Codeforces 427 D. Match amp; Catch

后缀数组....

在两个串中唯一出现的最小公共子串

D. Match & Catch

time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

Police headquarter is monitoring signal on different frequency levels. They have got two suspiciously encoded strings s1 and s2 from two different frequencies as signals. They are suspecting that these two strings are from two different criminals and they are planning to do some evil task.

Now they are trying to find a common substring of minimum length between these two strings. The substring must occur only once in the first string, and also it must occur only once in the second string.

Given two strings s1 and s2 consist of lowercase Latin letters, find the smallest (by length) common substring p of both s1 and s2, wherep is a unique substring in s1 and also in s2. See notes for formal definition of substring and uniqueness.

Input

The first line of input contains s1 and the second line contains s2 (1 ≤ |s1|, |s2| ≤ 5000). Both strings consist of lowercase Latin letters.

Output

Print the length of the smallest common unique substring of s1 and s2. If there are no common unique substrings of s1 and s2 print -1.

Sample test(s)

input

apple
pepperoni

output

2

input

lover
driver

output

1

input

bidhan
roy

output

-1

input

testsetses
teeptes

output

3

Note

Imagine we have string a = a1a2a3...a|a|, where |a| is the length of string a, and ai is the ith letter of the string.

We will call string alal + 1al + 2...ar (1 ≤ l ≤ r ≤ |a|) the substring [l, r] of the string a.

The substring [l, r] is unique in a if and only if there is no pair l1, r1 such that l1 ≠ l and the substring [l1, r1] is equal to the substring[l, r] in a.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>using namespace std;const int maxn=10100,INF=0x3f3f3f3f;int sa[maxn],rank[maxn],rank2[maxn],h[maxn],c[maxn],*x,*y,ans[maxn];
char str[maxn];bool cmp(int*r,int a,int b,int l,int n)
{if(r[a]==r[b]&&a+l<n&&b+l<n&&r[a+l]==r[b+l]) return true;return false;
}bool radix_sort(int n,int sz)
{for(int i=0;i<sz;i++) c[i]=0;for(int i=0;i<n;i++) c[x[y[i]]]++;for(int i=1;i<sz;i++) c[i]+=c[i-1];for(int i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];
}void get_sa(char c[],int n,int sz=128)
{x=rank,y=rank2;for(int i=0;i<n;i++) x[i]=c[i],y[i]=i;radix_sort(n,sz);for(int len=1;len<n;len*=2){int yid=0;for(int i=n-len;i<n;i++) y[yid++]=i;for(int i=0;i<n;i++) if(sa[i]>=len) y[yid++]=sa[i]-len;radix_sort(n,sz);swap(x,y);x[sa[0]]=yid=0;for(int i=1;i<n;i++){x[sa[i]]=cmp(y,sa[i],sa[i-1],len,n)?yid:++yid;}sz=yid+1;if(sz>=n) break;}for(int i=0;i<n;i++) rank[i]=x[i];
}void get_h(char str[],int n)
{int k=0; h[0]=0;for(int i=0;i<n;i++){if(rank[i]==0) continue;k=max(k-1,0);int j=sa[rank[i]-1];while(i+k<n&&j+k<n&&str[i+k]==str[j+k]) k++;h[rank[i]]=k;}
}int main()
{cin>>str;int sg=strlen(str);str[sg]=127;cin>>str+sg+1;int n=strlen(str);get_sa(str,n);get_h(str,n);int ans=INF;int s1=0,s2=0,last=-1;for(int i=1;i<n;i++){if(sa[i-1]<sg&&sa[i]<sg) continue;if(sa[i-1]>sg&&sa[i]>sg) continue;int pre=h[i-1];int next=h[i+1];if(h[i]>max(pre,next)){ans=min(ans,max(pre,next)+1);}}if(ans==INF) ans=-1;printf("%d\n",ans);return 0;
}