| # | User | Rating |
|---|---|---|
| 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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+3
Your profile picture looks great |
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0
I think my F would be a little better than std. You will find that all the edges are repeated the same number of times except for the number of times that the edges between the two $$$x$$$are repeated, so you can consider the inclusion repulsion, which is actually $$$dis_{x_1,x_1}+dis_{x_2,x_2}-2 \times dis_{x_1,x_2}$$$. This should make sense, and when you're done, you actually find that the point between the two $$$x$$$has been calculated exactly $$$n$$$. In fact, it is easy to prove that for $$$2 \times dis_{x_1,x_2}$$$the middle of the two $$$x$$$must not be counted because it is already looped, And then $$$dis_{x_1,x_1}+dis_{x_2,x_2}$$$actually understands that every point is on the edge between these two $$$x$$$, so repeat the $$$n$$$ calculation. So we've figured out the distance between any two points, and now we want to generate a tree, and in order to satisfy the distance that we figured out before, we need to minimize each edge, so we need to minimize the spanning tree. |
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+8
I really enjoyed this one, I'm a big fan of dX!! Although I didn't play this game, I like E and D very much. Personally, I think the difficulty of E is much less than that of D. It may also be my math problem. |
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+20
Personally, I think the problems F are not very difficult. They are very good. They do not involve advanced algorithms, but simple thinking problems |
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+1
I really want to see Jiangly vs. tourist too |
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+3
Personally, I think this E is very good. I can use BIT or segment tree, or even deduce the expression O(n) directly. I really like this problem |
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+7
omg green round |
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0
Good round, the difficulty of div2 is very moderate, very suitable for a novice like me, the E questions are not too difficult, I love it!!! |
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+6
I don't quite understand why there are many difficult questions to read recently, such as the D question, which is difficult to understand, but I can quickly pass after understanding |
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0
so unlucky |
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0
blue meow come and kiss |
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0
退钱退钱!!!推rating!!! |
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0
I don't know why this happened, I was in jail for an hour, what a fuck!!! |
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0
Thanks |
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0
Is there anyone E using bitset or set to do it, ask for a code |
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0
How long does it take to increase the rating? |
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0
How long does it take to increase the rating??? |
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0
SO cool! |
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0
What a nice photo! |
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