I wanted to go into details — but it would have been a full 30 minutes. Any suggestion/query is welcome.
I wanted to go into details — but it would have been a full 30 minutes. Any suggestion/query is welcome.
№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 3690 |
2 | jiangly | 3647 |
3 | Benq | 3581 |
4 | orzdevinwang | 3570 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | Radewoosh | 3509 |
8 | ecnerwala | 3486 |
9 | jqdai0815 | 3474 |
10 | gyh20 | 3447 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | maomao90 | 174 |
2 | awoo | 164 |
3 | adamant | 163 |
4 | TheScrasse | 159 |
5 | nor | 157 |
6 | maroonrk | 156 |
7 | -is-this-fft- | 152 |
8 | Petr | 146 |
8 | orz | 146 |
10 | BledDest | 145 |
Название |
---|
I don't think your solution can pass the system test. I think it will be TLE. In the worst case, updating the nodes' information can be O(n); So it's O(q*n)? Did I misunderstand? :D
updating will take O(log(n)) per query. Something like:
So, we are building the data structure for LCA incrementally after each query.
For more info check out "Another easy solution in <O(N logN, O(logN)>" section on TC