CF 322E - Ciel the Commander 树的点分治

【CF 322E - Ciel the Commander 树的点分治】的更多相关文章

CF 322E - Ciel the Commander 树的点分治

树链剖分可以看成是树的边分治,什么是点分治呢? CF322E - Ciel the Commander 题目:给出一棵树,对于每个节点有一个等级(A-Z,A最高),如果两个不同的节点有相同等级的父节点(节点不一定相同),则两个节点的路径上最少有一个比他们两的等级都高的节点.要求输出1~n的等级. 分析:考虑一条链的情况,如果把A放中间,把数列分成两段,对于每一段,中间放B,这样分成了四段,对于每一段继续这样分,显然26个字母可以安放2^26个位置. 对于树这种结构的话,如何安放? 类似于一条链的…

Codeforce 322E Ciel the Commander （点分治）

E. Ciel the Commander Now Fox Ciel becomes a commander of Tree Land. Tree Land, like its name said, has n cities connected by n - 1 undirected roads, and for any two cities there always exists a path between them. Fox Ciel needs to assign an officer…

Codeforces G. Ciel the Commander

题目描述: Ciel the Commander time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Now Fox Ciel becomes a commander of Tree Land. Tree Land, like its name said, has n cities connected by n - 1 undire…

Codeforces Round #190 (Div. 2) E. Ciel the Commander 点分治

E. Ciel the Commander Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.codeforces.com/contest/322/problem/E Description Now Fox Ciel becomes a commander of Tree Land. Tree Land, like its name said, has n cities connected by n - 1 undirected ro…

CodeForces 321C Ciel the Commander

Ciel the Commander Time Limit: 1000ms Memory Limit: 262144KB This problem will be judged on CodeForces. Original ID: 321C64-bit integer IO format: %I64d Java class name: (Any) Now Fox Ciel becomes a commander of Tree Land. Tree Land, like its na…

HDU4812 D Tree（树的点分治）

题目大概说给一棵有点权的树,输出字典序最小的点对,使这两点间路径上点权的乘积模1000003的结果为k. 树的点分治搞了.因为是点权过根的两条路径的LCA会被重复统计,而注意到1000003是质数,所以这个用乘法逆元搞一下就OK了.还有要注意“治”的各个实现,把时间复杂度“控制”在O(nlogn). WA了几次,WA在漏了点到子树根的路径,还有每次分治忘了清空数组. #include<cstdio> #include<cstring> #include<algorithm&g…

hdu 4670 树的点分治

思路:首先当然是要用树的点分治了.根节点为root,那么经过root的合法路径数求出来这题就解决了.因为我们可以用分治枚举根,最后将所有根的路径数加起来就是结果.当然这里的根不是整棵树的根,是子树根. 我们为每个节点分配一个长度为30的数组记录给定因数在每个节点权值出现的次数.如果某几个权值相乘的值Value的三次根仍是整数的话,那么Value在给定因数的所有幂一定是3的倍数.通过这个转换,我们将所有的幂都对3取余,结果还是一样. 在判断经过root的合法路径数时,我们进入其一个子树,将经过的路…