Theme Section⭐⭐（kmp！！）题意+思路_it's time for music! a lot of popular musicians ar-优快云博客

本文介绍了一个音乐节活动，主持人要求每位表演者歌曲中必须包含主题节段，且在歌曲开头、中间和结尾各出现一次。博主探讨如何计算每个歌曲主题节段的最大可能长度，以确保节目的创新性和难度。

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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
5
xy
abc
aaa
aaaaba
aaxoaaaaa
Sample Output
0
0
1
1
2

题意：给出一个字符串，求这个字符串前，中，后相同的部分长度为多少。

思路：首先判断符合前后的，可以直接用next数组，然后判断中间的。具体看代码

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
char s[1000010];
int ne[1000010];
int n,m,l;

void get_next()
{
	int j=-1,i=0;//j是在模板串中移动的指针，而i是作为ne的下标的位置 
	ne[0]=-1;
	while(i<l)
	{
		if(j==-1||s[i]==s[j])//相同继续下一个 
		{
			ne[++i]=++j;//ne 下标 从 1开始 
		}
		else
		{
			j=ne[j];//j的指针移动位置 
		}
	}
}
int kmp()
{
	int i,j;
	for(int i=ne[l];i;i=ne[i])//缩小循环的大小 
	{
		for(j=i+1;j<=l-i;j++)//中间部分的查找 
		{
			if(ne[j]==i)
			return i;
		}
	}
	return 0;
}
int main()
{
	int t;cin>>t;
	while(t--)
	{
		cin>>s;
		l=strlen(s);
		memset(ne,0,sizeof(ne));
		get_next();
		cout<<kmp()<<endl;
	}
	return 0;
}