POJ 2774-Long Long Message（后缀数组+高度数组-最大公共子串长度）

梦里梦外; 2022-06-17 08:57 122阅读 0赞

Long Long Message

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;4000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;131072K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;30192</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;12244</td> 
  </tr> 
  <tr> 
   <td colspan="3" align="center" style="color:rgb(14,173,0)"><strong>Case Time Limit:</strong>&nbsp;1000MS</td> 
  </tr> 
 </tbody> 
</table>

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

Source

[POJ Monthly--2006.03.26][],Zeyuan Zhu,"Dedicate to my great beloved mother."

## 题目意思： ##

计算两个字符串的最大公共子串长度。

## 解题思路： ##

后缀数组+高度数组模板题。

在两个串之间插入一个不会再串中出现的字符拼接成一个新串S'，计算高度数组并求出其分属于两个不同串S和T时的最大值。

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <queue>
    #include <algorithm>
    using namespace std;
    typedef long long ll;
    #define MAXN 200100
    int n,k;
    string s,t;
    int sa[MAXN],lcp[MAXN];
    int rank[MAXN*2],tmp[MAXN*2];
    void construct_lcp(string s,int sa[],int lcp[])//计算高度数组
    {
        int n=s.length();
        for(int i=0; i<=n; ++i)
            rank[sa[i]]=i;
        int h=0;
        lcp[0]=0;
        for(int i=0; i<n; ++i)
        {
            int j=sa[rank[i]-1];
            if(h>0) --h;//先减去首字母的长度
            for(; j+h<n&&i+h<n; ++h)
                if(s[j+h]!=s[i+h]) break;//保证前缀相同的前提下不断增加
            lcp[rank[i]-1]=h;
        }
    }
    bool compare_sa(int i,int j)//倍增法，比较rank
    {
        if(rank[i]!=rank[j]) return rank[i]<rank[j];
        else
        {
            int ri=i+k<=n?rank[i+k] :-1;
            int rj=j+k<=n?rank[j+k] :-1;
            return ri<rj;
        }
    }
    
    void construct_sa(string s,int sa[])//计算s的后缀数组
    {
        for(int i=0; i<=n; ++i)//初始长度为1，rank为字符编码
        {
            sa[i]=i;
            rank[i]=i<n?s[i] :-1;
        }
        for(k=1; k<=n; k*=2)//倍增法求后缀数组
        {
            sort(sa,sa+n+1,compare_sa);
            tmp[sa[0]]=0;
            for(int i=1; i<=n; ++i)
                tmp[sa[i]]=tmp[sa[i-1]]+(compare_sa(sa[i-1],sa[i])?1:0);
            for(int i=0; i<=n; ++i)
                rank[i]=tmp[i];
        }
    }
    bool contain(string s,int sa[],string t)
    {
        int a=0,b=s.length();
        while(b-a>1)
        {
            int c=(a+b)/2;
            if(s.compare(sa[c],t.length(),t)<0) a=c;
            else b=c;
        }
        return s.compare(sa[b],t.length(),t)==0;
    }
    void solve()
    {
        int sl=s.length();//原始串S的长度
        s+='\0'+t;//t串加入s串中
        n=s.length();
        construct_sa(s,sa);//计算后缀数组
        construct_lcp(s,sa,lcp);//计算高度数组
        int ans=0;
        for(int i=0; i<s.length(); ++i)//计算高度数组的最大值
            if((sa[i]<sl)!=(sa[i+1]<sl))//注意判断是否分属于两个不同字符串
                ans=max(ans,lcp[i]);
        cout<<ans<<endl;
    }
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("G:/cbx/read.txt","r",stdin);
        //freopen("G:/cbx/out.txt","w",stdout);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        cin>>s;
        cin>>t;
        solve();
        return 0;
    }

[POJ Monthly--2006.03.26]: http://poj.org/searchproblem?field=source&key=POJ+Monthly--2006.03.26