====== 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(