poj 2406 kmp求连续重复子串的个数

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 33119		Accepted: 13775

Description

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.
Source

Waterloo local 2002.07.01

#include<stdio.h>
#include<string.h>
#define N 1000005
char str
;
int next
;
void getnext(char *str)
{
int i,n,j;
n=strlen(str);
next[0]=-1; j=-1; i=0;
while(i<n)
{
if(j==-1||str[i]==str[j])
{
i++; j++;
next[i]=j;
}
else
j=next[j];
}
}
int main()
{
int n;
while(scanf("%s",str)!=EOF)
{
if(strcmp(str,".")==0)
break;
n=strlen(str);
getnext(str);
if(n%(n-next
)==0)
printf("%d\n",n/(n-next
));
else
printf("1\n");
}
return 0;
}

用后缀数组写的超时啦。

/*做法比较简单，穷举字符串 S 的长度 k，然后判断是否满足。判断的时候，
先看字符串 L 的长度能否被 k 整除，再看 suffix(1)和 suffix(k+1)的最长公共
前缀是否等于 n-k。在询问最长公共前缀的时候，suffix(1)是固定的，所以 RMQ
问题没有必要做所有的预处理，只需求出 height 数组中的每一个数到
height[rank[1]]之间的最小值即可。整个做法的时间复杂度为 O(n)。*/
#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
#define N 1000005
int t1
,t2
,x
,c
,sa
,s
,rank
,height
,b
;
char str
;
void build_sa(int *s,int n,int m)
{
int *x=t1,*y=t2,i,k;
for(i=0;i<m;i++) c[i]=0;
for(i=0;i<n;i++) c[x[i]=s[i]]++;
for(i=1;i<m;i++) c[i]+=c[i-1];
for(i=n-1;i>=0;i--) sa[--c[x[i]]]=i;
for(k=1;k<=n;k<<=1)
{
int p=0;
for(i=n-k;i<n;i++) y[p++]=i;
for(i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
for(i=0;i<m;i++) c[i]=0;
for(i=0;i<n;i++) c[x[y[i]]]++;
for(i=1;i<m;i++) c[i]+=c[i-1];
for(i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];
swap(x,y);
p=1; x[sa[0]]=0;
for(i=1;i<n;i++)
x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;
if(p>=n)
break;
m=p;
}
}
void getheight(int n)
{
int i,k=0,j;
for(i=0;i<=n;i++)
rank[sa[i]]=i;
for(i=0;i<n;i++)
{
if(k) k--;
j=sa[rank[i]-1];
while(s[j+k]==s[i+k])  k++;
height[rank[i]]=k;
}
}
int main()
{
int n,m,mi,i,k;
while(scanf("%s",str)!=EOF)
{
n=strlen(str);
for(i=0;str[i]!='\0';i++)
s[i]=str[i];
s
=0;
build_sa(s,n+1,200);
getheight(n);
//for(i=1;i<=n;i++)
// printf("high=%d\n",height[i]);
m=rank[0]; mi=height[m];
// printf("m=%d mi=%d\n",m,mi);
for(i=m;i>1;i--)
{
//printf("%d %d\n",i,height[i]);
if(height[i]<mi)
mi=height[i];
// printf("mi=%d\n",mi);
b[i-1]=mi;
}
mi=height[m+1];
for(i=m+1;i<=n;i++)
{
if(height[i]<mi)
mi=height[i];
b[i]=mi;
}
for(k=1;k<=n;k++)
{
m=rank[k];
if(b[m]==n-k)
break;
}
printf("%d\n",n/k);
}
return 0;
}