the problem is UVa 11157 — lazy frog
i understand that the subproblem is to find the minimax jump between two closest big stones. but how to prove that alternating jumps on the small stones is the best strategy ?
# | User | Rating |
---|---|---|
1 | tourist | 3985 |
2 | jiangly | 3814 |
3 | jqdai0815 | 3682 |
4 | Benq | 3529 |
5 | orzdevinwang | 3526 |
6 | ksun48 | 3517 |
7 | Radewoosh | 3410 |
8 | hos.lyric | 3399 |
9 | ecnerwala | 3392 |
9 | Um_nik | 3392 |
# | User | Contrib. |
---|---|---|
1 | cry | 169 |
2 | maomao90 | 162 |
2 | Um_nik | 162 |
4 | atcoder_official | 160 |
5 | djm03178 | 158 |
6 | -is-this-fft- | 157 |
7 | adamant | 155 |
8 | Dominater069 | 154 |
8 | awoo | 154 |
10 | luogu_official | 151 |
the problem is UVa 11157 — lazy frog
i understand that the subproblem is to find the minimax jump between two closest big stones. but how to prove that alternating jumps on the small stones is the best strategy ?
Name |
---|
Let us prove this by contradiction. Let this not be the optimal strategy and the maximum jump is Mi - 1 → Mi + 1. Using optimal solution frog will jump:
Mi - 1 → Mi frog must jump Mi - 2 → Mi + 1 — greater distance than Mi - 1 → Mi + 1;
Mi → Mi + 1 frog must jump Mi - 1 → Mi + 2 — greater distance than Mi - 1 → Mi + 1.
We got a contradiction, so your greedy is correct.