后缀数组(模板题) - 求最长公共子串 - poj 2774 Long Long Message

 

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Long Long Message

Time Limit: 4000MS		Memory Limit: 131072K
Total Submissions: 21228		Accepted: 8708
Case Time Limit: 1000MS

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.
2. All SMS has been appended to each other – (i+1)-th SMS comes directly after the i-th one – that is why those two messages are quite long.
3. His own SMS has been appended together, but possibly a great many redundancy characters appear leftwards and rightwards due to the broken computer.
E.g: if his SMS is “motheriloveyou”, either long message printed by that machine, would possibly be one of “hahamotheriloveyou”, “motheriloveyoureally”, “motheriloveyouornot”, “bbbmotheriloveyouaaa”, etc.
4. For these broken issues, the little cat has printed his original text twice (so there appears two very long messages). Even though the original text remains the same in two printed messages, the redundancy characters on both sides would be possibly different.

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:
The SMS in Byterland mobile service are charging in dollars-per-byte. That is why the little cat is worrying about how long could the longest original text be.

Why ask you to write a program? There are four resions:
1. The little cat is so busy these days with physics lessons;
2. The little cat wants to keep what he said to his mother seceret;
3. POJ is such a great Online Judge;
4. The little cat wants to earn some money from POJ, and try to persuade his mother to see the doctor :(

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
yeaphowmuchiloveyoumydearmother

Sample Output

27

---

 #

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=100010<<1;
//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=0; i<m; i++) Ws[i]=0;
     for(i=0; i<n; i++) Ws[x[i]=r[i]]++;
     for(i=1; i<m; i++) Ws[i]+=Ws[i-1];
     for(i=n-1; i>=0; i--) sa[--Ws[x[i]]]=i;
     for(j=1,p=1; p<n; j*=2,m=p)
     {
           for(p=0,i=n-j; i<n; i++) y[p++]=i;
           for(i=0; i<n; i++) if(sa[i]>=j) y[p++]=sa[i]-j;
           for(i=0; i<n; i++) wv[i]=x[y[i]];
           for(i=0; i<m; i++) Ws[i]=0;
           for(i=0; i<n; i++) Ws[wv[i]]++;
           for(i=1; i<m; i++) Ws[i]+=Ws[i-1];
           for(i=n-1; i>=0; i--) sa[--Ws[wv[i]]]=y[i];
           for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i<n; i++)
                 x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1: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=0;
     for(i=1; i<=n; i++) Rank[sa[i]]=i;
     for(i=0; i<n; height[Rank[i++]]=k)
           for(k?k--:0,j=sa[Rank[i]-1]; r[i+k]==r[j+k]; k++);
     // Unified
     for(int i=n;i>=1;--i) ++sa[i],Rank[i]=Rank[i-1];
}

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=0;i<len;++i) s1[i]-='a'-1;
           da(s1,sa,len+1,30);
           calheight(s1,sa,len);
           int ans=0;
           for(int i=2;i<=len;++i)
                 if((sa[i-1]-1<l1 && sa[i]-1>l1) || (sa[i-1]-1>l1 && sa[i]-1<l1))
                       ans=max(ans,height[i]);
           printf("%d\n",ans);
     }
     return 0;
}