Hello everyone!
Having found the SCC of a directed graph G, how can I contract each SCC to a single node?
Thanks.
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | cry | 164 |
1 | maomao90 | 164 |
3 | Um_nik | 163 |
4 | atcoder_official | 160 |
4 | adamant | 160 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
8 | Dominater069 | 154 |
8 | nor | 154 |
Hello everyone!
Having found the SCC of a directed graph G, how can I contract each SCC to a single node?
Thanks.
Name |
---|
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 :(