| # | User | Rating |
|---|---|---|
| 1 | tourist | 3985 |
| 2 | jiangly | 3741 |
| 3 | jqdai0815 | 3682 |
| 4 | Benq | 3529 |
| 5 | orzdevinwang | 3526 |
| 6 | ksun48 | 3489 |
| 7 | Radewoosh | 3483 |
| 8 | Kevin114514 | 3443 |
| 9 | ecnerwala | 3392 |
| 9 | Um_nik | 3392 |
| # | User | Contrib. |
|---|---|---|
| 1 | cry | 167 |
| 2 | Um_nik | 163 |
| 3 | maomao90 | 162 |
| 3 | atcoder_official | 162 |
| 5 | adamant | 158 |
| 5 | -is-this-fft- | 158 |
| 7 | awoo | 156 |
| 8 | djm03178 | 155 |
| 9 | TheScrasse | 154 |
| 10 | Dominater069 | 153 |
Let's you have a graph with n nodes and m bi-directional edges. (n, m <= 1e5). How would I get the cycles in this graph in an efficient way using DFS.
BTW, I've linked an image where I drew what I thought is a cycle. If it's not a cycle, please tell me what a cycle actually is.
I was working on a problem, and I managed to reduce it to the following question:
Given two arrays of integers a[0], and k[n], perform the following operation n times.
In the ith operation, find the rightmost element in a that is smaller than k[i],
Add k[i] to the right end of a.
Do 1 & 2 n times
I'm trying to use Segment Trees on this, but I'm not able to get the right construction.
Does anyone have a way to do this efficiently(under n^2)?
