spoj Repeats 后缀数组

A string s is called an (k,l)-repeat if s is obtained by concatenating k>=1 times some seed string t with length l>=1. For example, the string

s = abaabaabaaba

is a (4,3)-repeat with t = aba as its seed string. That is, the seed string t is 3 characters long, and the whole string s is obtained by repeating t 4 times.

Write a program for the following task: Your program is given a long string u consisting of characters ‘a’ and/or ‘b’ as input. Your program must find some (k,l)-repeat that occurs as substring within u with k as large as possible. For example, the input string

u = babbabaabaabaabab

contains the underlined (4,3)-repeat s starting at position 5. Since u contains no other contiguous substring with more than 4 repeats, your program must output the maximum k.

Input

In the first line of the input contains H- the number of test cases (H <= 20). H test cases follow. First line of each test cases is n - length of the input string (n <= 50000), The next n lines contain the input string, one character (either ‘a’ or ‘b’) per line, in order.

Output

For each test cases, you should write exactly one interger k in a line - the repeat count that is maximized.

Example

Input:
1
17
b
a
b
b
a
b
a
a
b
a
a
b
a
a
b
a
b

Output:
4

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxm = 50005;
int a[maxm], s[maxm], Height[maxm], tp[maxm], sa[maxm], Rank[maxm], dp[maxm][20];
int n, m;
char str[maxm];
int cmp(int *f, int x, int y, int w)
{
	return f[x] == f[y] && f[x + w] == f[y + w];
}
void Rsort()
{
	for (int i = 0;i <= m;i++) s[i] = 0;
	for (int i = 1;i <= n;i++) s[Rank[tp[i]]]++;
	for (int i = 1;i <= m;i++) s[i] += s[i - 1];
	for (int i = n;i >= 1;i--) sa[s[Rank[tp[i]]]--] = tp[i];
}
void suffix()
{
	for (int i = 1;i <= n;i++) Rank[i] = a[i], tp[i] = i;
	m = 30, Rsort();
	for (int i, w = 1, p = 1;p < n;w += w, m = p)
	{
		for (p = 0, i = n - w + 1;i <= n;i++) tp[++p] = i;
		for (i = 1;i <= n;i++) if (sa[i] > w) tp[++p] = sa[i] - w;
		Rsort(), swap(Rank, tp), Rank[sa[1]] = p = 1;
		for (i = 2;i <= n;i++) Rank[sa[i]] = cmp(tp, sa[i], sa[i - 1], w) ? p : ++p;
	}
	int j, k = 0;
	for (int i = 1;i <= n;Height[Rank[i++]] = k, tp[i] = 0)
		for (k = k ? k - 1 : k, j = sa[Rank[i] - 1];a[i + k] == a[j + k];k++);
}
void RMQ()
{
	for (int i = 1;i <= n;i++) dp[i][0] = Height[i];
	for (int j = 1;(1 << j) <= n;j++)
		for (int i = 1;i + (1 << j) - 1 <= n;i++)
			dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
}
int query(int x, int y)
{
	x = Rank[x], y = Rank[y];
	if (x > y) swap(x, y);
	int k = 0;
	while (1 << (k + 1) <= y - x) k++;
	return min(dp[x + 1][k], dp[y - (1 << k) + 1][k]);
}
int main()
{
	int i, j, k, sum, t;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &n);
		getchar();
		for (i = 1;i <= n;i++)
		{
			scanf("%c", &str[i]);
			a[i] = str[i] - 'a' + 1;
			getchar();
		}
		a[n + 1] = 0;
		suffix(), RMQ();
		int ans = 0, lcp, temp;
		for (int len = 1;len <= n;len++)
		{
			for (int i = 0;i + len <= n;i += len)
			{
				lcp = query(i, i + len);
				temp = lcp / len + 1;
				k = i - (len - lcp%len);
				if (k >= 0 && lcp%len)
					if (query(k, k + len) >= lcp) temp++;
				ans = max(ans, temp);
			}
		}
		printf("%dn", ans);
	}
	return 0;
}