SPOJ - REPEATS 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

since a (4, 3)-repeat is found starting at the 5th character of the input string.

题意：

求重复次数最多的连续重复子串出现的次数

思路：

罗穗骞论文和其他博主已经说的比较清楚了，不在赘述。

在此解释一下向左匹配是什么意思。

将i，j同时向左移动,如果s[i]仍然等于s[j]，则匹配成功。

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>

#define fuck(x) cerr<<#x<<" = "<<x<<endl;
#define debug(a, x) cerr<<#a<<"["<<x<<"] = "<<a[x]<<endl;
#define ls (t<<1)
#define rs ((t<<1)|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = ;
const int maxm = ;
const int inf = 0x3f3f3f3f;
const ll Inf = ;
const int mod = ;
const double eps = 1e-;
const double pi = acos(-);

char s[maxn];
int len, Rank[maxn], sa[maxn], tlen, tmp[maxn];
bool compare_sa(int i, int j) {
    if (Rank[i] != Rank[j]) { return Rank[i] < Rank[j]; }
    //如果以i开始,长度为k的字符串的长度,已经超出了字符串尾,那么就赋值为-1
    //这是因为,在前面所有数据相同的情况下,字符串短的字典序小.
    int ri = i + tlen <= len ? Rank[i + tlen] : -inf;
    int rj = j + tlen <= len ? Rank[j + tlen] : -inf;
    return ri < rj;
}
void construct_sa() {
    //初始的RANK为字符的ASCII码
    for (int i = ; i <= len; i++) {
        sa[i] = i;
        Rank[i] = i < len ? s[i] : -inf;
    }
    for (tlen = ; tlen <= len; tlen *= ) {
        sort(sa, sa + len + , compare_sa);
        tmp[sa[]] = ;
        //全新版本的RANK,tmp用来计算新的rank
        //将字典序最小的后缀rank计为0
        //sa之中表示的后缀都是有序的,所以将下一个后缀与前一个后缀比较,如果大于前一个后缀,rank就比前一个加一.
        //否则就和前一个相等.
        for (int i = ; i <= len; i++) {
            tmp[sa[i]] = tmp[sa[i - ]] + (compare_sa(sa[i - ], sa[i]) ?  : );
        }
        for (int i = ; i <= len; i++) {
            Rank[i] = tmp[i];

        }
    }
}
int height[maxn];
void construct_lcp() {
//    for(int i=0;i<=n;i++){Rank[sa[i]]=i;}
    int h = ;
    height[] = ;
    for (int i = ; i < len; i++) {//i为后缀数组起始位置
        int j = sa[Rank[i] - ];//获取当前后缀的前一个后缀(排序后)
        if (h > )h--;
        for (; j + h < len && i + h < len; h++) {
            if (s[j + h] != s[i + h])break;
        }
        height[Rank[i]] = h;
    }
}

int st[maxn][];
void rmq_init(){
    for(int i=;i<=len;i++){
        st[i][]=height[i];
    }
    int l=;
    for(int i=;l<=len;i++){
        for(int j=;j+l/<=len;j++){
            st[j][i]=min(st[j][i-],st[j+l/][i-]);
        }
        l<<=;
    }
}
int ask_min(int i,int j){
    int k=int(log(j-i+1.0)/log(2.0));
    return min(st[i][k],st[j-(<<k)+][k]);
}
int lcp(int a,int b)
{
    a=Rank[a],b=Rank[b];
    if(a>b)
        swap(a,b);
    return ask_min(a+,b);
}

int main() {
//    ios::sync_with_stdio(false);
//    freopen("in.txt", "r", stdin);

    int T;
    scanf("%d",&T);
    while (T--){
        scanf("%d",&len);
        for(int i=;i<len;i++){
            scanf("%s",s+i);
        }
        construct_sa();
        construct_lcp();
        rmq_init();

        int ans=,ans1=;
        for(int i=;i<=len;i++){//i是长度
            for(int j=;j+i<len;j+=i){
                ans=lcp(j,j+i);
                int k=j-(i-ans%i);
                ans=ans/i+;
                int p=ans-;
                if(k>=&&lcp(k,k+i)>=i){ans++;}
                ans1=max(ans,ans1);
                j+=p*i;
            }
        }
        printf("%d\n",ans1);
    }

    return ;
}