POJ - 2406 Power Strings （后缀数组 最大重复次数）

Power Strings

 

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

题目链接：http://poj.org/problem?id=2406

题目大意：给出一个字符串L是一个连续重复串，问最大重复次数是多少

思路：字符串L是由子串S重复R次得到的，枚举S的长度k，判断S是否满足条件。首先判断L的长度是否能被S的长度k整除，再看suffix(1)和suffix(k+1)的LCP是否等于n-k。线性预处理出所有后缀与suffi(1)的LCP

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll long long
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
const int N=2000010;
int wa[N*3],wb[N*3],wv[N*3],wss[N*3];//所有相关数组都要开到3倍
int rak[N*3],h[N*3],a[N*3],sa[N*3];
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++) wss[i]=0;
    for(i=0;i<n;i++) wss[wv[i]]++;
    for(i=1;i<m;i++)wss[i]+=wss[i-1];
    for(int i=n-1;i>=0;i--) b[--wss[wv[i]]]=a[i];
}
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++];
}
void da(int a[],int sa[],int rak[],int h[],int n,int m)
{
	for(int i=n;i<3*n;i++) a[i]=0;
    dc3(a,sa,n+1,m);
	int i,j,k=0;
	for(int i=0;i<=n;i++) rak[sa[i]]=i;
	for(int i=0;i<n;i++)
	{
		if(k) k--;
		int j=sa[rak[i]-1];
		while(a[i+k]==a[j+k])k++;
		h[rak[i]]=k;
	}
}
char s[N];
int b[N];
int main()
{
    while(~scanf("%s",s))
    {
        if(s[0]=='.')break;
        int n=strlen(s);
        for(int i=0;i<n;i++)a[i]=s[i];
        a[n]=0;
        da(a,sa,rak,h,n,127);
        int x=rak[0];
        b[x]=n;
        for(int i=x;i>=2;i--) b[i-1]=min(b[i],h[i]);
        for(int i=x+1;i<=n;i++) b[i]=min(b[i-1],h[i]);
        int ans=0;
        for(int i=1;i<n;i++)
        {
            if(n%i!=0) continue;
            if(b[rak[i]]==n-i)
            {
                ans=n/i;
                break;
            }
        }
        if(ans==0) ans=1;
        printf("%d\n",ans);
    }
    return 0;
}

 

来源：https://blog.csdn.net/chimchim04/article/details/99693282