hello, in this sum question I have been trying to implement a modified dijsktra but fail. I just want to know if modified dijsktra work here.
please don't give further hints
→ Обратите внимание
→ 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 | 162 |
| 2 | adamant | 148 |
| 3 | Um_nik | 146 |
| 4 | Dominater069 | 143 |
| 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: 27.05.2026 03:55:41 (f1).
Десктопная версия, переключиться на мобильную.
При поддержке
I am not sure what you mean by modified Dijkstra,
but I did solve it with Dijkstra.
thanks, i will try more .. modified dijsktra means it has slight modification .. okay sorry if it confuses
Good luck :)
so, earlier I was trying to just use some kind of hash to track the weights. but then I realised, I could just stotre the nodes of shortest path and take 2 consecutive nodes and mark the precomputed hashng of edges with those 2 vertices as true.. then just print indices of vertices which are false after getting all shortest paths..
what did u do
My solution doesn't have hashing actually.
What I did was: do a All-Pairs Shortest Path by using Dijkstra V times.
Then I enumerate through each edge, and see if the shortest path between any pairs of nodes pass that edge.
The whole solution works in O(V3 + E * V2).
The complexity is kind of lousy, but we have 8 minutes to work with.