D. Match & Catch
time limit per test1 second
memory limit per test512 megabytes
inputstandard input
outputstandard 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, where p 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.
Examples
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.
题意：给出两个串 (s1,s2) ，求两个字符串的一个子串，使得该子串在 (s1,s2) 中分别只出现过一次，且长度最短，输出最短长度。
题解：对于在各自串中只出现过一次的子串，那么以该子串为前缀的两个后缀就是 (sa) 数组中相邻位置，假设两个后缀分别为 (sa[i],sa[i+1])，要保证该子串不在其他地方出现，那么必须满足(height[i-1] < height[i] > height[i+1]) ，不过 (height[i]) 并不是最短长度，最短长度应该是 (min(height[i-1]+1,height[i+1]+1)) .
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define fi first
#define se second
#define dbg(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<endl;
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const int inf=0x3fffffff;
const ll mod=1000000007;
const int maxn=1e4+100;
char s1[maxn],s2[maxn];
/*
*suffix array
*倍增算法 O(n*logn)
*待排序数组长度为n,放在0~n-1中，在最后面补一个0
*build_sa( ,n+1,m+1); //注意是n+1,m是s数组中的最大值;
*getHeight(,n);
*例如：
*n = 8;
*num[] = { 1, 1, 2, 1, 1, 1, 1, 2, $ };注意num最后一位为0，其他大于0
*rank[] = { 4, 6, 8, 1, 2, 3, 5, 7, 0 };rank[0~n-1]为有效值，rank[n]必定为0无效值
*sa[] = { 8, 3, 4, 5, 0, 6, 1, 7, 2 };sa[1~n]为有效值，sa[0]必定为n是无效值
*height[]= { 0, 0, 3, 2, 3, 1, 2, 0, 1 };height[2~n]为有效值
*
*/
int sa[maxn];//SA数组，表示将S的n个后缀从小到大排序后把排好序的
//的后缀的开头位置顺次放入SA中
int t1[maxn],t2[maxn],c[maxn];//求SA数组需要的中间变量，不需要赋值
int rk[maxn],height[maxn];
//待排序的字符串放在s数组中，从s[0]到s[n-1],长度为n,且最大值小于m,
//除s[n-1]外的所有s[i]都大于0，r[n-1]=0
//函数结束以后结果放在sa数组中
void build_sa(int s[],int n,int m)
{
int i,j,p,*x=t1,*y=t2;
//第一轮基数排序，如果s的最大值很大，可改为快速排序
for(i=0;i<m;i++)c[i]=0;
for(i=0;i<n;i++)c[x[i]=s[i]]++;
for(i=1;i<m;i++)c[i]+=c[i-1];
for(i=n-1;i>=0;i--)sa[--c[x[i]]]=i;
for(j=1;j<=n;j<<=1)
{
p=0;
//直接利用sa数组排序第二关键字
for(i=n-j;i<n;i++)y[p++]=i;//后面的j个数第二关键字为空的最小
for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;
//这样数组y保存的就是按照第二关键字排序的结果
//基数排序第一关键字
for(i=0;i<m;i++)c[i]=0;
for(i=0;i<n;i++)c[x[y[i]]]++;
for(i=1;i<m;i++)c[i]+=c[i-1];
for(i=n-1;i>=0;i--)sa[--c[x[y[i]]]]=y[i];
//根据sa和x数组计算新的x数组
swap(x,y);
p=1;x[sa[0]]=0;
for(i=1;i<n;i++)
x[sa[i]]=y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;
if(p>=n)break;
m=p;//下次基数排序的最大值
}
}
void getHeight(int s[],int n)
{
int i,j,k=0;
for(i=0;i<=n;i++) rk[sa[i]]=i;
for(i=0;i<n;i++)
{
if(k)k--;
j=sa[rk[i]-1];
while(s[i+k]==s[j+k])k++;
height[rk[i]]=k;
}
}
int a[maxn];
int main()
{
scanf("%s%s",s1,s2);
int l1=(int)strlen(s1),l2=(int)strlen(s2);
rep(i,0,l1) a[i]=s1[i]-'a'+1;
a[l1]=27;
rep(i,0,l2) a[i+l1+1]=s2[i]-'a'+1;
int l=l1+l2+1;
a[l]=0;
build_sa(a,l+1,30);
getHeight(a,l);
int mi=1e9;
rep(i,2,l+1)
{
if(i>2&&height[i-1]>=height[i]) continue;
if(i<l&&height[i+1]>=height[i]) continue;
if(sa[i-1]<l1&&sa[i]>l1)
{
int t=1e9;
if(i==2) t=height[i+1]+1;
else if(i==l) t=height[i-1]+1;
else t=max(height[i+1]+1,height[i-1]+1);
mi=min(mi,min(height[i],t));
}
if(sa[i-1]>l1&&sa[i]<l1)
{
int t=1e9;
if(i==2) t=height[i+1]+1;
else if(i==l) t=height[i-1]+1;
else t=max(height[i+1]+1,height[i-1]+1);
mi=min(mi,min(height[i],t));
}
}
if(mi==1e9) puts("-1");
else printf("%d
",mi);
return 0;
}