2013年7月4日
POJ 1323 Game Prediction (贪心)
摘要： Game PredictionTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 8866Accepted: 4225DescriptionSuppose there are M people, including you, playing a special card game. At the beginning, each play... 阅读全文
posted @ 2013-07-04 20:18 码代码的猿猿 阅读(112) 评论(0) 推荐(0) 编辑
HDOJ 1677 Nested Dolls(LIS+Dilworth)
摘要： Nested DollsTime Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1805 Accepted Submission(s): 501Problem DescriptionDilworth is the world’s most p... 阅读全文
posted @ 2013-07-04 08:15 码代码的猿猿 阅读(332) 评论(0) 推荐(0) 编辑
POJ 1548 Robots (Dilworth)
摘要： RobotsTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 3621Accepted: 1643DescriptionYour company provides robots that can be used to pick up litter from fields after sporting events and concer... 阅读全文
posted @ 2013-07-04 07:14 码代码的猿猿 阅读(224) 评论(0) 推荐(0) 编辑
POJ 1065 Wooden Sticks
摘要： Wooden SticksTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 16262Accepted: 6748DescriptionThere is a pile of n wooden sticks. The length and weight of each stick are known in advance. The st... 阅读全文
posted @ 2013-07-04 01:50 码代码的猿猿 阅读(182) 评论(0) 推荐(0) 编辑
偏序集的Dilworth定理
摘要： 定理1 令（X,≤）是一个有限偏序集，并令r是其最大链的大小。则X可以被划分成r个但不能再少的反链。其对偶定理称为Dilworth定理：定理2 令（X,≤）是一个有限偏序集，并令m是反链的最大的大小。则X可以被划分成m个但不能再少的链。说白了就是 链的最少划分数=反链的最长长度 阅读全文
posted @ 2013-07-04 00:09 码代码的猿猿 阅读(225) 评论(0) 推荐(0) 编辑