标签：spi ons 后缀 init ann diff pre dha nes
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, where p 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.
好长时间没有写过后缀数组的问题了，看了一下午的课本回顾了一下。。。。
这个题就是属于后缀数组的基础的问题，感觉就是套模板。下面贴一下后缀数组的模板。
const int maxn=1e4+5;
int t[maxn],t2[maxn];
int sa[maxn],c[maxn];
char s[maxn];
int n;
void build(int m)
{
int *x=t,*y=t2;
for(int i=0;i<m;i++) c[i]=0;
for(int i=0;i<n;i++) c[x[i]=s[i]]++;
for(int i=1;i<m;i++) c[i]+=c[i-1];
for(int i=n-1;i>=0;i--) sa[--c[x[i]]]=i;
for(int k=1;k<=n;k<<=1)
{
int p=0;
for(int i=n-k;i<n;i++) y[p++]=i;
for(int i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
//基数排序第一关键字
for(int i=0;i<m;i++) c[i]=0;
for(int i=0;i<n;i++) c[x[y[i]]]++;
for(int i=1;i<m;i++) c[i]+=c[i-1];
for(int i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];
swap(x,y);
p=1; x[sa[0]]=0;
for(int i=1;i<n;i++) x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;
if(p>=n) break;
m=p;
}
}
int rank[maxn],height[maxn];
void getheight()
{
int k=0;
for(int i=0;i<n;i++) rank[sa[i]]=i;
for(int i=0;i<n;i++)
{
if(k) k--;
if(rank[i]==0)
{
height[rank[i]]=0;
continue;
}
int j=sa[rank[i]-1];
while(s[i+k]==s[j+k]) k++;
height[rank[i]]=k;
}
}
标签：spi ons 后缀 init ann diff pre dha nes
原文地址：http://www.cnblogs.com/Heilce/p/6544705.html
