2012 ACM-ICPC World Finals

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Problem A

Unsolved.

Problem B

Solved by zerol. 01:36 (+)

Problem C

Solved by ultmaster. 02:20 (+)

Problem D

Solved by kblack. 01:16 (+)

Problem E

Solved by zerol & ultmaster. 04:18 (+2)

Problem F

Unsolved.

Problem G

Unsolved.

Problem H

Unsolved.

Problem I

Unsolved.

Problem J

Unsolved.

Problem K

Solved by kblack. 03:04 (+)

题意:若干堆从小到大的的碟子,拆开和重新叠上,最少的步数整理成一堆。

题解:先分析答案的计算,假装原始堆是颜色,那么使用的步数等于 $2*(C+1)-n-1$,其实 $C$ 是从上到下颜色的变化次数。对于同一种大小的碟子,变化数已经固定了,关键是不同的大小之间是否能减少变化数,相邻每对长度都有一些可以用,除了一种大小全是同一种颜色外,两边的颜色都要不同(相同并不会更好),对于可以用的颜色序列正反搞一搞(超过三种的一定好),还剩能用的都是好的。

Problem L

Unsolved.