[SPOJ8222]Substrings

[SPOJ8222]Substrings

试题描述

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

String S consists of at most 250000 lowercase latin letters.

输出

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

输入示例

ababa

输出示例

数据规模及约定

见“输入”

题解

构造后缀自动机，然后用每个节点 i 的 right_i 集合大小更新 f[Max[i]]，然后再用每个 f[i] 更新 f[i-1]。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
	return x * f;
}

#define maxn 500010
#define maxa 26

int n;
int ToT, rt, last, ch[maxn][maxa], par[maxn], Max[maxn], siz[maxn];
void extend(int x) {
	int p = last, np = ++ToT; Max[np] = Max[p] + 1; siz[np] = 1;
	last = np;
	while(p && !ch[p][x]) ch[p][x] = np, p = par[p];
	if(!p){ par[np] = rt; return ; }
	int q = ch[p][x];
	if(Max[q] == Max[p] + 1){ par[np] = q; return ; }
	int nq = ++ToT; Max[nq] = Max[p] + 1;
	memcpy(ch[nq], ch[q], sizeof(ch[q]));
	par[nq] = par[q];
	par[q] = nq;
	par[np] = nq;
	while(p && ch[p][x] == q) ch[p][x] = nq, p = par[p];
	return ;
}

int f[maxn];
char Str[maxn];
int sa[maxn], Ws[maxn];
void build() {
	for(int i = 1; i <= ToT; i++) Ws[n-Max[i]]++;
	for(int i = 1; i <= n; i++) Ws[i] += Ws[i-1];
	for(int i = ToT; i; i--) sa[Ws[n-Max[i]]--] = i;
	for(int i = 1; i <= ToT; i++) siz[par[sa[i]]] += siz[sa[i]];
	return ;
}

int main() {
	scanf("%s", Str); n = strlen(Str);
	rt = last = ToT = 1;
	for(int i = 0; i < n; i++) extend(Str[i] - 'a');

	build();
	for(int i = 1; i <= ToT; i++) f[Max[i]] = max(f[Max[i]], siz[i]);
	for(int i = n; i > 1; i--) f[i-1] = max(f[i], f[i-1]);

	for(int i = 1; i <= n; i++) printf("%d\n", f[i]);

	return 0;
}