SPOJ 题目 8222 NSUBSTR - Substrings（后缀自动机+DP求子串出现最大次数）

NSUBSTR - Substrings

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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
题目大意:问这个字符串长度为1~len的子串出现的最大次数分别是多少
ac代码



#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
#define N 510005
struct sam
{
sam *pre,*son[26];
int len,g;
}que
,*root,*tail,*b
;
int tot;
void add(int c,int l)
{
sam *p=tail,*np=&que[tot++];
np->len=l;
tail=np;
while(p&&p->son[c]==NULL)
{
p->son[c]=np;
p=p->pre;
}
if(p==NULL)
np->pre=root;
else
{
sam *q=p->son[c];
if(p->len+1==q->len)
np->pre=q;
else
{
sam *nq=&que[tot++];
*nq=*q;
nq->len=p->len+1;
np->pre=q->pre=nq;
while(p&&p->son[c]==q)
{
p->son[c]=nq;
p=p->pre;
}
}
}
}
char str[N>>1];
int dp[N>>1];
int main()
{
while(scanf("%s",str)!=EOF)
{
int len=strlen(str);
tot=0;
root=tail=&que[tot++];
int i,j;
for(i=0;i<len;i++)
add(str[i]-'a',i+1);
int cnt[N>>1];
memset(cnt,0,sizeof(cnt));
for(i=0;i<tot;i++)
cnt[que[i].len]++;
for(i=1;i<=len;i++)
cnt[i]+=cnt[i-1];
for(i=0;i<tot;i++)
b[--cnt[que[i].len]]=&que[i];
for(i=0;i<len;i++)
{
(root=root->son[str[i]-'a'])->g++;
//root->g++;
}
memset(dp,0,sizeof(dp));
for(i=tot-1;i>=0;i--)
{
dp[b[i]->len]=max(dp[b[i]->len],b[i]->g);
if(b[i]->pre)
b[i]->pre->g+=b[i]->g;
}
for(i=len-1;i>0;i--)
dp[i]=max(dp[i],dp[i+1]);
for(i=1;i<=len;i++)
printf("%d\n",dp[i]);
}
}