# | User | Rating |
---|---|---|
1 | jiangly | 3640 |
2 | Benq | 3593 |
3 | tourist | 3572 |
4 | orzdevinwang | 3561 |
5 | cnnfls_csy | 3539 |
6 | ecnerwala | 3534 |
7 | Radewoosh | 3532 |
8 | gyh20 | 3447 |
9 | Rebelz | 3409 |
10 | Geothermal | 3408 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 173 |
2 | adamant | 164 |
3 | awoo | 161 |
4 | TheScrasse | 160 |
5 | nor | 159 |
6 | maroonrk | 156 |
7 | SecondThread | 152 |
8 | pajenegod | 146 |
9 | BledDest | 144 |
10 | Um_nik | 143 |
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.