Does there exist a way an efficient method to find the longest path in a DAG among all possible paths? I know how to solve this in $$$O(V*(V+E))$$$ by considering each vertex as the source, and then using topo sort, and iterating over the vertices in that order and update the distances accordingly. Any suggestions/help would be appreciated or maybe we can prove there doesn't exist a faster method. Thanks!
→ Обратите внимание
→ Трансляции
→ Лидеры (рейтинг)
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | jiangly | 3631 |
| 4 | Kevin114514 | 3574 |
| 5 | maroonrk | 3521 |
| 6 | strapple | 3515 |
| 7 | Radewoosh | 3461 |
| 8 | tourist | 3428 |
| 9 | turmax | 3378 |
| 10 | Um_nik | 3376 |
| Страны | Города | Организации | Всё → |
→ Лидеры (вклад)
| № | Пользователь | Вклад |
|---|---|---|
| 1 | Qingyu | 161 |
| 2 | adamant | 146 |
| 3 | Um_nik | 145 |
| 4 | Dominater069 | 142 |
| 5 | errorgorn | 140 |
| 6 | cry | 138 |
| 7 | Proof_by_QED | 136 |
| 8 | YuukiS | 135 |
| 9 | chromate00 | 134 |
| 10 | soullless | 133 |
→ Найти пользователя
→ Прямой эфир
Codeforces (c) Copyright 2010-2026 Михаил Мирзаянов
Соревнования по программированию 2.0
Время на сервере: 31.05.2026 08:51:54 (g1).
Десктопная версия, переключиться на мобильную.
При поддержке
Auto comment: topic has been updated by peoplePleaser (previous revision, new revision, compare).
Auto comment: topic has been updated by peoplePleaser (previous revision, new revision, compare).
The algorithm you described should be $$$O(V+E)$$$, because you can just relax each edge once in your topo-sort order.
It's $$$O(V+E)$$$ if we fix the source vertex, right? but we need to find the longest path over all possible paths.
Just initialize all vertices to have value 0 instead of -INF; alternatively, make a super-source that connects to all vertices.
Thanks, I thought of this solution. But how to handle the case when all weights are negative? We will always end up with a value of zero, right?
You can store longest non-empty path, in which case you relax $$$val[v] = \max(val[v], \max(val[u], 0) + e)$$$
Thanks got it!
My bad! I am such a fool, if all weights are negative, the answer is the largest (least negative) weight edge.