Difference between revisions of "2012 ACM-ICPC World Finals"
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(Created page with "== Problem A == Unsolved. == Problem B == Solved by zerol. 01:36 (+) == Problem C == Solved by ultmaster. 02:20 (+) == Problem D == Solved by kblack. 01:16 (+) == Prob...") |
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Solved by kblack. 03:04 (+) | Solved by kblack. 03:04 (+) | ||
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| + | 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。 | ||
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| + | 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 == | == Problem L == | ||
Unsolved. | Unsolved. | ||
Revision as of 11:52, 9 March 2019
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.