687. Repeats spoj （后缀数组 重复次数最多的连续重复子串）

687. Repeats

Problem code: REPEATS

 

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 <iostream>
#include <stdio.h>
#include <math.h>
#include <vector>
#include <string.h>
using namespace std;
#define N 50002
char a[N];
int c[N],d[N],e[N],sa[N],height[N],n,b[N],m,dp[N][16];
int cmp(int *r,int a,int b,int l)
{
    return r[a]==r[b]&&r[a+l]==r[b+l];
}
void da()
{
    int i,j,p,*x=c,*y=d,*t;
    memset(b,0,sizeof(b));
    for(i=0; i<n; i++)b[x[i]=a[i]]++;
    for(i=1; i<m; i++)b[i]+=b[i-1];
    for(i=n-1; i>=0; i--)sa[--b[x[i]]]=i;
    for(j=1,p=1; p<n; j*=2,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++)e[i]=x[y[i]];
        for(i=0; i<m; i++)b[i]=0;
        for(i=0; i<n; i++)b[e[i]]++;
        for(i=1; i<m; i++)b[i]+=b[i-1];
        for(i=n-1; i>=0; i--)sa[--b[e[i]]]=y[i];
        for(t=x,x=y,y=t,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++;
    }
}
void callheight()
{
    int i,j,k=0;
    b[0]=0;
    for(i=1; i<n; i++)b[sa[i]]=i;
    for(i=0; i<n-1; height[b[i++]]=k)
        for(k?k--:0,j=sa[b[i]-1]; a[i+k]==a[j+k]; k++);
}
int fun(int i,int j)
{
    i=b[i];
    j=b[j];
    if(i>j)swap(i,j);
    i++;
    int k=(int)(log(j-i+1.0)/log (2.0));
    return min(dp[i][k],dp[j-(1<<k)+1][k]);
}
void initrmq()
{
    int i,j;
    memset(dp,0,sizeof(dp));
    for(i=0; i<=n; i++)
        dp[i][0]=height[i];
    for(j=1; (1<<j)<=n; j++)
        for(i=0; i+(1<<j)<=n; i++)
            dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
int main()
{
    int t,i,j,r;
    scanf("%d",&t);
    for(r=0; r<t; r++)
    {
        scanf("%d",&n);
        for(i=0; i<n; i++)a[i]=getchar(),a[i]=getchar();
        a[n++]='\0';
        m=100;
        da();
        callheight();
        initrmq();
        int max=1;
        for(i=1; i<n/2; i++)
        {
            for(j=0; j+i<n; j+=i)
            {
                int k=fun(j,j+i);
                int kk=k/i+1;
                int tt=i-k%i;
                tt=j-tt;
                if (tt>=0&&k%i!=0)
                    if(fun(tt,tt+i)>=k)
                        kk++;
                if(max<kk)
                {
                    max=kk;
                }
            }
        }
        printf("%d\n",max);
    }
}