【SPOJ】8222. Substrings（后缀自动机）

http://www.spoj.com/problems/NSUBSTR/

题意：给一个字符串S，令F(x)表示S的所有长度为x的子串中，出现次数的最大值。求F(1)..F(Length(S))

这题做法：

首先建立字符串的后缀自动机。

所以对于一个节点$s$，长度范围为$[Min(s), Max(s)]$，出现次数是$|Right(s)|$，那么我们用$|Right(s)|$去更新$f(Max(s))$，最后用$f(i)$更新$f(i-1)$即可。

 

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }

struct sam {
	static const int N=1000005;
	int c[N][26], l[N], f[N], root, last, cnt;
	sam() { cnt=0; root=last=++cnt; }
	void add(int x) {
		int now=last, a=++cnt; last=a;
		l[a]=l[now]+1;
		for(; now && !c[now][x]; now=f[now]) c[now][x]=a;
		if(!now) { f[a]=root; return; }
		int q=c[now][x];
		if(l[q]==l[now]+1) { f[a]=q; return; }
		int b=++cnt;
		memcpy(c[b], c[q], sizeof c[q]);
		l[b]=l[now]+1;
		f[b]=f[q];
		f[q]=f[a]=b;
		for(; now && c[now][x]==q; now=f[now]) c[now][x]=b;
	}
	void build(char *s) {
		int len=strlen(s);
		rep(i, len) add(s[i]-'a');
	}
}a;

const int N=250005;
char s[N];
int f[N], r[1000005], t[N], b[1000005];
void getans(int len) {
	int *l=a.l, *p=a.f, cnt=a.cnt;
	for(int now=a.root, i=0; i<len; ++i) now=a.c[now][s[i]-'a'], ++r[now];
	for1(i, 1, cnt) ++t[l[i]];
	for1(i, 1, len) t[i]+=t[i-1];
	for1(i, 1, cnt) b[t[l[i]]--]=i;
	for3(i, cnt, 1) r[p[b[i]]]+=r[b[i]];
	for1(i, 1, cnt) f[l[i]]=max(f[l[i]], r[i]);
	for3(i, len, 1) f[i]=max(f[i+1], f[i]);
	for1(i, 1, len) printf("%d
", f[i]);
}
int main() {
	scanf("%s", s);
	a.build(s);
	getans(strlen(s));
	return 0;
}

　　

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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