| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 4009 |
| 2 | jiangly | 3823 |
| 3 | Benq | 3738 |
| 4 | Radewoosh | 3633 |
| 5 | jqdai0815 | 3620 |
| 6 | orzdevinwang | 3529 |
| 7 | ecnerwala | 3446 |
| 8 | Um_nik | 3396 |
| 9 | ksun48 | 3390 |
| 10 | gamegame | 3386 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | cry | 167 |
| 2 | Um_nik | 163 |
| 2 | maomao90 | 163 |
| 4 | atcoder_official | 161 |
| 5 | adamant | 159 |
| 6 | -is-this-fft- | 158 |
| 7 | awoo | 157 |
| 8 | TheScrasse | 154 |
| 9 | nor | 153 |
| 9 | Dominater069 | 153 |
The editorial for D1 ( Easy version ) , I am facing difficulties understanding the editorial solution ...
Problem LINK : here
Note that because we guarantee that the root is a query, when we are computing the answer for any node 'v' in this DFS, we can assume that either 'v' or some vertex not in the subtree of 'v' has already been queried.
can someone elaborate little ?
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