【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\n", 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