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题目大意：给你一个串让你求出重复次数最多的连续重复子串的重复次数。

解题思路：论文上给出的解答是：

这还没完，因为经过这两个点的情况还不完备，应还可以假设起点在 [ i*j-i+1, i*j-d]，其中 d = i-L/i (d = i-L%i)其意义为根据已知的匹配长度，可以将起点往前移动的范围，太靠后将不能够构造出比之前更好的解。如果要求出某个最多的连续重复子串的最小字典序子需要枚举所有起点，但如果只是要的到最多的重复次数或者任意最多的连续重复子串，那么只需要枚举i*j-d处的起点即可，因为后面的起点若能够得到最优的结果，那么i*j-d处也一定能得到，且答案一样均为L/i+2。(这样保证了第0倍不是从起点开始的解的情况。)

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 <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 1000000007

#define Read() freopen("autocomplete.in","r",stdin)
#define Write() freopen("autocomplete.out","w",stdout)
#define Cin() ios::sync_with_stdio(false)

using namespace std;


inline int read()
{
    char ch;
    bool flag = false;
    int a = 0;
    while(!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')));
    if(ch != '-')
    {
        a *= 10;
        a += ch - '0';
    }
    else
    {
        flag = true;
    }
    while(((ch = getchar()) >= '0') && (ch <= '9'))
    {
        a *= 10;
        a += ch - '0';
    }
    if(flag)
    {
        a = -a;
    }
    return a;
}
void write(int a)
{
    if(a < 0)
    {
        putchar('-');
        a = -a;
    }
    if(a >= 10)
    {
        write(a / 10);
    }
    putchar(a % 10 + '0');
}

const int maxn = 50050;

int wa[maxn], wb[maxn], wv[maxn], ws1[maxn];
int sa[maxn];

int cmp(int *r, int a, int b, int l)
{
    return r[a] == r[b] && r[a+l] == r[b+l];
}

void da(int *r, int *sa, int n, int m)
{
    int i, j, p, *x = wa, *y = wb;
    for(i = 0; i < m; i++) ws1[i] = 0;
    for(i = 0; i < n; i++) ws1[x[i] = r[i]]++;
    for(i = 1; i < m; i++) ws1[i] += ws1[i-1];
    for(i = n-1; i >= 0; i--) sa[--ws1[x[i]]] = i;

    for(j = 1, p = 1; p < n; j <<= 1, m = p)
    {
        for(p = 0, i = n-j; i < n; i++) y[p++] = i;
        for(i = 0; i < n; i++)
            if(sa[i] >= j) y[p++] = sa[i]-j;
        for(i = 0; i < n; i++) wv[i] = x[y[i]];
        for(i = 0; i < m; i++) ws1[i] = 0;
        for(i = 0; i < n; i++) ws1[wv[i]]++;
        for(i = 1; i < m; i++) ws1[i] += ws1[i-1];
        for(i = n-1; i >= 0; i--) sa[--ws1[wv[i]]] = y[i];
        for(swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
            x[sa[i]] = cmp(y, sa[i-1], sa[i], j)?p-1:p++;
    }
}


int rank[maxn], height[maxn];
void calheight(int *r, int *sa, int n)
{
    int i, j, k = 0;
    for(i = 1; i <= n; i++) rank[sa[i]] = i;
    for(int i = 0; i < n; height[rank[i++]] = k)
        for(k?k--:0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
    return ;
}

int dp[maxn][30];


void RMQ(int len)
{
    for(int i = 1; i <= len; i++)
        dp[i][0] = height[i];
    for(int j = 1; 1<<j <= maxn; j++)
    {
        for(int i = 1; i+(1<<j)-1 <= len; i++)
            dp[i][j] = min(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
    }
}
int lg[maxn];

int querry(int l, int r)
{
    int k = lg[r-l+1];
    return min(dp[l][k], dp[r-(1<<k)+1][k]);
}


void init()
{
    lg[0] = -1;
    for (int i = 1; i < maxn; ++i)
        lg[i] = lg[i>>1] + 1;
}
int seq[maxn];

void Del1(int n)
{
    int Max = 0;
    int pos = 0;
    for(int i = 0; i < n; i++)
    {
        int l = rank[i];
        int r = rank[2*n-i+1];
        if(l > r) swap(l, r);
        int tmp = querry(l+1, r);
        if(tmp*2 > Max)
        {
            Max = tmp*2;
            pos = i;
        }
        l = rank[i];
        r = rank[2*n-i];
        if(l > r) swap(l, r);
        tmp = querry(l+1, r);
        if(tmp*2-1 > Max)
        {
            Max = tmp*2-1;
            pos = i;
        }
    }
    for (int i = pos-Max/2; Max--; ++i)
        putchar(seq[i]);
    puts("");
    return ;
}

void Del(int n)
{
    int Max = 0;
    for(int i = 1; i <= n/2; i++)
    {
        for(int j = 0; i*j+i < n; j++)
        {
            int l = rank[i*j];
            int r = rank[i*j+i];
            if(l > r) swap(l, r);
            int x = querry(l+1, r);
            int y = 0;
            int xp = i - x%i;
            if(xp && j)
            {
                l = rank[i*j-xp];
                r = rank[i*j+i-xp];
                if(l > r) swap(l, r);
                y = querry(l+1, r);
            }
            Max = max(Max, max(x/i, y/i)+1);
        }
    }
    cout<<Max<<endl;
}

char str[10];

int main()
{
    init();
    int T;
    cin >>T;
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i = 0; i < n; i++)
        {
             scanf("%s", str);
             seq[i] = str[0]-'a'+1;
        }
        seq[n] = 0;
        da(seq, sa, n+1, 5);
        calheight(seq, sa, n);
        RMQ(n);
        Del(n);
    }
    return 0;
}

SPOJ 687. Repeats(后缀数组求最长重复子串)

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原文地址：http://blog.csdn.net/xu12110501127/article/details/43153497