POJ-2774

Long Long Message

Time Limit: 4000MS		Memory Limit: 131072K
Total Submissions: 23163		Accepted: 9509
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

Source

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

/**
          题意：求两个字符串的最长公共子串
          做法：后缀数组
          求两个串的最长公共子串等价于求两个串的最长公共前缀的最大值
          所以把两个字符串合并成一个字符串并且对其进行da()；
          sa[]数组    表示排第几的是谁
          Rank[]      表示谁排第几
          heigth[]    表示suffix(sa[i])和suffix(sa[i-])之间的公共前缀的个数
**/
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <queue>
#define maxn 220010
                   //           100000.
int t1[maxn],t2[maxn],c[maxn];
int sa[maxn],height[maxn],rank[maxn];
int a[maxn],b[maxn];
using namespace std;
bool cmp(int *r,int a,int b,int l)
{
    return r[a] == r[b] &&r[a+l] == r[b+l];
}
void da(int str[],int sa[],int rank[],int heigth[],int n,int m)
{
    n++;
    ///桶式排序
    int i,j,p,*x = t1,*y = t2;
    for(i=0; i<m; i++) c[i] = 0;  ///初始化
    for(i=0; i<n; i++) c[x[i] = str[i]] ++;///每个字符串出现的次数
    for(i=1; i<m; i++) c[i] += c[i-1];      ///字符串中各个字符的个数
    for(i=n-1; i>=0; i--)
        sa[--c[x[i]]] = i;            ///下一轮的第二关键字
    for(j=1; j<=n; j<<=1)
    {
        p=0;
        for(i=n-j; i<n; i++) y[p++] = i;        ///n-j 因为每次排序都是对字符串开始的长度为2^k的字符串进行排序的 
                                                                                /// 所以n-j是拍不到的部分 并且是对第二关键字的排序
        for(i=0; i<n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;              ///只有满足sa[i] >= j 才能作为y数组的第二关键字存储第一关键字
        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];
        swap(x,y);
        p=1;
        x[sa[0]] = 0;
        for(i=1; i<n; i++)
            x[sa[i]] = cmp(y,sa[i-1],sa[i],j) ?p-1:p++;
        if(p >= n) break;
        m = p;
    }
    int k=0;
    n--;
    for(i=0; i<=n; i++)
        rank[sa[i]] = i;
    for(i=0; i<n; i++)
    {
        if(k) k--;
        j = sa[rank[i]-1];
        while(str[i+k] == str[k+j]) k++;
        heigth[rank[i]] = k;
    }
}
char str[maxn];
char str1[maxn];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
    while(~scanf("%s",str))
    {
             //       cout<<str<<endl;
        scanf("%s",str1);
        int len = strlen(str);
        int n =0;
        for(int i=0; i<len; i++)
        {
            a[n++] = str[i] -'a'+1;
        }
        a[n++] = 28;
        len = strlen(str1);
        for(int i=0; i<len; i++)
        {
            a[n++] = str1[i] - 'a'+1;
        }
        da(a,sa,rank,height,n,30);
        int maxx=0,pos=0;
        len=strlen(str);
        for(int i=2; i<n; i++)
            if(height[i]>maxx)
            {
                if(0<=sa[i-1]&&sa[i-1]<len&&len<sa[i])
                    maxx=height[i];
                if(0<=sa[i]&&sa[i]<len&&len<sa[i-1])
                    maxx=height[i];
            }
        printf("%d\n",maxx);
    }
}