How to solve this kattis problem? It is a multi-source but not once, one by one. You should count the number of vertices with the distance at least k after each source is added. Restarting dijkstra every time is not enough.
Thanks.
№ | 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 → |
№ | Пользователь | Вклад |
---|---|---|
1 | cry | 164 |
1 | maomao90 | 164 |
3 | Um_nik | 163 |
4 | atcoder_official | 160 |
5 | -is-this-fft- | 158 |
6 | adamant | 157 |
6 | awoo | 157 |
8 | TheScrasse | 154 |
8 | nor | 154 |
10 | djm03178 | 153 |
How to solve this kattis problem? It is a multi-source but not once, one by one. You should count the number of vertices with the distance at least k after each source is added. Restarting dijkstra every time is not enough.
Thanks.
Название |
---|
Take advantage of K being small. Maybe incorporate into state somehow?
Define nodes based on two parameters (location, distance from alien base). Run a modified Dijkstra where you find the minimum input id of an alien base that can reach each node. This works because if two alien bases can both reach a town with the same distance D, then for all other nodes reachable from this node, the base with smaller id will always get there earlier.
Srry, I don't understand your solution. Can you describe little bit more about it? Or provide code if you have written it. Thanks.
https://ideone.com/ER1jNn
Thanks, I got it. By the way, isn't there more standard (closer to standard dijkstra) solution?