UCV2013B — Alice in Amsterdam, I mean Wonderland
This is a graph problem from spoj
Can anyone tell me how to solve it??
Link :https://www.spoj.com/problems/UCV2013B/
Thanks in advance
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | Kevin114514 | 3603 |
| 4 | jiangly | 3583 |
| 5 | turmax | 3559 |
| 6 | tourist | 3541 |
| 7 | strapple | 3515 |
| 8 | ksun48 | 3461 |
| 9 | dXqwq | 3436 |
| 10 | Otomachi_Una | 3413 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | Qingyu | 157 |
| 2 | adamant | 153 |
| 3 | Um_nik | 147 |
| 4 | Proof_by_QED | 146 |
| 5 | Dominater069 | 145 |
| 6 | errorgorn | 142 |
| 7 | cry | 139 |
| 8 | YuukiS | 135 |
| 9 | TheScrasse | 134 |
| 10 | chromate00 | 133 |
UCV2013B — Alice in Amsterdam, I mean Wonderland
This is a graph problem from spoj
Can anyone tell me how to solve it??
Link :https://www.spoj.com/problems/UCV2013B/
Thanks in advance
| Название |
|---|



Solve using Floyd-Warshall algorithm with a special case of containing a negative cycle.
https://cp-algorithms.com/graph/all-pair-shortest-path-floyd-warshall.html
If there exists a negative cycle, running the algorithm two times will produce lower distance than running it once (or as the blog says, the distance from a vertex to itself will be negative).
I advise you read the algorithm first, then handle the case with negative cycles.