spoj687 后缀数组重复次数最多的连续重复子串

REPEATS - Repeats

no tags

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

题意:

求重复次数最多的连续重复子串。

思路:

可以先枚举重复子串的长度L,然后求长度为 L 的子串最多能连续出现几次。

假设在原字符串中连续出现 2 次，记这个子字符串为 S，那么 S 肯定包括了字符 r[0], r[L], r[L*2],

r[L*3], ……中的某相邻的两个。所以只须看字符 r[L*i]和 r[L*(i+1)]往前和

往后各能匹配到多远，记这个总长度为 K，那么这里连续出现了 K/L+1 次。最后

看最大值是多少。

对于后面的那部分可以根据height[]直接求得。对于前面的那部分

先求后面部分减去长度L的x个子串后，是否还有。如果还有那么这一部分肯定属于前面。

这样L - (后面部分长度)%L就是前面部分的长度tp，然后判断 r[L* i] - tp 和 r[L* i] - tp + L的公共前缀，

如果长度大于等于r[L*i]和r[L*(i+1)]的公共长度，那么前面一定只是有1.

/*

 * Author:  sweat123

 * Created Time:  2016/6/29 15:28:42

 * File Name: main.cpp

 */

#include<set>

#include<map>

#include<queue>

#include<stack>

#include<cmath>

#include<string>

#include<vector>

#include<cstdio>

#include<time.h>

#include<cstring>

#include<iostream>

#include<algorithm>

#define INF 1<<30

#define MOD 1000000007

#define ll long long

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define pi acos(-1.0)

using namespace std;

const int MAXN = ;

int wa[MAXN],wb[MAXN],wc[MAXN],n,height[MAXN],Rank[MAXN],r[MAXN],sa[MAXN];

char s[MAXN];

void da(int *r,int *sa,int n,int m){

    int *x = wa,*y = wb;

    for(int i = ; i < m; i++)wc[i] = ;

    for(int i = ; i < n; i++)wc[x[i] = r[i]] ++;

    for(int i = ; i < m; i++)wc[i] += wc[i-];

    for(int i = n - ; i >= ; i--)sa[--wc[x[i]]] = i;

    for(int k = ,p = ; p < n; m = p,k <<= ){

        p = ;

        for(int i = n - k; i < n; i++)y[p++] = i;

        for(int i = ; i < n; i++)if(sa[i] >= k)y[p++] = sa[i] - k;

        for(int i = ; i < m; i++)wc[i] = ;

        for(int i = ; i < n; i++)wc[x[y[i]]] ++;

        for(int i = ; i < m; i++)wc[i] += wc[i-];

        for(int i = n - ; i >= ; i--)sa[--wc[x[y[i]]]] = y[i];

        swap(x,y);

        p = ;

        x[sa[]] = ;

        for(int i = ; i < n; i++){

            x[sa[i]] = (y[sa[i]] == y[sa[i-]] && y[sa[i]+k] == y[sa[i-]+k])?p-:p++;

        }

    }

}

void calheight(int *r,int *sa,int n){

    for(int i = ; i <= n; i++)Rank[sa[i]] = i;

    int j,k = ;

    for(int i = ; i < n; height[Rank[i++]] = k){

        for(k?k--:,j = sa[Rank[i]-]; r[i+k]==r[j+k]; k++);

    }

}

int dp[MAXN][];

void RMQ(){

    for(int i = ; i <= n; i++){

        dp[i][] = height[i];

    }

    for(int i = ; i < ; i++){

        for(int j = ; j + ( << i) -  <= n; j++){

            dp[j][i] = min(dp[j][i-],dp[(j+(<<(i-)))][i-]);

        }

    }

}

int getnum(int x,int y){

    x = Rank[x];

    y = Rank[y];

    if(x > y)swap(x,y);

    x += ;

    int k = (int)((log(y - x + ) * 1.0) / log(2.0));

    return min(dp[x][k],dp[y - (<<k) + ][k]);

}

void solve(){

    int ans = ;

    for(int i = ; i < n; i++){

        for(int j = ; j + i < n; j += i){

            int ret = getnum(j,j+i);

            int t = ret / i + ;// behind r[i] can repeat t times

            int tp = j - (i - ret % i);

            if(tp >=  && (ret % i != ) && getnum(tp,tp+i) >= ret){

                t ++;

            }

            ans = max(t,ans);

        }

    }

    printf("%d\n",ans);

}

int main(){

    int t;

    scanf("%d",&t);

    while(t--){

        scanf("%d",&n);

        getchar();

        for(int i = ; i < n; i++){

            scanf("%c",&s[i]);

            r[i] = s[i];

            getchar();

        }

        r[n] = ;

        da(r,sa,n+,);

        calheight(r,sa,n);

        RMQ();

        solve();

    }

    return ;

}