spoj687. Repeats

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:117babbabaabaabaababOutput:4

后缀数组先穷举长度L，然后求长度为L 的子串最多能连续出现几次。首先连续出现1 次是肯定可以的，所以这里只考虑至少2 次的情况。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。如图7 所示。穷举长度L 的时间是n，每次计算的时间是n/L。所以整个做法的时间复杂度是O(n/1+n/2+n/3+……+n/n)=O(nlogn)。

因为是隔段枚举，所以，有可能在k%l!=0的时候，这样，会有可能，向前移几格，可能刚好能多加一个，所以，再加一个判断就可以了！

#include <iostream>#include <stdio.h>#include <math.h>#include <string.h>using namespace std;#define maxn 200500int wa[maxn],wb[maxn],ws[maxn],wv[maxn],wsd[maxn],r[maxn],ans[maxn];char str[maxn];#define M 100010int dp[M][18];#define F(x) ((x)/3+((x)%3==1?0:tb))#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)int c0(int *r,int a,int b){return r[a]==r[b]&&r[a+1]==r[b+1]&&r[a+2]==r[b+2];}int c12(int k,int *r,int a,int b){if(k==2) return r[a]<r[b]||r[a]==r[b]&&c12(1,r,a+1,b+1);else return r[a]<r[b]||r[a]==r[b]&&wv[a+1]<wv[b+1];}void sort(int *r,int *a,int *b,int n,int m){    int i;    for(i=0;i<n;i++) wv[i]=r[a[i]];    for(i=0;i<m;i++) wsd[i]=0;    for(i=0;i<n;i++) wsd[wv[i]]++;    for(i=1;i<m;i++) wsd[i]+=wsd[i-1];    for(i=n-1;i>=0;i--) b[--wsd[wv[i]]]=a[i];    return;}void dc3(int *r,int *sa,int n,int m){    int i,j,*rn=r+n,*san=sa+n,ta=0,tb=(n+1)/3,tbc=0,p;    r[n]=r[n+1]=0;    for(i=0;i<n;i++) if(i%3!=0) wa[tbc++]=i;    sort(r+2,wa,wb,tbc,m);    sort(r+1,wb,wa,tbc,m);    sort(r,wa,wb,tbc,m);    for(p=1,rn[F(wb[0])]=0,i=1;i<tbc;i++)        rn[F(wb[i])]=c0(r,wb[i-1],wb[i])?p-1:p++;    if(p<tbc) dc3(rn,san,tbc,p);    else for(i=0;i<tbc;i++) san[rn[i]]=i;    for(i=0;i<tbc;i++) if(san[i]<tb) wb[ta++]=san[i]*3;    if(n%3==1) wb[ta++]=n-1;    sort(r,wb,wa,ta,m);    for(i=0;i<tbc;i++) wv[wb[i]=G(san[i])]=i;    for(i=0,j=0,p=0;i<ta && j<tbc;p++)        sa[p]=c12(wb[j]%3,r,wa[i],wb[j])?wa[i++]:wb[j++];    for(;i<ta;p++) sa[p]=wa[i++];    for(;j<tbc;p++) sa[p]=wb[j++];    return;}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,*t;    for(i=0;i<m;i++) wsd[i]=0;    for(i=0;i<n;i++) wsd[x[i]=r[i]]++;    for(i=1;i<m;i++) wsd[i]+=wsd[i-1];    for(i=n-1;i>=0;i--) sa[--wsd[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++) wv[i]=x[y[i]];        //技牢快快        for(i=0;i<m;i++) wsd[i]=0;        for(i=0;i<n;i++) wsd[wv[i]]++;        for(i=1;i<m;i++) wsd[i]+=wsd[i-1];        for(i=n-1;i>=0;i--) sa[--wsd[wv[i]]]=y[i];        //技牢快快        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++;    }    return;}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(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;}void makermq(int n,int b[]){    int i,j;    for(i=1;i<=n;i++)        dp[i][0]=b[i];    for(j=1;(1<<j)<=n;j++)        for(i=1;i+(1<<j)-1<=n;i++)            dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);}int rmq(int s,int v){    s=rank[s],v=rank[v];    if(s>v)swap(s,v);    s++;    int k=(int)(log((v-s+1)*1.0)/log(2.0));    return min(dp[s][k],dp[v-(1<<k)+1][k]);}int solve(int n){    int i,k,tmp,j,maxx=1,l,t;    for(l=1;l<n;l++)    for(i=0;i+l<n;i=i+l){        k=rmq(i,i+l);        t=k/l+1;        tmp=i-(l-k%l);        if(tmp>=0&&(k%l!=0)&&rmq(tmp,tmp+l)>=k)        t++;        if(t>maxx){            maxx=t;        }    }    printf("%d\n",maxx);}int main(){    int n,n1,i,tcase;    scanf("%d",&tcase);    while(tcase--){        scanf("%d",&n);        getchar();        char c;        for(i=0;i<n;i++){            c=getchar();            getchar();            r[i]=(int)c+1;        }        int l=n;        r[n]=0;        int m=300;        //da(r,ans,n+1,m);        dc3(r,ans,n+1,m);        calheight(r,ans,n);        makermq(n,height);        int maxx=solve(n);    }    return 0;}