【字符串入门专题1】【hdu4763】【未ac题】【扩展kmp exkmp】-优快云博客

本文介绍了一种算法，用于确定音乐节上表演曲目的最大可能主题段落长度。通过分析给定歌曲的结构，该算法能够找出符合特定格式要求的主题部分，即‘EAEBE’格式，其中E为主题段，而A和B可以是任意长度的其他部分。文章提供了实现这一功能的代码示例。

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Theme Section

Problem Description

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

引用书上的解释：假设歌曲长度为l，主旋律长度为n，将歌曲长度看做一个长为l的字符串s。主旋律长度l一定满足一下两个条件，1:next[l-n]==n，即该字符串以长度为n的前缀为后缀。2:max{next[n]...next[l-n-1]} > n ,即s[n...l-n-1]的某个长度为n的子串等于长度为n的前缀。首先求出next数组，然后就可以依此减少n的值，查看能否满足前两个条件,判断l是否为一个可行解。

AC代码：but 自己没有看懂先留着

#include<string.h>
#include<stdio.h>
#define N 11000000

int l,n;
int a[N];
char s[N];

int main()
{
	int t,i,j,p,maxl;
	scanf("%d",&t);
	while(t --)
	{
		getchar();
		scanf("%s",s);
		l = strlen(s);
		a[0] = l;
		i = 1;
		j = 0;
		p = -1;
		while(i < l)//求出next数组 
		{
			if(p == -1)
			{
				j = 0;
				do
					p++;
				while(s[i+p]==s[j+p]);
				a[i] = p;
			}
			else if(a[j] < p)
				a[i] = a[j];
			else if(a[j] > p)
			{
				j = 0;
				a[i] = p;
			}
			else 
			{
				j = 0;
				while(s[i+p] == s[j+p])
					p++;
				a[i] = p;
			}
			i++;
			j++;
			p--;
		}
		a[l] = 0;
		n = l/3;
		maxl = a[l-n-n];
		for(i = n; i < l-n-n; i ++)
			if(maxl < a[i])
				maxl = a[i];
		while(maxl<n||a[l-n]!=n)//将n从小到大进行比较 
		{
			n --;
			if(maxl < a[n]) maxl = a[n];
			if(maxl < a[l-n-n]) maxl = a[l-n-n];
			if(maxl < a[l-n-n-1]) maxl = a[l-n-n-1];
		}
		printf("%d\n",n);
	}
	return 0;
}

转载于:https://www.cnblogs.com/hellocheng/p/7350054.html