# | User | Rating |
---|---|---|
1 | jiangly | 3846 |
2 | tourist | 3799 |
3 | orzdevinwang | 3706 |
4 | jqdai0815 | 3682 |
5 | ksun48 | 3590 |
6 | Ormlis | 3533 |
7 | Benq | 3468 |
8 | Radewoosh | 3463 |
9 | ecnerwala | 3451 |
9 | Um_nik | 3451 |
# | User | Contrib. |
---|---|---|
1 | cry | 166 |
2 | -is-this-fft- | 161 |
3 | Qingyu | 160 |
3 | Dominater069 | 160 |
5 | atcoder_official | 158 |
6 | adamant | 155 |
7 | Um_nik | 152 |
8 | djm03178 | 151 |
8 | luogu_official | 151 |
10 | awoo | 148 |
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.