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Language:
Default Long Long Message
Description The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes to him these days: his mother is getting ill. Being worried about spending so much on railway tickets (Byterland is such a big country, and he has to spend 16 shours on train to his hometown), he decided only to send SMS with his mother.
The little cat lives in an unrich family, so he frequently comes to the mobile service center, to check how much money he has spent on SMS. Yesterday, the computer of service center was broken, and printed two very long messages. The brilliant little cat soon found out: 1. All characters in messages are lowercase Latin letters, without punctuations and spaces. You are given those two very long messages, and you have to output the length of the longest possible original text written by the little cat. Background: Why ask you to write a program? There are four resions: Input Two strings with lowercase letters on two of the input lines individually. Number of characters in each one will never exceed 100000.
Output A single line with a single integer number – what is the maximum length of the original text written by the little cat.
Sample Input yeshowmuchiloveyoumydearmotherreallyicannotbelieveit Sample Output 27 |
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Mean:
给你两个字符串s1和s2，输出这两个字符串的最长公共子串长度。
analyse:
求最长公共子序列的方法很多，这里用后缀数组实现。
后缀数组怎么求最长公共子序列呢？
在后缀数组中，height数组:height[i]保存的是字典序排名相邻的两个后缀子串的最长公共前缀。
将s2接到s1后面，然后中间用一个未出现的字符隔开，再求height数组。
这两个字符串的最长公共子串必定存在于合并后的S串的最长公共前缀之中。
只需要寻找分别来自于s1串和来自于s2串的两个前缀的height的最大值，即得答案。
Time complexity：O(nlogn)
Source code：
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-05-09-21.22
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;
const int MAXN=<<;
//以下为倍增算法求后缀数组
int wa[MAXN],wb[MAXN],wv[MAXN],Ws[MAXN];
int cmp(int *r,int a,int b,int l)
{return r[a]==r[b]&&r[a+l]==r[b+l];}
/**< 传入参数：str,sa,len+1,ASCII_MAX+1 */
void da(const char *r,int *sa,int n,int m)
{
int i,j,p,*x=wa,*y=wb,*t;
for(i=; i<m; i++) Ws[i]=;
for(i=; i<n; i++) Ws[x[i]=r[i]]++;
for(i=; i<m; i++) Ws[i]+=Ws[i-];
for(i=n-; i>=; i--) sa[--Ws[x[i]]]=i;
for(j=,p=; p<n; j*=,m=p)
{
for(p=,i=n-j; i<n; i++) y[p++]=i;
for(i=; i<n; i++) if(sa[i]>=j) y[p++]=sa[i]-j;
for(i=; i<n; i++) wv[i]=x[y[i]];
for(i=; i<m; i++) Ws[i]=;
for(i=; i<n; i++) Ws[wv[i]]++;
for(i=; i<m; i++) Ws[i]+=Ws[i-];
for(i=n-; i>=; i--) sa[--Ws[wv[i]]]=y[i];
for(t=x,x=y,y=t,p=,x[sa[]]=,i=; i<n; i++)
x[sa[i]]=cmp(y,sa[i-],sa[i],j)?p-:p++;
}
return;
}
int sa[MAXN],Rank[MAXN],height[MAXN];
/**< str,sa,len */
void calheight(const char *r,int *sa,int n)
{
int i,j,k=;
for(i=; i<=n; i++) Rank[sa[i]]=i;
for(i=; i<n; height[Rank[i++]]=k)
for(k?k--:,j=sa[Rank[i]-]; r[i+k]==r[j+k]; k++);
// Unified
for(int i=n;i>=;--i) ++sa[i],Rank[i]=Rank[i-];
}
char s1[MAXN],s2[MAXN];
int main()
{
while(~scanf("%s%s",s1,s2))
{
int l1=strlen(s1);
strcat(s1,"{");
strcat(s1,s2);
int len=strlen(s1);
for(int i=;i<len;++i) s1[i]-='a'-;
da(s1,sa,len+,);
calheight(s1,sa,len);
int ans=;
for(int i=;i<=len;++i)
if((sa[i-]-<l1 && sa[i]->l1) || (sa[i-]->l1 && sa[i]-<l1))
ans=max(ans,height[i]);
printf("%d\n",ans);
}
return ;
}