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 a bit harder, for some 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 at most 10 lines, each line consists of no more than 100000 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
aaaajfaaaa
Output:
2
Solution
提出一个串做SAM
做完后，将剩下的所有串与SAM进行匹配
每次匹配时记录每个节点以它为结尾最长能够匹配多长，然后将这次匹配的结果与保存的最后的答案取min（我们要保证剩下的所有的串与第一个串匹配的是同一段）
全部匹配完后在每个节点取max就是答案了
#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=10+5,MAXS=100000+10,inf=0x3f3f3f3f;
int n,las=1,tot=1,len[MAXS<<1],fa[MAXS<<1],ch[MAXS<<1][30],ans,st[MAXS<<1],rk[MAXS<<1],cnt[MAXS],now[MAXS<<1];
char s[MAXN][MAXS];
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=' ')
{
if(x<0)putchar('-'),x=-x;
if(x>9)write(x/10);
putchar(x%10+'0');
if(ch!=' ')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];
}
}
}
inline int match(int w)
{
memset(now,0,sizeof(now));
for(register int i=1,j=1,res=0,lt=strlen(s[w]+1);i<=lt;++i)
{
int c=s[w][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(now[j],res);
}
for(register int i=tot;i>=1;--i)chkmax(now[fa[rk[i]]],now[rk[i]]);
for(register int i=1;i<=tot;++i)chkmin(st[i],now[i]);
}
int main()
{
while(scanf("%s",s[++n]+1)!=EOF);
n--;
for(register int i=1,lt=strlen(s[1]+1);i<=lt;++i)extend(s[1][i]-'a'+1);
for(register int i=1;i<=tot;++i)st[i]=len[i];
for(register int i=1;i<=tot;++i)cnt[len[i]]++;
for(register int i=1,lt=strlen(s[1]+1);i<=lt;++i)cnt[i]+=cnt[i-1];
for(register int i=1;i<=tot;++i)rk[cnt[len[i]]--]=i;
for(register int i=2;i<=n;++i)match(i);
for(register int i=2;i<=tot;++i)chkmax(ans,st[i]);
write(ans,'
');
return 0;
}