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;
}