Theme Section

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4961    Accepted Submission(s): 2478

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

0

0

1

1

2

Source

2013 ACM/ICPC Asia Regional Changchun Online

#include <bits/stdc++.h>
using namespace std;

const int mn = 1000010;

char ch[mn], t1[mn], t2[mn];
int len;

int nx[mn];
void cal_next(char ch[])
{
	int len = strlen(ch);
	nx[0] = -1;
	int k = -1;
	for (int i = 1; i < len; i++)
	{
		while (k != -1 && ch[k + 1] != ch[i])
			k = nx[k];
		if (ch[k + 1] == ch[i])
			k++;
		nx[i] = k;
	}
}

int KMP(char a[], char b[])
{
	cal_next(b);

	int n = strlen(a), m = strlen(b);
	int k = -1;
	for (int i = 0; i < n; i++)
	{
		while (k > -1 && b[k + 1] != a[i])
			k = nx[k];
		if (b[k + 1] == a[i])
			k++;
		if (k == m - 1)
			return i - m + 1;
	}
	return -1;
}

int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%s", ch);

		len = strlen(ch);
		cal_next(ch);

		int ans = 0;
		for (int i = nx[len - 1]; i != -1; i = nx[i])
		{
			if (3 * (i + 1) > len)  // 保证不重叠
				continue; 

			/// 每次按 nx[i] 减小长度查询
			// 在 中间一段字符串 查询是否存在 前nx位字符
			for (int j = i + 1; j < len - i - 1; j++)
				t1[j - (i + 1)] = ch[j];
			t1[len - 2 * (i + 1)] = '\0';
			for (int j = 0; j <= i; j++)
				t2[j] = ch[j];
			t2[i + 1] = '\0';

			if (KMP(t1, t2) != -1)
			{
				ans = i + 1;
				break;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}