# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3831 |
3 | Radewoosh | 3646 |
4 | jqdai0815 | 3620 |
4 | Benq | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | gamegame | 3386 |
10 | ksun48 | 3373 |
# | User | Contrib. |
---|---|---|
1 | cry | 164 |
1 | maomao90 | 164 |
3 | Um_nik | 163 |
4 | atcoder_official | 160 |
5 | -is-this-fft- | 158 |
6 | awoo | 157 |
7 | adamant | 156 |
8 | TheScrasse | 154 |
8 | nor | 154 |
10 | Dominater069 | 153 |
Name |
---|
Consider a graph with 2 * n vertexes, vertex number i corresponds to situation when there are i people that have already been to finals. We have an arc (i, j) in this graph when we can send k of i experienced people in team and after it we have j experienced people (i.e. j = n + i - 2 * k, 0<=k <= i). It's obvious that each infinite path in this graph corresponds to some coach's strategy. One can prove, that one of optimal strategies is repetition of simple cycle in such a graph. That means you should find a cycle with lowest average weight in this graph. You can do it using combination of binary search and Bellman-Ford algorithm for finding cycles of negative weight.
By the way contestants found an O(n^2) solution, by unfortunately nobody have explained me why does it work yet. It would be great if somebody tells us it.