| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | ecnerwala | 3649 |
| 2 | Benq | 3581 |
| 3 | orzdevinwang | 3570 |
| 4 | Geothermal | 3569 |
| 4 | cnnfls_csy | 3569 |
| 6 | tourist | 3565 |
| 7 | maroonrk | 3531 |
| 8 | Radewoosh | 3521 |
| 9 | Um_nik | 3482 |
| 10 | jiangly | 3468 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | maomao90 | 174 |
| 2 | awoo | 164 |
| 3 | adamant | 161 |
| 4 | TheScrasse | 159 |
| 5 | nor | 158 |
| 6 | maroonrk | 156 |
| 7 | -is-this-fft- | 151 |
| 8 | SecondThread | 147 |
| 9 | orz | 146 |
| 10 | pajenegod | 145 |
I've been trying to solve this question, but I keep on getting a TLE. At first I had an O(V*(E+V)) approach (link). Then I optimized my code so that I store the results of other nodes while calculating the answer of some node. (link). I'm still getting an TLE.
Can anyone please suggest another optimization/approach for this question so that I can get an AC?
Auto comment: topic has been updated by ameyanator (previous revision, new revision, compare).
Urm anyone please?!
Either adding
cout.tie(0)or usingscanf/printfmay work :).UPD: No, it's not any of them, I tried.
Yeh I had tried these and still TLE.
Your solution's complexity seems to be O(V3).
Yes you are right and that is why I'm getting a TLE. :(.
Update Solved the question. All it required was an optimization — Inside our bfs we need to consider only those nodes that the current node stems out to. We can preprocess this or use adjacency list representation.