spoj8222

Posted 2020-10-06 weeping

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题目：

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

 

 思路：

　　先求出每个状态的endpos集合大小，用cnt记录，用cnt[s]去更新f[len[s]],再用f[i+1]去更新f[i].

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 char ss[300004];
 6 int ans,f[300001<<1];
 7 
 8 struct SAM
 9 {
10     static const int MAXN = 300001<<1;//大小为字符串长度两倍
11     static const int LetterSize = 26;
12 
13     int tot, last, ch[MAXN][LetterSize], fa[MAXN], len[MAXN];
14     int sum[MAXN], tp[MAXN], cnt[MAXN]; //sum,tp用于拓扑排序，tp为排序后的数组
15 
16     void init( void)
17     {
18         last = tot = 1;
19         len[1] = 0;
20         memset(ch,0,sizeof ch);
21         memset(fa,0,sizeof fa);
22         memset(cnt,0,sizeof cnt);
23     }
24 
25     void add( int x)
26     {
27         int p = last, np = last = ++tot;
28         len[np] = len[p] + 1, cnt[last] = 1;
29         while( p && !ch[p][x]) ch[p][x] = np, p = fa[p];
30         if( p == 0)
31             fa[np] = 1;
32         else
33         {
34             int q = ch[p][x];
35             if( len[q] == len[p] + 1)
36                 fa[np] = q;
37             else
38             {
39                 int nq = ++tot;
40                 memcpy( ch[nq], ch[q], sizeof ch[q]);
41                 len[nq] = len[p] + 1, fa[nq] = fa[q], fa[q] = fa[np] = nq;
42                 while( p && ch[p][x] == q)  ch[p][x] = nq, p = fa[p];
43             }
44         }
45     }
46 
47     void toposort( void)
48     {
49         for(int i = 1; i <= len[last]; i++)   sum[i] = 0;
50         for(int i = 1; i <= tot; i++)   sum[len[i]]++;
51         for(int i = 1; i <= len[last]; i++)   sum[i] += sum[i-1];
52         for(int i = 1; i <= tot; i++)   tp[sum[len[i]]--] = i;
53         for(int i = tot; i; i--)   cnt[fa[tp[i]]] += cnt[tp[i]];
54     }
55 } sam;
56 
57 
58 int main(void)
59 {
60     //freopen("in.acm","r",stdin);
61     sam.init();
62     scanf("%s",ss);
63     for(int i=0,len=strlen(ss);i<len;i++)  sam.add(ss[i]-‘a‘);
64     sam.toposort();
65     for(int i=1;i<=sam.tot;i++)    f[sam.len[i]]=max(f[sam.len[i]],sam.cnt[i]);
66     for(int i=sam.len[sam.last];i;i--) f[i]=max(f[i],f[i+1]);
67     for(int i=1;i<=sam.len[sam.last];i++)   printf("%d\n",f[i]);
68     return 0;
69 }