№ | Пользователь | Рейтинг |
---|---|---|
1 | jiangly | 3976 |
2 | tourist | 3815 |
3 | jqdai0815 | 3682 |
4 | ksun48 | 3614 |
5 | orzdevinwang | 3526 |
6 | ecnerwala | 3514 |
7 | Benq | 3482 |
8 | hos.lyric | 3382 |
9 | gamegame | 3374 |
10 | heuristica | 3357 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | cry | 169 |
2 | -is-this-fft- | 165 |
3 | atcoder_official | 161 |
3 | Um_nik | 161 |
5 | djm03178 | 157 |
6 | Dominater069 | 156 |
7 | adamant | 154 |
8 | luogu_official | 152 |
9 | awoo | 151 |
10 | TheScrasse | 147 |
Название |
---|
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.