| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3382 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 155 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
| 4 | AmShZ | 145 |
| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 134 |
| 9 | maroonrk | 132 |
| 10 | cry | 130 |
| Name |
|---|



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.