题意:给一棵树,有三个操作:①询问两点$(x,y)$之间的距离②把$x$和原来的父亲断开并连到它的$h$级祖先,作为新父亲最右的儿子③询问与根节点距离为$k$的点中最右的点是哪个点 用出栈入栈序$s_{1\cdots 2n}$来维护整棵树,入栈记$1$出栈记$-1$,那么一个节点$x$的深度就是$\sum\limits_{i=1}^{in_x}s_x$ 每个平衡树节点记1.这个节点是出栈还是入栈2.子树和3.最大前缀和4.最小前缀和,那么我们就可以在平衡树上二分找到最右的深度为$d$的节点(注意…

目录 @description@ @solution@ @accepted code@ @details@ @description@ 给定一棵 n 个点的树,每个点的儿子是有序的. 现给定 m 次操作,每次操作是下列三种中的一种: (1)给定 u, v,询问 u, v 之间的距离. (2)给定 v, h,断开 v 到父亲的边,将 v 这棵子树加入到它的第 h 个祖先的最后一个儿子. (3)给定 k,询问在当前这棵树上 dfs 后得到 dfs 序中,最后一个深度为 k 的点的编号. Input…

题目 题目大意 给你一棵树,接下来对这棵树进行三种操作: 1.询问两点之间的距离. 2.让某个点变为它原来的第\(h\)个祖先的最后一个儿子. 3.求\(dfs\)序中最后一个深度为\(k\)的点. 正解 第一种是Cold_Chair大爷提出来的\(LCT\)维护\(ETT\)的做法. 具体怎样就不说了--据说代码5000+ 第二种就直接是\(ETT\)了(其实这是一道ETT的板题啊) 对于入栈点(记作\(l\))打个\(+1\),对于出栈点(记作\(r\))打个\(-1\),维护前缀和.这个前…

You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9596    Accepted Submission(s): 4725 Problem Description Many geometry(几何)problems were designed in the ACM/I…

This article is from blog of Amazon CTO Werner Vogels. -------------------- Today is a very exciting day as we release Amazon DynamoDB, a fast, highly reliable and cost-effective NoSQL database service designed for internet scale applications. Dynamo…

You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10204    Accepted Submission(s): 5042 Problem Description Many geometry(几何)problems were designed in the ACM/…

Problem Description Many geometry(几何)problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is ve…

Revision History Date Issue Description Author 15/May/2015 1.0 Finish most of the designed function. Only the Windows application is finished. litianpeng.yanwenxiongandyuxuehui 21/May/2015 V1.1 Finish all of the function on windows store and windowsp…

A Computer Graphics Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 336    Accepted Submission(s): 277 Problem Description In this problem we talk about the study of Computer Graphics.…

称号: You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 145 Accepted Submission(s): 100   Problem Description Many geometry(几何)problems were designed in the ACM/IC…