题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5405 题意: 给你一棵n个节点的树,有点权.        要求支持两种操作: 操作1:更改某个节点的权值. 操作2:给定u,v, 求 Σw[i][j]   i , j 为任意两点且i到j的路径与u到v的路径相交. 解法: 这是一个大树剖题. 容易发现对于一个询问,答案为总点权和的平方 减去 去掉u--v这条链后各个子树的点权和的平方的和. 开两棵线段树,tag1记录点权和,tag2记录某点的所有轻…

题意 \(n\) 个节点的树,每个点有点权,\(m\) 次操作,操作分两种,修改一个节点的点权,对于一个 \((u,v)\) ,询问 \(\displaystyle\sum_{i=1}^n\sum_{j=1}^n f(i,j)\) 的值,其中如果路径 \((i,j)\) 与路径 \((u,v)\) 有公共点,\(f(i,j)=w_iw_j\)( \(w_i\) 表示节点 \(i\) 的点权),否则 \(f(i,j)=0\) . \(1\leq n,m \leq 10^5​\) 思路 先讲一下用…

Sometimes Naive Problem Description   Rhason Cheung had a naive problem, and asked Teacher Mai for help. But Teacher Mai thought this problem was too simple, sometimes naive. So she ask you for help. She has a tree with n vertices, numbered from 1 to…

Problem Description You are given a tree with N nodes which are numbered by integers 1..N. Each node is associated with an integer as the weight. Your task is to deal with M operations of 4 types: 1.Delete an edge (x, y) from the tree, and then add a…

Box Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) [Problem Description] There are N boxes on the ground, which are labeled by numbers from 1 to N. The boxes are magical, the size of each one can be enlarged or r…

GCD Tree Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 415    Accepted Submission(s): 172 Problem Description Teacher Mai has a graph with n vertices numbered from 1 to n. For every edge(u,v),…

Tree Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 920    Accepted Submission(s): 388 Problem Description You are given a tree with N nodes which are numbered by integers 1..N. Each node is a…

Happy King Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 821    Accepted Submission(s): 179 Problem Description There are n cities and n−1 roads in Byteland, and they form a tree. The citie…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4010 题意; 先给你一棵树,有 \(4\) 种操作: 1.如果 \(x\) 和 \(y\) 不在同一棵树上则在\(x-y\)连边. 2.如果 \(x\) 和 \(y\) 在同一棵树上并且 \(x!=y\) 则把 \(x\) 换为树根并把 \(y\) 和 \(y\) 的父亲分离. 3.如果 \(x\) 和 \(y\) 在同一棵树上则 \(x\) 到 \(y\) 的路径上所有的点权值\(+w\). 4…

lca的做法还是非常明显的.简单粗暴, 只是不是正解.假设树是长链就会跪,直接变成O(n).. 最后跑的也挺快,出题人还是挺阳光的.. 动态树的解法也是听别人说能ac的.预计就是放在splay上剖分一下,做法还是比較复杂的.,, 来一发lca: #include <stdio.h> #include <iostream> #include <algorithm> #include <sstream> #include <stdlib.h> #in…