SPOJREPEATS-Repeats

REPEATS - Repeats

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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

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

题意：给定一个字符串，求重复次数最多的连续重复子串的重复次数

解题思路：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。当枚举到合适的子串长度时，在枚举r[L*i]和r[L*(i+1)]的过程中，必然可以出现r[L*i]在第一个循环节里，而r[L*(i+1)]在第二个循环节里的这种情况，如果此时r[L*i]是第一个循环节的首字符，这样直接用公共前缀k除以i并向下取整就可以得到最后结果。但如果r[L*i]如果不是首字符，这样算完之后结果就有可能偏小，因为r[L*i]前面可能还有少许字符也能看作是第一个循环节里的。先算出从r[L*i]开始，除匹配了k/i个循环节，还剩余了几个字符，剩余的自然是k%i个字符。如果说r[L*i]的前面还有i-k%i个字符完成比配的话，这样就相当于还可以再匹配出一个循环节，所以只要检查一下从r[L*i-(i-k%i)]和r[L*i-(i-k%i)+i]开始是否有i-k%i个字符能够完成匹配即可，也就是检查这两个后缀的最长公共前缀是否比i-k%i大即可

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <stack>#include <queue>#include <vector>#include <map>using namespace std;const int N=200010;const int INF=0x7FFFFFFF;struct Sa{    char s[N];    int rk[2][N],sa[N],h[N],w[N],now,n;    int rmq[N][20],lg[N],bel[N];    bool GetS()    {        scanf("%d",&n);        for(int i=1;i<=n;i++)            scanf("%s",s+i);        return true;    }    void getsa(int z,int &m)    {        int x=now,y=now^=1;        for(int i=1; i<=z; i++) rk[y][i]=n-i+1;        for(int i=1,j=z; i<=n; i++)            if(sa[i]>z) rk[y][++j]=sa[i]-z;        for(int i=1; i<=m; i++) w[i]=0;        for(int i=1; i<=n; i++) w[rk[x][rk[y][i]]]++;        for(int i=1; i<=m; i++) w[i]+=w[i-1];        for(int i=n; i>=1; i--) sa[w[rk[x][rk[y][i]]]--]=rk[y][i];        for(int i=m=1; i<=n; i++)        {            int *a=rk[x]+sa[i],*b=rk[x]+sa[i-1];            rk[y][sa[i]]=*a==*b&&*(a+z)==*(b+z)?m-1:m++;        }    }    void getsa(int m)    {        now=rk[1][0]=sa[0]=s[0]=0;        n=strlen(s+1);        for(int i=1; i<=m; i++) w[i]=0;        for(int i=1; i<=n; i++) w[s[i]]++;        for(int i=1; i<=m; i++) rk[1][i]=rk[1][i-1]+(bool)w[i];        for(int i=1; i<=m; i++) w[i]+=w[i-1];        for(int i=1; i<=n; i++) rk[0][i]=rk[1][s[i]];        for(int i=1; i<=n; i++) sa[w[s[i]]--]=i;        rk[1][n+1]=rk[0][n+1]=0;        for(int x=1,y=rk[1][m]; x<=n&&y<=n; x<<=1) getsa(x,y);        for(int i=1,j=0; i<=n; h[rk[now][i++]]=j?j--:j)        {            if(rk[now][i]==1) continue;            int k=n-max(sa[rk[now][i]-1],i);            while(j<=k&&s[sa[rk[now][i]-1]+j]==s[i+j]) ++j;        }    }    void getrmq()    {        h[n+1]=h[1]=lg[1]=0;        for(int i=2; i<=n; i++)            rmq[i][0]=h[i],lg[i]=lg[i>>1]+1;        for(int i=1; (1<<i)<=n; i++)        {            for(int j=2; j<=n; j++)            {                if(j+(1<<i)>n+1) break;                rmq[j][i]=min(rmq[j][i-1],rmq[j+(1<<i-1)][i-1]);            }        }    }    int lcp(int x,int y)    {        int l=min(rk[now][x],rk[now][y])+1,r=max(rk[now][x],rk[now][y]);        return min(rmq[l][lg[r-l+1]],rmq[r-(1<<lg[r-l+1])+1][lg[r-l+1]]);    }    void work()    {        getsa(300);        getrmq();        int ans=1;        for(int L=1;L<=n;L++)        {            for(int i=1;i+L<=n;i+=L)            {                int R=lcp(i,i+L);                ans=max(ans,R/L+1);                if(i>L-R%L)                    ans=max(lcp(i-L+R%L,i+R%L)/L+1,ans);            }        }        printf("%d\n",ans);    }}sa;int main(){    int t;    scanf("%d",&t);    while(t--)    {        sa.GetS();        sa.work();    }    return 0;}