CF1207G.Indie Album
普及/提高-
通过率:0%
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题目描述
Mishka's favourite experimental indie band has recently dropped a new album! Songs of that album share one gimmick. Each name si is one of the following types:
- 1 c — a single lowercase Latin letter;
- 2 j c — name sj ( 1≤j<i ) with a single lowercase Latin letter appended to its end.
Songs are numbered from 1 to n . It's guaranteed that the first song is always of type 1 .
Vova is rather interested in the new album but he really doesn't have the time to listen to it entirely. Thus he asks Mishka some questions about it to determine if some song is worth listening to. Questions have the following format:
- i t — count the number of occurrences of string t in si (the name of the i -th song of the album) as a continuous substring, t consists only of lowercase Latin letters.
Mishka doesn't question the purpose of that information, yet he struggles to provide it. Can you please help Mishka answer all Vova's questions?
输入格式
The first line contains a single integer n ( 1≤n≤4⋅105 ) — the number of songs in the album.
Each of the next n lines contains the desciption of the i -th song of the album in the following format:
- 1 c — si is a single lowercase Latin letter;
- 2 j c — si is the name sj ( 1≤j<i ) with a single lowercase Latin letter appended to its end.
The next line contains a single integer m ( 1≤m≤4⋅105 ) — the number of Vova's questions.
Each of the next m lines contains the desciption of the j -th Vova's question in the following format:
- i t ( 1≤i≤n , 1≤∣t∣≤4⋅105 ) — count the number of occurrences of string t in si (the name of the i -th song of the album) as a continuous substring, t consists only of lowercase Latin letters.
It's guaranteed that the total length of question strings t doesn't exceed 4⋅105 .
输出格式
For each question print a single integer — the number of occurrences of the question string t in the name of the i -th song of the album as a continuous substring.
输入输出样例
输入#1
20 1 d 2 1 a 2 2 d 2 3 a 2 4 d 2 5 a 2 6 d 2 7 a 1 d 2 9 o 2 10 k 2 11 i 2 12 d 2 13 o 2 14 k 2 15 i 2 1 o 2 17 k 2 18 i 2 15 i 12 8 da 8 dada 8 ada 6 dada 3 dada 19 doki 19 ok 16 doki 15 doki 9 d 1 a 20 doki
输出#1
4 3 3 2 0 1 1 2 1 1 0 2
说明/提示
Song names of the first example:
- d
- da
- dad
- dada
- dadad
- dadada
- dadadad
- dadadada
- d
- do
- dok
- doki
- dokid
- dokido
- dokidok
- dokidoki
- do
- dok
- doki
- dokidoki
Thus the occurrences for each question string are:
- string "da" starts in positions [1,3,5,7] in the name "dadadada";
- string "dada" starts in positions [1,3,5] in the name "dadadada";
- string "ada" starts in positions [2,4,6] in the name "dadadada";
- string "dada" starts in positions [1,3] in the name "dadada";
- no occurrences of string "dada" in the name "dad";
- string "doki" starts in position [1] in the name "doki";
- string "ok" starts in position [2] in the name "doki";
- string "doki" starts in positions [1,5] in the name "dokidoki";
- string "doki" starts in position [1] in the name "dokidok";
- string "d" starts in position [1] in the name "d";
- no occurrences of string "a" in the name "d";
- string "doki" starts in positions [1,5] in the name "dokidoki".