I have built a brute force solution to this problem but of course time limit exceeded monster has appeared!
please can anyone explain a solution or just give a hint for this problem, i have read the editorial but i didn't understand what is written.
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | Kevin114514 | 3603 |
| 4 | jiangly | 3583 |
| 5 | turmax | 3559 |
| 6 | tourist | 3541 |
| 7 | strapple | 3515 |
| 8 | ksun48 | 3461 |
| 9 | dXqwq | 3436 |
| 10 | Otomachi_Una | 3413 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | Qingyu | 157 |
| 2 | adamant | 153 |
| 3 | Um_nik | 147 |
| 3 | Proof_by_QED | 147 |
| 5 | Dominater069 | 145 |
| 6 | errorgorn | 142 |
| 7 | cry | 139 |
| 8 | YuukiS | 135 |
| 9 | TheScrasse | 134 |
| 10 | chromate00 | 133 |
I have built a brute force solution to this problem but of course time limit exceeded monster has appeared!
please can anyone explain a solution or just give a hint for this problem, i have read the editorial but i didn't understand what is written.
recently i have been studying graph theory from competitive programming book ,in applications of DFS there was articulation points and bridges . He defines other than visited array a another one here is the statement.
[This algorithm maintains two numbers: dfs_num(u) and dfs_low(u). Here, dfs_num(u) stores the iteration counter when the vertex u is visited for the first time and not just for distinguishing DFS_WHITE versus DFS_GRAY/DFS_BLACK. The other number dfs_low(u) stores the lowest dfs_num reachable from DFS spanning sub tree of u. Initially dfs_low(u) = dfs_num(u) when vertex u is first visited. Then, dfs_low(u) can only be made smaller if there is a cycle (some back edges exist). Note that we do not update dfs_low(u) with back edge (u, v) if v is a direct parent of u. ]
please anyone give an explanation for this part.
| Название |
|---|


