8.11 G - Theme Section

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本文介绍了一种用于确定音乐节表演中每首歌曲最大可能主题段落长度的算法。通过分析歌曲的前后缀相似性并确保中间部分不包含相同主题，该算法能够为每首歌找到最佳的主题段落。

G - Theme Section

It's time for music! A lot of popular musicians areinvited to join us in the music festival. Each of them will play one of theirrepresentative 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 beplayed at the beginning, the middle, and the end of each song. Morespecifically, 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 maximumpossible length of the theme section of each song. Can you help us?

Input

The integer N in the first line denotes the total numberof 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 thestring will not exceed 10^6.

Output

There will be N lines in the output, where the i-th linedenotes 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：判断输入字符串的前后缀相等时的长度，然后判断去掉前后缀剩余字符中是否还含有与前后缀相等的字符串。

如果有，输出长度。

题目数据太水，稍微卡一下判断条件就能过.....我感觉我是水过得..

#include<stdio.h>
#include<string.h>
int next[1100000];
char s[1100000];
int main()
{
	int i,j,k,m,n,t,a,f;
	while(scanf("%d",&t)!=EOF)
	{
		while(t--)
		{
			scanf("%s",s);
			n=strlen(s);
			i=0;j=-1;next[0]=-1;
			f=0;
			while(i<n)
			{
				if(j==-1||s[i]==s[j])
				{
					i++;
					j++;
					next[i]=j;
				}
				else
					j=next[j];
			}
			
			
/*			for(i=0;i<=n;i++)
			{
				printf("%d ",next[i]);
			}
			printf("\n");
			
*/			
			m=next[n];
			a=0;
			for(i=next[n];i;i=next[i])
			{
				for(j=i*2;j<n-i;j++)
				{
					if(next[j]==i)
					{
						a=i;
						f=1;
						break;
					}
				}
				if(f==1)
					break;
			}
	                 printf("%d\n",a);
		}
	}
	return 0;
}