题目描述
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.
输入输出格式
输入格式：
The input contains at most 10 lines, each line consists of no more than 100000 lowercase letters, representing a string.
输出格式：
The length of the longest common substring. If such string doesn‘t exist, print "0" instead.
输入输出样例
输出样例#1：
2
大概就是要你求最多10个串的最长公共子串，裸上后缀自动机。。。。
#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define ll long long
#define maxn 4000005
using namespace std;
int a[maxn],c[maxn],tot;
int base=1,cnt=1,pre=1,n;
int f[maxn],ch[maxn][26];
int l[maxn],siz[maxn],ans=0;
char s[maxn];
inline void ins(int x){
int p=pre,np=++cnt;
pre=np,l[np]=l[p]+1;
siz[np]=base;
for(;p&&!ch[p][x];p=f[p]) ch[p][x]=np;
if(!p) f[np]=1;
else{
int q=ch[p][x];
if(l[q]==l[p]+1) f[np]=q;
else{
int nq=++cnt;
l[nq]=l[p]+1;
memcpy(ch[nq],ch[q],sizeof(ch[q]));
f[nq]=f[q];
f[q]=f[np]=nq;
for(;ch[p][x]==q;p=f[p]) ch[p][x]=nq;
}
}
}
inline void build(){
n=strlen(s),tot+=n;
for(int i=0;i<n;i++) ins(s[i]-‘a‘);
}
inline void solve(){
base--;
for(int i=1;i<=cnt;i++) c[l[i]]++;
for(int i=tot;i>=0;i--) c[i]+=c[i+1];
for(int i=1;i<=cnt;i++) a[c[l[i]]--]=i;
for(int i=1;i<=cnt;i++){
int now=a[i];
siz[f[now]]|=siz[now];
if(siz[now]==base) ans=max(ans,l[now]);
}
}
int main(){
while(scanf("%s",s)==1) build(),base<<=1;
solve();
printf("%d\n",ans);
return 0;
}
