Theme Section( 扩展 KMP )

待我称王封你为后i 2023-06-01 06:12 26阅读 0赞

*  Theme Section( 扩展 KMP )
    
     *  题目
     *  题意
     *  思路
     *  题解

# Theme Section( 扩展 KMP ) #

## 题目 ##

It's time for music! A lot of popular musicians are invited to join us in the music festival. Each of them will play one of their representative songs. To make the programs more interesting and challenging, the hosts are going to add some constraints to the rhythm of the songs, i.e., each song is required to have a 'theme section'. The theme section shall be played at the beginning, the middle, and the end of each song. More specifically, given a theme section E, the song will be in the format of ' EAEBE', where section A and section B could have arbitrary number of notes. Note that there are 26 types of notes, denoted by lower case letters 'a' - 'z'.

To get well prepared for the festival, the hosts want to know the maximum possible length of the theme section of each song. Can you help us?  
**Input**  
The integer N in the first line denotes the total number of songs in the festival. Each of the following N lines consists of one string, indicating the notes of the i-th (1 <= i <= N) song. The length of the string will not exceed 10^6.

**Output**  
There will be N lines in the output, where the i-th line denotes the maximum possible length of the theme section of the i-th song.

**Sample Input**  
6  
xy  
abc  
aaa  
aaaaba  
aaxoaaaaa  
aaaaabaaa  
**Sample Output**  
0  
0  
1  
1  
2  
2

## 题意 ##

题意是在一个字符串中找出一个前缀一个后缀和一个中间的子串，互相不重叠并且三部分完全一样

## 思路 ##

既然是三个部分不重叠，那么每个部分就不可能超过全长的 **1/3** 。可以先看后面的再看前面的再看中间的，所以首先后面的直接从 **大于等于 2/3** 的地方开始判断。

*  先用extend数组判断字符串后缀是否等于前缀。如果相等的话，由extend的性质得 **nex\[i\]+i==len**
 *  然后就是中间的，当然注意从哪开始到哪结束，易得 范围是 **\[nex\[i\],i)** 。当然还要完全一样，所以中间的长度可能不够（跳过），可能一样长（符合条件），或者更长（符合条件）的情况。

## 题解 ##

#include <iostream> #include <cstring> using namespace std; char str[1000050]; int nex[1000050]; int len; void getNext() { int p = 0, a = 0; nex[0] = len; for (int i = 1; i < len; i++) { if (i >= p || i + nex[i - a] >= p) { if (i >= p) p = i; while (p < len && str[p] == str[p - i]) p++; nex[i] = p - i; a = i; } else nex[i] = nex[i - a]; } } int main() { int T; cin >> T; while (T--) { cin >> str; len = strlen(str); getNext(); int s = len / 3; int res = 0; for (int i =len- s; i < len; i++) { if (nex[i] + i == len) { for (int j = nex[i]; j < i; j++) { if (nex[j] >= nex[i]) res = max(res, nex[i]); } } } cout << res << endl; } }

转载于:https://www.cnblogs.com/tttfu/p/11307694.html