bzoj2588: Spoj 10628. Count on a tree（树上第k大）（主席树）

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SPOJ 10628 Count on a tree（Tarjan离线LCA+主席树求树上第K小）

COT - Count on a tree #tree You are given a tree with N nodes.The tree nodes are numbered from 1 to N.Each node has an integer weight. We will ask you to perform the following operation: u v k : ask for the kth minimum weight on the path from node u …

计蒜客 38229.Distance on the tree-1.树链剖分(边权)+可持久化线段树(区间小于等于k的数的个数)+离散化+离线处理 or 2.树上第k大(主席树)+二分+离散化+在线查询 (The Preliminary Contest for ICPC China Nanchang National Invitational 南昌邀请赛网络赛)

Distance on the tree DSM(Data Structure Master) once learned about tree when he was preparing for NOIP(National Olympiad in Informatics in Provinces) in Senior High School. So when in Data Structure Class in College, he is always absent-minded about…

BZOJ 2588: Spoj 10628. Count on a tree [树上主席树]

2588: Spoj 10628. Count on a tree Time Limit: 12 Sec Memory Limit: 128 MBSubmit: 5217 Solved: 1233[Submit][Status][Discuss] Description 给定一棵N个节点的树,每个点有一个权值,对于M个询问(u,v,k),你需要回答u xor lastans和v这两个节点间第K小的点权.其中lastans是上一个询问的答案,初始为0,即第一个询问的u是明文. Input 第一…

BZOJ 2588: Spoj 10628. Count on a tree 树上跑主席树

2588: Spoj 10628. Count on a tree Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://www.lydsy.com/JudgeOnline/problem.php?id=2588 Description 给定一棵N个节点的树,每个点有一个权值,对于M个询问(u,v,k),你需要回答u xor lastans和v这两个节点间第K小的点权.其中lastans是上一个询问的答案,初始为0,即第一个询问的u是明文. Inp…

Count on a tree（SPOJ COT + 树上第k大 + 主席树 + LCA）

题目链接:https://www.spoj.com/problems/COT/en/ 题目: 题意: 给你一棵有n个节点的树,求节点u到节点v这条链上的第k大. 思路: 我们首先用dfs进行建题目给的树,然后在dfs时进行主席树的update操作.众所周知,主席树采用的是前缀和思想,区间第k大是与前一个树添加新的线段树,而树上第k大则是与父亲节点添加新的线段树,因而在此思想上此题的答案为sum[u] + sum[v] - sum[lca(u,v)] - sum[fa[lca(u,v)].求第k大…