POJ2774 后缀数组

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.
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

题意：求两个字符串最大公共长度。

思路：后缀数组。（摘自罗穗骞的国家集训队论文）字符串的任何一个子串都是这个字符串的某个后缀的前缀。求 A 和 B 的最长公共子串等价于求 A 的后缀和 B 的后缀的最长公共前缀的最大值。如果枚举A和 B 的所有的后缀，那么这样做显然效率低下。由于要计算 A 的后缀和 B 的后缀的最长公共前缀，所以先将第二个字符串写在第一个字符串后面，中间用一个没有出现过的字符隔开，再求这个新的字符串的后缀数组。观察一下，看看能不能从这个新的字符串的后缀数组中找到一些规律。以 A=“ aaaba ”，B=“ abaa ”为例，如图 8 所示。

    

    那么是不是所有的 height 值中的最大值就是答案呢？不一定！有可能这两个后缀是在同一个字符串中的，所以实际上只有当suffix(sa[i-1])和suffix(sa[i]) 不是同一个字符串中的两个后缀时，height[i]才是满足条件的。而这其中的最大值就是答案。记字符串 A 和字符串 B 的长度分别为|A|和|B|。求新的字符串的后缀数组和 height 数组的时间是 O(|A|+|B|) ，然后求排名相邻 但原来不在同一个字符串中的两个后缀的height值的最大值，时间也是O(|A|+|B|)，所以整个做法的时间复杂度为 O(|A|+|B|) 。时间复杂度已经取到下限，由此看出，这是一个非常优秀的算法。

 

/*
后缀数组就是套模板求先应得数组，这题用到了两个数组，分别是sa[],height[];
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。
*/

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int MAX = 2e5+10;
char str[MAX];
int s[MAX], sa[MAX], t[MAX], t2[MAX], c[MAX], n;
int rank1[MAX], height[MAX];

void build_sa(int m) {
    int *x = t, *y = t2;
    for(int i = 0; i < m; i ++) c[i] = 0;
    for(int i = 0; i < n; i ++) c[x[i]=s[i]] ++;
    for(int i = 1; i < m; i ++) c[i] += c[i-1];
    for(int i = n-1; i >= 0; i --) sa[--c[x[i]]] = i;
    for(int k = 1; k <= n; k <<= 1) {
        int p = 0;
        for(int i = n-k; i < n; i ++) y[p++] = i;
        for(int i = 0; i < n; i ++) if(sa[i] >= k) y[p++] = sa[i] - k;
        for(int i = 0; i < m; i ++) c[i] = 0;
        for(int i = 0; i < n; i ++) c[x[y[i]]]++;
        for(int i = 1; i < m; i ++) c[i] += c[i-1];
        for(int i = n-1; i >= 0; i --) sa[--c[x[y[i]]]] = y[i];
        swap(x,y);
        p = 1;
        x[sa[0]] = 0;
        for(int 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 k = 0;
    for(int i = 0; i < n; i ++) rank1[sa[i]] = i;
    for(int i = 0; i < n; i ++) {
        if(k) k--;
        int j = sa[rank1[i]-1];
        while(s[i+k] == s[j+k]) k++;
        height[rank1[i]] = k;
    }
}
int main() {
    n = 0;
    int l1, l2;
    scanf("%s",str);
    l1 = strlen(str);
    for(int i = 0; i < l1; i ++) s[n++] = str[i] - 'a' + 1;
    s[n++] = 28;
    scanf("%s",str);
    l2 = strlen(str);
    for(int i = 0; i < l2; i ++) s[n++] = str[i] - 'a' + 1;
    s[n++] = 0;
    build_sa(30);
    getHeight();
    int maxn = 0;
    for(int i = 2; i < n; i ++) {
        if(maxn < height[i]) {
            if(sa[i] >= 0 && sa[i] < l1 && sa[i-1] > l1)
                maxn = height[i];
            if(sa[i-1] >= 0 && sa[i-1] < l1 && sa[i] > l1)
                maxn = height[i];
        }
    }
    printf("%d
",maxn);
    return 0;
}