后缀数组....
在两个串中唯一出现的最小公共子串
1 second
512 megabytes
standard input
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.
The first line of input contains s1 and
the second line contains s2 (1 ≤ |s1|, |s2| ≤ 5000).
Both strings consist of lowercase Latin letters.
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.
apple
pepperoni
2
lover
driver
1
bidhan
roy
-1
testsetses
teeptes
3
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;
}