DP - Max Value from Circle


http://www.lofter.com/postentry?from=search&permalink=1cb3058e_2d358ca
说有n个节点n条边组成一个圈,每个节点上面有一个数,边上有一个+或*,如果消掉某条边,其相邻两个节点就用这个运算符合并。这样一路消边到底,问用什么过程能让最后得到的数最大(dp题) 

这题我竟然还是看着妹子的提示,知道用动归,然后竟然想了n久。还是功力和妹子比差的有点远,继续努力吧。

解法:
假设列子为:妹子的动态规划题目,记录下解题思路 - silver9886@126 - silver9886@126的博客,那么dp的状态转移方程为:
dp(new) = {[1,max(2,3,4)],[2,max(1,3,4)],[3,max(1,2,4)],[4,max(1,2,3)]}
我们以[1,max(2,3,4)]做进一步展开,其他的类似
[1,max(2,3,4)]=max{1+max(2,3,4),1 * max(2,3,4)}=1+max(2,3,4)=1+max{2*max(3,4),4+max(2,3)}(这里注意没有3的原因是3和1是不相邻的,因此3和1是无法发生作用关系的)=1+max{2*(3+4),4+2*3}=1+14 = 15.
其他的式子类似。
这里注意有些中间的计算过程是一样的,因此,如果可以,将中间计算过程存储起来,在后续中不需计算了。这也是动归快的原因。
http://lixinzhang.github.io/qu-jian-xing-dp.html

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