摘要：
"题目链接" 依旧是树形dp啦，一样的找根节点然后自下而上更新即可 设$dp_{i,0}$表示第i个不设，$dp_{i,1}$表示第i个设一个，容易得到其状态转移 $dp_{i,0} = \sum{dp_{j,1}}(j为i的儿子节点)$ $dp_{i,1} = 1 + \sum{min(dp_{j 阅读全文
摘要：
"题目链接" 经典的树形dp，最大独立集，对于每个点就有2个状态，选/不选 设$dp_{i,0}$表示不选第i个，$dp_{i,1}$表示选第i个，容易得到其状态转移 $dp_{i,0} = \sum{max(dp_{j,0}, dp_{j,1})}(j为i的儿子节点)$ $dp_{i,1} = r 阅读全文
摘要：
There are two processes to switch, when one run:io instruction, switch on other process. After ios, the first process's state is ready, which must swi 阅读全文