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2020-2021:teams:mian:nowcoder_training:2020_multi-university_training_contest_10
2020牛客暑期多校训练营（第十场）
Results
Summary
- Solved 7 out of 12 problems
- Rank 32/1041 in official records
- Solved 8 out of 12 afterwards
| # | Who | = | Penalty | A | B | C | D | E | F | G | H | I | J | Dirt |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 11 | 大吉大利，今晚吃 mian(); | 3 | 534 | +1 02:01 | +1 01:07 | -2 | -3 | +2 04:26 | 57% 4/7 |
Member Distribution
| Solved | A | B | C | D | E | F | G | H | I | J |
|---|---|---|---|---|---|---|---|---|---|---|
| Pantw | √ | |||||||||
| Withinlover | √ | |||||||||
| Gary | √ |
(√ for solved, O for upsolved, - for tried but not solved)
Solutions
A
半天也没搞懂怎么证得，尽量选2，不能选2就选一次3
B
C
D
E
二分然后暴力从后往前推就好了
F
G
H
I
J
考虑 DP。对每个结点，考虑其所有儿子的子树，算出它们与对应点的儿子的子树的权值 G[i][j]，算出来之后把它们建成二分图，再直接在二分图上跑一个最小费用最大流即可。
Comments
ptw:
- 抢榆枋而止
Gary:
- C题思路上有问题，平常都是类似题解的方式讨论，今天脑子抽了
- 在一次提醒自己写板子
Withinlover:
- 注 意 肠 道 卫 生
- 现在看E题挺显然的不知道自己当时在懵啥（
2020-2021/teams/mian/nowcoder_training/2020_multi-university_training_contest_10.txt · 最后更改: 2020/08/12 15:43 由 withinlover
