spoj 687 Repeats//后缀数组

SPOJ Problem Set (classical)

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

 

 

/*这个博客的后缀数组的RMQ写的相当快http://hi.baidu.com/fhnstephen/blog/item/870da9ee3651404379f0555f.html就拿这个当模板好了，仰慕博客主人*/#include<math.h>#include<stdio.h>using namespace std;const int maxn=50010;const int inf=5000000;int w[maxn],wa[maxn],wb[maxn],wv[maxn];int sa[maxn],rank[maxn],height[maxn];int a[maxn],f[maxn][20],n,ft[maxn];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++) w[i]=0; for (i=0; i<n; i++) w[x[i]=r[i]]++; for (i=1; i<m; i++) w[i]+=w[i-1]; for (i=n-1; i>=0; i--) sa[--w[x[i]]]=i; for (p=1,j=1; p<n; m=p,j*=2) { 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<m; i++) w[i]=0; for (i=0; i<n; i++) w[wv[i]=x[y[i]]]++; for (i=1; i<m; i++) w[i]+=w[i-1]; for (i=n-1; i>=0; i--) sa[--w[wv[i]]]=y[i]; for (t=x,x=y,y=t,x[sa[0]]=0,p=1,i=1; i<n; i++) x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++; } return;}void cal(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;}int nmin(int a,int b){ return a<b?a:b;}void rmq(int n){ int i,j; for (i=1; i<=n; i++) f[i][0]=height[i]; for (j=1; j<20; j++) for (i=1; i+(1<<j)-1<=n; i++) f[i][j]=nmin(f[i][j-1],f[i+(1<<j-1)][j-1]); return;}int lcp(int a,int b){ int x=rank[a],y=rank[b]; if (x>y) { int t=x; x=y; y=t; } x++; int t=ft[y-x+1]; return nmin(f[x][t],f[y-(1<<t)+1][t]);}int main(){ //freopen("in.txt","r",stdin); int i,testcase=0; char x; for (i=0; i<maxn; i++) ft[i]=int(double(log(i))/log(2.00)); scanf("%d/n",&testcase); while (testcase--) { scanf("%d/n",&n); for (i=0; i<n; i++) { scanf("%c/n",&x); a[i]=x; } a[n]=0; da(a,sa,n+1,128); cal(a,sa,n); rmq(n); int k,max=0,r=0,t; for(int l=1; l<n; l++)//枚举长度 for(int i=0; i+l<n; i+=l) { k=lcp(i,i+l); r=k/l+1;//注意为什么是k/l+1 t=i-(l-k%l); if (t>=0&&k%l!=0) if (lcp(t,t+l)>=k) r++; if (r>max) max=r; //printf("l=%d r=%d i=%d i+l=%d/n",l,r,i,i+l); } printf("%d/n",max); } return 0;}