Spoj NSUBSTR - Substrings

题目描述

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|.

输入输出格式

输入格式：

 

String S consists of at most 250000 lowercase latin letters.

 

输出格式：

 

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

 

输入输出样例

输入样例#1： 

ababa

输出样例#1： 

3
2
2
1
1


对于每个节点x，i<=max{x}的f(i)都可以被size[x]更新，所以差分打个标记就可以了hhhh
今天刷后缀自动机真过瘾hhhh

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define maxn 1000005
using namespace std;
int f[maxn],ch[maxn][26];
int l[maxn],siz[maxn],n;
int tag[maxn],cnt=1,pre=1;
int a[maxn],c[maxn];
char s[maxn];

inline void ins(int x){
	int p=pre,np=++cnt; 
	pre=np,l[np]=l[p]+1;
	siz[np]=1;
	
	for(;p&&!ch[p][x];p=f[p]) ch[p][x]=np;
	if(!p) f[np]=1;
	else{
		int q=ch[p][x];
		if(l[q]==l[p]+1) f[np]=q;
		else{
			int nq=++cnt;
			l[nq]=l[p]+1;
			memcpy(ch[nq],ch[q],sizeof(ch[q]));
			f[nq]=f[q];
			f[q]=f[np]=nq;
			for(;ch[p][x]==q;p=f[p]) ch[p][x]=nq;
		}
	}
}

inline void build(){
	for(int i=0;i<n;i++) ins(s[i]-'a');
	for(int i=1;i<=cnt;i++) c[l[i]]++;
	for(int i=n;i>=0;i--) c[i]+=c[i+1];
	for(int i=1;i<=cnt;i++) a[c[l[i]]--]=i;
}

inline void solve(){
	for(int i=1;i<=cnt;i++){
		int now=a[i];
		siz[f[now]]+=siz[now];
		tag[l[now]]=max(tag[l[now]],siz[now]);
	}
	for(int i=n;i;i--) tag[i]=max(tag[i],tag[i+1]);
}

int main(){
	scanf("%s",s);
	n=strlen(s);
	build();
	solve();
	for(int i=1;i<=n;i++) printf("%d\n",tag[i]);
	return 0;
}