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其实还有种简单些的方法……In Reply To:考虑树的基本环路系统……然后大概是这样…… Posted by:frkstyc at 2007-10-06 21:55:52 dp求每个子树连到子树外的新边的数量。假设为f(n)。 f(n)=n点出发的新边数-lca为n的新边数*2+所有子节点p的f(p) 然后很容易就可以树形dp出结果。。 Followed by: Post your reply here: |
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