8222. Substrings

Problem code: NSUBSTR

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:

给出一个字符串(len <= 2.5e5),求出这个串中,长度为 i (1 <= i <= len) 并且出现次数最多的串,并且输出出现次数.

思路:

根据Spoj1811所提出的这个性质,我们现在SAM上自己匹配自己.得到一个出事的状态.

不难发现.当这个点的后继(parnet关系)出现了之后,这点所表示的字符串也同样的出现了.

那么这个点所表示的字符串也出现了.

那么这个点所表示的字符串出现了多少次呢?

其实是这个点的后缀节点的和.......

然后我们就能够统计这个东西了.

我有个地方没有弄明白,就是,这个点所能够接受的字串有若干个....为什么只用判断这个最长的串呢?

一点初步的想法: 因为par表示这个点的所有的串的最长公共后缀,所以我们能够保证par至少 sz[x] 次,

虽然不知到为什么这个par表示的其他串不用计算,但是似乎在par更新par_par 的时候会计算进去.

根据意识流大法......好像是这么回事....下次再补上吧.

代码也就顺理成章了.

 1 #include<cstdlib>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 using namespace std;
 6 const int maxn = (int)2.55e5, sigma = 26;
 7 char str[maxn];
 8 int cnt[maxn];
 9 int id[maxn];
10 int ans[maxn];
11 int cmp(int,int);
12 struct Sam{
13     int ch[maxn * 2][sigma],par[maxn * 2],stp[maxn * 2];
14     int sz,last;
15     void init(){
16         sz = last = 1;
17         memset(ch,0,sizeof(ch));
18         memset(par,0,sizeof(par));
19         memset(stp,0,sizeof(stp));
20     }
21     void add(int c){
22         stp[++sz] = stp[last] + 1;
23         int p = last, np = sz;
24         for(; !ch[p][c]; p = par[p]) ch[p][c] = np;
25         if(p == 0) par[np] = 1;
26         else{
27             int q = ch[p][c];
28             if(stp[q] != stp[p] + 1){
29                 stp[++sz] = stp[p] + 1;
30                 int nq = sz;
31                 memcpy(ch[nq],ch[q],sizeof(ch[q]));
32                 par[nq] = par[q];
33                 par[q] = par[np] = nq;
34                 for(; ch[p][c] == q; p = par[p]) ch[p][c] = nq;
35             }
36             else par[np] = q;
37         }
38         last = np;
39     }
40     void ins(char *pt){
41         int i;
42         init();
43         for(i = 0; pt[i]; ++i) add(pt[i] - ‘a‘);
44     }
45     void solve(char *pt){
46         memset(id,0,sizeof(id)); memset(cnt,0,sizeof(cnt)); memset(ans,0,sizeof(ans));
47         int i,p = 1;
48         for(i = 1; i <= sz; ++i) id[i] = i;
49         sort(id+1,id+sz+1,cmp);
50         for(i = 0; pt[i]; ++i)
51             ++cnt[p = ch[p][pt[i] - ‘a‘]];
52         for(i = 1; i <= sz; ++i){
53             ans[stp[id[i]]] = max(ans[stp[id[i]]], cnt[id[i]]);
54             if(par[id[i]] != 0) cnt[par[id[i]]] += cnt[id[i]];
55         }
56         for(i = 0; pt[i]; ++i) printf("%d\n",ans[i + 1]);
57     }
58 }sam;
59 int cmp(int x,int y){
60     return sam.stp[x] > sam.stp[y];
61 }
62 int main()
63 {
64     freopen("substr.in","r",stdin);
65     freopen("substr.out","w",stdout);
66     while(scanf("%s",str) != EOF){
67         sam.ins(str);
68         sam.solve(str);
69     }
70     return 0;
71 }