| # | User | Rating |
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| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | Kevin114514 | 3611 |
| 4 | jiangly | 3583 |
| 5 | strapple | 3515 |
| 6 | tourist | 3470 |
| 7 | Radewoosh | 3415 |
| 8 | Um_nik | 3376 |
| 9 | maroonrk | 3361 |
| 10 | XVIII | 3345 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 162 |
| 2 | adamant | 148 |
| 3 | Um_nik | 146 |
| 4 | Dominater069 | 143 |
| 5 | errorgorn | 141 |
| 6 | cry | 138 |
| 7 | Proof_by_QED | 136 |
| 8 | YuukiS | 135 |
| 9 | chromate00 | 134 |
| 10 | soullless | 133 |
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+62
Maybe because I am not good at math and interactive. |
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+90
Only I think F much easier than E1 and E1 much easier than E2? |
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+8
The author is not from china, is other place this problem? |
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+12
The input and output format not same. |
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+17
I don't think this will make difference. |
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+20
I think the decisive factor should be how many people AC the problem not rely on their effort in the contest. |
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+34
But need quite code capability. |
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+33
E and F, it is natural to think of dp. And one using data structure another using FFT to optimize the transition. |
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+51
I mean compare with other div2 contest, this one is quite boring. |
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+71
No difficulty of thinking, but coding. A boring competition. |
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+47
By the way, the discrimination of this round is not good, maybe the reason is that the difficulty in thinking in problem E and F is insufficient. |
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+31
For problem F, if we enumerate one of the biggest height node, then the contribution of node i (i is not the biggest node we determine) is max(0,h[i]-(the maximum h of the node except the node in the subtree of the biggest node when the root is i and i itself). We first determine the root of the tree, then my solution is to calculate up[i] and down[i], it means if the biggest node in the subtree of i, what the contribution will be for node father[i] and if the biggest node not in the subtree of i, what the contribution will be for node i. Then we can easily calculate the answer. |
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