Substrings(SPOJ8222) (sam(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

题意：

给定一个字符串\(S\)，令\(F(x)\)表示\(S\)中所有长度为\(x\)的子串出现最多的次数，

题解：

把串塞进一个后缀自动机，在\(parent\)树上反向拓扑，求出每个长度时的最大值。

#include<bits/stdc++.h>
using namespace std;
const int N=2000010;
char s[N];
int a[N],c[N],ans[N];
struct SAM{
	int last,cnt;
	int size[N],ch[N][26],fa[N<<1],l[N<<1];
	void ins(int c){
		int p=last,np=++cnt;last=np;l[np]=l[p]+1;
		for(;p&&!ch[p][c];p=fa[p])ch[p][c]=np;
		if(!p)fa[np]=1;
		else{
			int q=ch[p][c];
			if(l[p]+1==l[q])fa[np]=q;
			else{
				int nq=++cnt;l[nq]=l[p]+1;
				memcpy(ch[nq],ch[q],sizeof ch[q]);
				fa[nq]=fa[q];fa[q]=fa[np]=nq;
				for(;ch[p][c]==q;p=fa[p])ch[p][c]=nq;
			}
		}
		size[np]=1;
	}
	void build(char s[]){
		int len=strlen(s+1);
		last=cnt=1;
		for(int i=1;i<=len;++i)ins(s[i]-'a');
	}
	void calc(int len){
		for(int i=1;i<=cnt;++i)c[l[i]]++;
		for(int i=1;i<=cnt;++i)c[i]+=c[i-1];
		for(int i=1;i<=cnt;++i)a[c[l[i]]--]=i;
		for(int i=cnt;i;--i){
			int p=a[i];
			size[fa[p]]+=size[p];
			ans[l[p]]=max(ans[l[p]],size[p]);
		}
		for(int i=1;i<=len;++i){
			printf("%d\n",ans[i]);
		}
	}
}sam;
int main(){
	cin>>s+1;
	sam.build(s);
	sam.calc(strlen(s+1));
}