How can I find all the simple cycles in an undirected graph?
In the picture, there are 3 simple cycle:
1-2-3
1-2-4-5
1-3-2-4-5
How can I print all of them?
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→ Трансляции
→ Лидеры (рейтинг)
| № | Пользователь | Рейтинг |
|---|---|---|
| 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 |
| Страны | Города | Организации | Всё → |
→ Лидеры (вклад)
| № | Пользователь | Вклад |
|---|---|---|
| 1 | cry | 164 |
| 1 | maomao90 | 164 |
| 3 | Um_nik | 163 |
| 4 | atcoder_official | 160 |
| 5 | adamant | 159 |
| 6 | -is-this-fft- | 158 |
| 7 | awoo | 157 |
| 8 | TheScrasse | 154 |
| 8 | Dominater069 | 154 |
| 8 | nor | 154 |
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Auto comment: topic has been updated by Parvej (previous revision, new revision, compare).
Just try to backtracking/bruteforces. Each vertices during the visit will be marked (not go to that vertice again), else unmark it. The complexity will be $$$O(n!)$$$ time
SPyofgame I think time complexity will be O ((N^2) * (N!)) as for each path there will be N^2 computation right? Correct me if I am wrong. Thanks in advance.
I dont think we need $$$O(n^2)$$$ for each $$$O(n!)$$$ path since you can try to implement directly or something similar.
By the way, $$$O(n^2)$$$ is also considerably small compared to $$$O(n!)$$$. So unless the time limit is small, you can try to improve the complexity by changing implementation or even a bit of heuristic — which can provide better complexity and constant.