【刷题】SPOJ 1811 LCS - Longest Common Substring

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

Solution

做字符串题根SPOJ打交道很多啊

SAM模板，并get新技能，一个串的SAM与另一个串的匹配

将一个串的SAM建好之后，枚举另一个串的字符，如果可以直接匹配就直接匹配，如果直接匹配不了，那么将SAM的指针不停地往上跳，使得代表字符串的长度越来越小，以便能够匹配

#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
const int MAXN=250000+10;
int n1,n2,tot=1,las=1,ch[MAXN<<1][30],len[MAXN<<1],fa[MAXN<<1],size[MAXN<<1],ans;
char s1[MAXN],s2[MAXN];
template<typename T> inline void read(T &x)
{
	T data=0,w=1;
	char ch=0;
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
	x=data*w;
}
template<typename T> inline void write(T x,char ch='\0')
{
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);
	putchar(x%10+'0');
	if(ch!='\0')putchar(ch);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
inline void extend(int c)
{
	int p=las,np=++tot;
	las=np;
	len[np]=len[p]+1;
	while(p&&!ch[p][c])ch[p][c]=np,p=fa[p];
	if(!p)fa[np]=1;
	else
	{
		int q=ch[p][c];
		if(len[q]==len[p]+1)fa[np]=q;
		else
		{
			int nq=++tot;
			fa[nq]=fa[q];
			memcpy(ch[nq],ch[q],sizeof(ch[nq]));
			len[nq]=len[p]+1;
			fa[np]=fa[q]=nq;
			while(p&&ch[p][c]==q)ch[p][c]=nq,p=fa[p];
		}
	}
	size[np]=1;
}
int main()
{
	scanf("%s%s",s1+1,s2+1);
	n1=strlen(s1+1),n2=strlen(s2+1);
	for(register int i=1;i<=n1;++i)extend(s1[i]-'a'+1);
	for(register int i=1,j=1,res=0,c;i<=n2;++i)
	{
		c=s2[i]-'a'+1;
		if(ch[j][c])res++,j=ch[j][c];
		else
		{
			while(j&&!ch[j][c])j=fa[j];
			if(!j)res=0,j=1;
			else res=len[j]+1,j=ch[j][c];
		}
		chkmax(ans,res);
	}
	write(ans,'\n');
	return 0;
}