step+1;

			a[np].pre=a[q].pre=nq;

			for(;a[p].ch[c]==q;p=a[p].pre) a[p].ch[c]=nq;

		}else a[np].pre=q;

	}	

	last=np;

}

int main()

{

//	freopen("spojLCS.in","r",stdin);

	scanf("%s%s",s1+1,s2+1);int n=strlen(s1+1),m=strlen(s2+1);

	For(i,n) insert(s1[i]-'a');

	int now=0,ans=0,len=0;

	For(i,m)

	{

		char c=s2[i]-'a';

		while (now&&!a[now].ch[c]) now=a[now].pre,len=a[now].step;

		if (a[now].ch[c]) now=a[now].ch[c],len++;

		ans=max(ans,len);

	}

	printf("%d\n",ans);

	

	return 0;

}