Difference between revisions of "ACM-ICPC 2015 Asia Tsukuba Regional"
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Solved by kblack. 04:22 (+9) | Solved by kblack. 04:22 (+9) | ||
+ | |||
+ | 题意:$p$ 个进程,$r$ 种资源,判断从哪一次资源分配后无法靠安排之后的顺序解决死锁。 | ||
+ | |||
+ | 题解:无法避免死锁的点存在二分性,因为能跑完的进程一定先跑完更好,通过类似 [https://acm.ecnu.edu.cn/wiki/index.php?title=2018_Multi-University,_HDU_Day_7#Problem_K] 的方式贪心判断是否能全部跑完,需要注意一些边界情况。 | ||
== Problem G == | == Problem G == | ||
Solved by zerol. 03:31 (+) | Solved by zerol. 03:31 (+) |
Revision as of 10:51, 7 October 2018
Problem A
Solved by zerol. 00:15 (+)
温暖的签到。
Problem B
Solved by zerol & ultmaster. 00:24 (+)
题意:求最小距离,使得两块板之间可以塞下题给的所有木棍(不能交换顺序)。
题解:贪心往左放即可。依次计算每根木棍最左可以放到哪儿。
Problem C
Solved by ultmaster. 01:39 (+4)
Problem D
Solved by kblack. 01:42 (+)
题意:在一个矩形内,每个人有一个直角的视野,在墙上挂钟使得每个人都能至少看到一个钟。
题解:直角视野对应周长上的一个区间,然后就是一个在环上丽娃河装路灯的问题。我们枚举第一个钟的位置,之后贪心地尽量靠后安装。
Problem E
Solved by ultmaster. 03:50 (+)
Problem F
Solved by kblack. 04:22 (+9)
题意:$p$ 个进程,$r$ 种资源,判断从哪一次资源分配后无法靠安排之后的顺序解决死锁。
题解:无法避免死锁的点存在二分性,因为能跑完的进程一定先跑完更好,通过类似 [1] 的方式贪心判断是否能全部跑完,需要注意一些边界情况。
Problem G
Solved by zerol. 03:31 (+)