poj2774 Long Long Message

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

yeshowmuchiloveyoumydearmotherreallyicannotbelieveityeaphowmuchiloveyoumydearmother

Sample Output

27

题意：求两个字符串的最长公共串长度

思路：利用后缀数组知识求得sa和height两个数组，求得height数组的最大值就是所求的值

sa[i]表示所有后缀按字典数排序后以s[i]
           开始的后缀排在第i位。height[i]表示字典数为i和i-1后缀的
           的最长串的前缀。

另外如果我们连接两个原始串,中间不加区分符号的话,会出现什么情况呢?比如串aaa和串aaaaaaa,那么就会算的他们的后缀LCP可能为6了,这明显是错的.所以要分隔符

而且在串的最后加上个0

ac：代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;


const int mx=200100;
char st[mx];
int s[mx],sa[mx],t[mx],t2[mx],c[mx],n;
int rank[mx],height[mx];


void build_sa(int m)
{
    int i,*x=t,*y=t2;
    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 (int k=1;k<=n;k<<=1)
    {
        int p=0;
        for (i=n-k;i<n;i++) y[p++]=i;
        for (i=0;i<n;i++) if (sa[i]>=k) y[p++]=sa[i]-k;
        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];
        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]+k]==y[sa[i]+k]?p-1:p++;
        if (p>=n) break;
        m=p;
    }
}


void getHeight()
{
    int i,j,k=0;
    for (i=0;i<n;i++) rank[sa[i]]=i;
    for (i=0;i<n;i++)
    {
        if (k) k--;
        int j=sa[rank[i]-1];
        while (s[i+k]==s[j+k]) k++;
        height[rank[i]]=k;
    }
}


int main()
{
   n=0;
   scanf("%s",st);
   int len1=strlen(st);
   for(int i=0;i<len1;i++)
   s[n++]=st[i]-'a'+1;
   s[n++]=28;//随便，只要不是字母和0就行
   scanf("%s",st);
   int len2=strlen(st);
   for(int i=0;i<len2;i++)
   s[n++]=st[i]-'a'+1;
   s[n++]=0;
   build_sa(29);
   getHeight();
   int max=0;
   for(int i=2;i<=n;i++)
   {
       if(height[i]>max)
       {
           if (sa[i]>=0&&sa[i]<len1&&sa[i-1]>len1)
            max=height[i];
            if (sa[i-1]>=0&&sa[i-1]<len1&&sa[i]>len1)
            max=height[i];
       }
   }
   printf("%d\n",max);
    return 0;
}