Hello everyone!
Having found the SCC of a directed graph G, how can I contract each SCC to a single node?
Thanks.
→ Обратите внимание
До соревнования
Codeforces Round #956 (Div. 2) and ByteRace 2024
38:25:13
Зарегистрироваться »
Codeforces Round #956 (Div. 2) and ByteRace 2024
38:25:13
Зарегистрироваться »
*есть доп. регистрация
→ Трансляции
→ Лидеры (рейтинг)
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 3803 |
| 2 | jiangly | 3707 |
| 3 | Benq | 3627 |
| 4 | ecnerwala | 3584 |
| 5 | orzdevinwang | 3573 |
| 6 | Geothermal | 3569 |
| 6 | cnnfls_csy | 3569 |
| 8 | Radewoosh | 3542 |
| 9 | jqdai0815 | 3532 |
| 10 | gyh20 | 3447 |
| Страны | Города | Организации | Всё → |
→ Лидеры (вклад)
| № | Пользователь | Вклад |
|---|---|---|
| 1 | awoo | 163 |
| 2 | adamant | 161 |
| 2 | maomao90 | 161 |
| 4 | maroonrk | 152 |
| 5 | -is-this-fft- | 151 |
| 6 | nor | 149 |
| 7 | SecondThread | 147 |
| 7 | atcoder_official | 147 |
| 9 | TheScrasse | 146 |
| 10 | Petr | 144 |
→ Найти пользователя
→ Прямой эфир
Codeforces (c) Copyright 2010-2024 Михаил Мирзаянов
Соревнования по программированию 2.0
Время на сервере: 06.07.2024 03:09:48 (h2).
Десктопная версия, переключиться на мобильную.
При поддержке
Suppose you know C[v] — which component contains node v.
Now you scan through edges of the original graph, let current one be x->y.
Then you should add an edge C[x]->C[y] in compressed graph if C[x] is not equal to C[y].
Yes, and you should also remember to mark which edges have already been added so you don't double count any edge. You can use a boolean array or an edge set
What's so bad about keeping a multi-edge in DAG? :P
If we have to do some traversal in the compressed graph we are going to waste some operations while scanning the adjacency list of each vertex :P
I'd rather use disjoint sets. It doesn't require too much memory and I think it's faster than an edge set :D and are VERY easy to code
How can i use disjoint sets for this purpose? I have never seen this application, i usually use DSU for Kruskal's only :P
Nvm, bullsh*t in rev1
Awesome! It's really simple, you are right, this is the easiest/most efficient option.
But if you have edges X->Y, Y->Z, X->Z, won't you miss one of them this way?
omg you are right, I remember I used disjoint sets but it wasn't this way. Nvm thanks!
Other valid option would be using a hash function so you can check in O(1) whether the edge exists or not! It should be easy to come up with a function that wont generate collisions since amount of edges is usually small :)
EDIT : never used unordered set, but this may actually be the implementation of what i said above
UPD: I am sorry, there is nothing correct here :(
Won't it be more efficient supposing there won't be collisions? I'm assuming you will use map data structure to associate integers to each edge
I am sorry, I have no idea what I was thinking at that moment but I suppose it was something wrong :(