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;
}
