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
since a (4, 3)-repeat is found starting at the 5th character of the input string.
方法还是论文上说的方法，我用2个后缀数组，另外一个是字符串反转求出来的，然后分别LCP
/*************************************************************************
> File Name: SPOJ-687.cpp
> Author: ALex
> Mail: [email protected]
> Created Time: 2015年04月07日 星期二 18时16分49秒
************************************************************************/
#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>
using namespace std;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
class SuffixArray
{
public:
static const int N = 52000;
int init[N];
int X[N];
int Y[N];
int Rank[N];
int sa[N];
int height[N];
int buc[N];
int LOG[N];
int dp[N][20];
int size;
void clear()
{
size = 0;
}
void insert(int n)
{
init[size++] = n;
}
bool cmp(int *r, int a, int b, int l)
{
return (r[a] == r[b] && r[a + l] == r[b + l]);
}
void getsa(int m = 256) //m一般为最大值+1
{
init[size] = 0;
int l, p, *x = X, *y = Y, n = size + 1;
for (int i = 0; i < m; ++i)
{
buc[i] = 0;
}
for (int i = 0; i < n; ++i)
{
++buc[x[i] = init[i]];
}
for (int i = 1; i < m; ++i)
{
buc[i] += buc[i - 1];
}
for (int i = n - 1; i >= 0; --i)
{
sa[--buc[x[i]]] = i;
}
for (l = 1, p = 1; l <= n && p < n; m = p, l *= 2)
{
p = 0;
for (int i = n - l; i < n; ++i)
{
y[p++] = i;
}
for (int i = 0; i < n; ++i)
{
if (sa[i] >= l)
{
y[p++] = sa[i] - l;
}
}
for (int i = 0; i < m; ++i)
{
buc[i] = 0;
}
for (int i = 0; i < n; ++i)
{
++buc[x[y[i]]];
}
for (int i = 1; i < m; ++i)
{
buc[i] += buc[i - 1];
}
for (int i = n - 1; i >= 0; --i)
{
sa[--buc[x[y[i]]]] = y[i];
}
int i;
for (swap(x, y), x[sa[0]] = 0, p = 1, i = 1; i < n; ++i)
{
x[sa[i]] = cmp(y, sa[i - 1], sa[i], l) ? p - 1 : p++;
}
}
}
void getheight()
{
int h = 0, n = size;
for (int i = 0; i <= n; ++i)
{
Rank[sa[i]] = i;
}
height[0] = 0;
for (int i = 0; i < n; ++i)
{
if (h > 0)
{
--h;
}
int j =sa[Rank[i] - 1];
for (; i + h < n && j + h < n && init[i + h] == init[j + h]; ++h);
height[Rank[i] - 1] = h;
}
}
//预处理每一个数字的对数，用于rmq，常数优化
void initLOG()
{
LOG[0] = -1;
for (int i = 1; i < N; ++i)
{
LOG[i] = (i & (i - 1)) ? LOG[i - 1] : LOG[i - 1] + 1;
}
}
void initRMQ()
{
initLOG();
int n = size;
int limit;
for (int i = 0; i < n; ++i)
{
dp[i][0] = height[i];
}
for (int j = 1; j <= LOG[n]; ++j)
{
limit = n - (1 << j);
for (int i = 0; i <= limit; ++i)
{
dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
}
}
}
int LCP(int a, int b)
{
int t;
a = Rank[a];
b = Rank[b];
if (a > b)
{
swap(a, b);
}
--b;
t = LOG[b - a + 1];
return min(dp[a][t], dp[b - (1 << t) + 1][t]);
}
}SA1, SA2;
char buf[52000];
char str[10];
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
SA1.clear();
SA2.clear();
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i)
{
scanf("%s", str);
SA1.insert((int)str[0]);
buf[i] = str[0];
}
buf[n] = ‘\0‘;
for (int i = n - 1; i >= 0; --i)
{
SA2.insert((int)buf[i]);
}
SA1.getsa((int)(‘b‘) + 1);
SA2.getsa((int)(‘b‘) + 1);
SA1.getheight();
SA2.getheight();
SA1.initRMQ();
SA2.initRMQ();
int ans = 1, s, ans_L = 0;
for (int L = 1; L <= n; ++L)
{
for (int i = 0; i < n; ++i)
{
int a = L * i;
int b = L * (i + 1);
if (b >= n)
{
break;
}
int l1 = SA1.LCP(a, b);
int l2 = SA2.LCP(n - a - 1, n - b - 1);
int k = l1 + l2;
if (k > 0)
{
--k;
}
int cnt = k / L + 1;
if (cnt > ans)
{
ans = cnt;
s = a;
ans_L = L;
}
}
}
printf("%d\n", ans);
}
return 0;
}