SPOJ 8222 Substrings 后缀自动机

You are given a string S which consists of 250000 lowercase latin letters at most. We define F(x) as the maximal number of times that some string with length x appears in S. For example for string 'ababa' F(3) will be 2 because there is a string 'aba' that occurs twice. Your task is to output F(i) for every i so that 1<=i<=|S|.

Input

String S consists of at most 250000 lowercase latin letters.

Output

Output |S| lines. On the i-th line output F(i).

Example

Input:
ababa

Output:
3
2
2
1
1

没什么可说的

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define maxn 500005
using namespace std;
char s[maxn];
int len;
struct data {
    int son[maxn][26],link[maxn],step[maxn],last,cnt,size[maxn];
    int v[maxn],pos[maxn],f[maxn];
    data() {last=cnt=1;}
    void extend(int c) {
        int p=last,np=last=++cnt;step[np]=step[p]+1;size[np]=1;
        while(p&&!son[p][c]) son[p][c]=np,p=link[p];
        if(!p) link[np]=1;
        else  {
            int q=son[p][c];
            if(step[q]==step[p]+1) link[np]=q;
            else {
                int nq=++cnt;
                memcpy(son[nq],son[q],sizeof(son[q]));
                link[nq]=link[q];
                link[q]=link[np]=nq;
                step[nq]=step[p]+1;
                while(p&&son[p][c]==q) son[p][c]=nq,p=link[p];
            }
        }
    }
    void pre() {
        for(int i=1;i<=cnt;i++) v[step[i]]++;
        for(int i=1;i<=cnt;i++) v[i]+=v[i-1];
        for(int i=cnt;i;i--) pos[v[step[i]]--]=i;
        for(int i=cnt;i;i--) size[link[pos[i]]]+=size[pos[i]];
        for(int i=1;i<=cnt;i++) f[step[i]]=max(f[step[i]],size[i]);
        for(int i=len;i;i--) f[i]=max(f[i],f[i+1]);
        for(int i=1;i<=len;i++) printf("%d
",f[i]);
    }
}sam;
int main() {
    scanf("%s",s+1);
    len=strlen(s+1);
    for(int i=1;i<=len;i++) sam.extend(s[i]-'a');
    sam.pre();
    return 0;
}