poj 2406 Power Strings （字尾陣列）
阿新 • • 發佈：2018-12-24
Power Strings
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 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 |
題目大意：給出一個長度為L的字串，已知字串由某個字串S迴圈R次得到，求R得最大值。
標解是kmp
這題字尾陣列會TLE，我只是用來練習字尾陣列而已。
還是說一下字尾陣列的做法：列舉字串S的長度K，先判斷是否可以整除，如果可以整除，再判斷suffix(1)與suffix(k+1)的最長公共字首是否等於n-k。因為這道題suffix(1)是固定的，所以可以不用rmq，直接預處理出height陣列中的每個數到height[rank[1]]的最小值即可。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 1000003
using namespace std;
int n,m,len,sa[N],rank[N],height[N],xx[N],yy[N],*x,*y;
int mn[N],a[N],b[N],p;
char s[N];
int cmp(int i,int j,int l)
{
return y[i]==y[j]&&(i+l>len?-1:y[i+l])==(j+l>len?-1:y[j+l]);
}
void get_SA()
{
x=xx; y=yy; m=30;
for (int i=1;i<=len;i++) b[x[i]=a[i]]++;
for (int i=1;i<=m;i++) b[i]+=b[i-1];
for (int i=len;i>=1;i--) sa[b[x[i]]--]=i;
for (int k=1;k<=len;k<<=1){
p=0;
for (int i=len-k+1;i<=len;i++) y[++p]=i;
for (int i=1;i<=len;i++)
if (sa[i]>k) y[++p]=sa[i]-k;
for (int i=1;i<=m;i++) b[i]=0;
for (int i=1;i<=len;i++) b[x[y[i]]]++;
for (int i=1;i<=m;i++) b[i]+=b[i-1];
for (int i=len;i>=1;i--) sa[b[x[y[i]]]--]=y[i];
swap(x,y); p=2; x[sa[1]]=1;
for (int i=2;i<=len;i++)
x[sa[i]]=cmp(sa[i-1],sa[i],k)?p-1:p++;
if (p>len) break;
m=p+1;
}
p=0;
for (int i=1;i<=len;i++) rank[sa[i]]=i;
for (int i=1;i<=len;i++){
if (rank[i]==1) continue;
int j=sa[rank[i]-1];
while (i+p<=len&&j+p<=len&&s[i+p]==s[j+p]) p++;
height[rank[i]]=p;
p=max(0,p-1);
}
}
int main()
{
freopen("a.in","r",stdin);
while (true) {
scanf("%s",s+1);
len=strlen(s+1);
if (s[1]=='.') break;
memset(b,0,sizeof(b));
memset(sa,0,sizeof(sa));
memset(rank,0,sizeof(rank));
memset(height,0,sizeof(height));
for (int i=1;i<=len;i++) a[i]=s[i]-'a'+1;
get_SA();
int pos=rank[1]; mn[pos]=1000000000;
for (int i=pos+1;i<=len;i++) mn[i]=min(mn[i-1],height[i]);
for (int i=pos-1;i>=1;i--) mn[i]=min(mn[i+1],height[i+1]);
int ans=len;
for (int i=1;i<=len;i++){
if (len%i) continue;
if (mn[rank[i+1]]==len-i) {
ans=i;
break;
}
}
printf("%d\n",len/ans);
}
}