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日常娱乐：倍增初探

2016-04-12 16:58:28| 分类：娱乐 | 标签： |举报 |字号大中小订阅

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新坑_(:зゝ∠)_

补作业ing

进度：3/5

BZOJ 1787: [Ahoi2008]Meet 紧急集合：

题意：给定树上的三个点，求一个点使得三个点到该点的路径和最短。

设三点为x,y,z，k1=lca(x,y),k2=lca(x,z),k3=lca(y,z);

由lca(lca(x,y),z)唯一，可知k1,k2,k3中必定有两个相等。考虑实际意义可得，相同的lca节点总路程必定长于剩余的lca节点总路程。（差为两个lca节点的路径长度）

∴剩余的lca节点即所求节点。路径和求出lca(原节点,ans节点)后dfs即可。

代码w

BZOJ 1602: [Usaco2008 Oct]牧场行走：

把上面一题最后一步求路径和做一遍就可以了2333.

代码w

BZOJ 1699: [Usaco2007 Jan]Balanced Lineup排队

怎么搞都可以程度的RBQ...不...RMQ.

代码w

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