Query on a tree You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. We will ask you to perfrom some instructions of the following form: CHANGE i ti : change the cost of the i-th edge to ti or Q…

第一次写树剖~ #include<iostream> #include<cstring> #include<cstdio> #define L(u) u<<1 #define R(u) u<<1|1 using namespace std; ; ],next1[MAX*],tov[MAX*],val[MAX*],tot,n; int fa[MAX],w[MAX],son[MAX],depth[MAX],tot2,size[MAX]; ],tree…

题目链接:http://www.spoj.com/problems/QTREE/en/ QTREE - Query on a tree #tree You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. We will ask you to perfrom some instructions of the following form:…

传送门:Problem QTREE https://www.cnblogs.com/violet-acmer/p/9711441.html 题解: 树链剖分的模板题,看代码比看文字解析理解来的快~~~~~~~ AC代码献上: #include<iostream> #include<cstdio> #include<cmath> #include<cstring> using namespace std; #define ls(x) ((x)<<1…

[题目分析] 垃圾vjudge又挂了. 树链剖分裸题. 垃圾spoj,交了好几次,基本没改动却过了. [代码](自带常数,是别人的2倍左右) #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define maxn 20005 int T,n,fr[maxn],h[maxn],to[maxn],ne[maxn]…

题意:给一棵树,每次更新某条边或者查询u->v路径上的边权最大值. 解法:做过上一题,这题就没太大问题了,以终点的标号作为边的标号,因为dfs只能给点分配位置,而一棵树每条树边的终点只有一个. 询问的时候,在从u找到v的过程中顺便查询到此为止的最大值即可. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath&…

题目链接 给一棵树, 每条边有权值, 两种操作, 一种是将一条边的权值改变, 一种是询问u到v路径上最大的边的权值. 树链剖分模板. #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <set&g…

You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. We will ask you to perfrom some instructions of the following form: CHANGE i ti : change the cost of the i-th edge to tior QUERY a b : ask fo…

树链剖分整理 树链剖分就是把树拆成一系列链,然后用数据结构对链进行维护. 通常的剖分方法是轻重链剖分,所谓轻重链就是对于节点u的所有子结点v,size[v]最大的v与u的边是重边,其它边是轻边,其中size[v]是以v为根的子树的节点个数,全部由重边组成的路径是重路径,根据论文上的证明,任意一点到根的路径上存在不超过logn条轻边和logn条重路径. 这样我们考虑用数据结构来维护重路径上的查询,轻边直接查询. 通常用来维护的数据结构是线段树,splay较少见. 具体步骤 预处理 第一遍dfs 求…

2588: Spoj 10628. Count on a tree Time Limit: 12 Sec Memory Limit: 128 MB 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 第一行两个整数N,M. 第二行有N个整数,其中第i个整数表示点i的权值. 后面N-1行每行两个整数(x,y)…

You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. We will ask you to perfrom some instructions of the following form: CHANGE i ti : change the cost of the i-th edge to ti or QUERY a b : ask f…

传送门 题意 给出一棵树,每条边都有权值,有两种操作: 把第p条边的权值改为x 询问x,y路径上的权值最大的边 code #include<cstdio> #include<algorithm> #include<cstring> #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 using namespace std; ; struct Edge { int to,nxt,w; }e[]; in…

POJ3237 Tree 树链剖分 边权 传送门:http://poj.org/problem?id=3237 题意: n个点的,n-1条边 修改单边边权 将a->b的边权取反 查询a->b边权最大值 题解: 修改边权就查询点的深度大的点,用大的点去存这条边的边权,其余的就和点权的是一样的了 取反操作用线段树维护,区间最大值取反就是区间最小值,区间最小值取反就是区间最大值 所以维护两颗线段树即可,lazy标记表示覆盖单边的边权 代码: #include <set> #include…

Hdu 5274 Dylans loves tree (树链剖分模板) 题目传送门 #include <queue> #include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #include <vector> #define ll long…

  Query on a tree Time Limit: 851MS   Memory Limit: 1572864KB   64bit IO Format: %lld & %llu Submit Status Description You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. We will ask you to per…

You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. We will ask you to perfrom some instructions of the following form: CHANGE i ti : change the cost of the i-th edge to tior QUERY a b : ask fo…

https://vjudge.net/problem/SPOJ-QTREE 题意: 给出一棵树,树上的每一条边都有权值,现在有查询和更改操作,如果是查询,则要输出u和v之间的最大权值. 思路: 树链剖分的模板题. 树链剖分简单来说,就是把树分成多条链,然后再将这些链映射到数据结构上处理(线段树,树状数组等等). 具体的话可以看看这个http://blog.sina.com.cn/s/blog_6974c8b20100zc61.html #include<iostream> #include&l…

QTREE - Query on a tree #number-theory You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. We will ask you to perfrom some instructions of the following form: CHANGE i ti : change the cost of t…

375. Query on a tree Problem code: QTREE You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. We will ask you to perfrom some instructions of the following form: CHANGE i ti : change the cost of…

这道题是树链剖分的裸题,正在学LCT,用LCT写了,发现LCT代码比树链剖分还短点(但我的LCT跑极限数据用的时间大概是kuangbin大神的树链剖分的1.6倍,所以在spoj上是850ms卡过的). 收获: 1.边转换成点(即若存在边(u,v),则新加一个点z代表边,将z连接u和v,z的点权就是(u,v)的边权,非边点的权设为-oo),然后对边权的统计就变成了对点权的统计(这是LCT中处理边信息的通法之一). 2.若要连接两个点u,v,先让它们分别称为根,然后将其中一个的path-parent…

Description 给出一个树,每条边有边权,支持两种操作,询问 \(u,v\) 路径上边权最大值,修改第 \(i\) 条边的边权,\(n\leqslant 10^4,T\leqslant 10\) Sol 树链剖分. 基于边的树链剖分,对于一个点,可能有许多儿子,但是它只能有一个父亲,给它编号表示它到它父亲的边,只需要修改查询的是最后一步就可以了. Code #include<cstdio> #include<vector> #include<iostream>…

4353: Play with tree Time Limit: 20 Sec  Memory Limit: 256 MBSubmit: 31  Solved: 19[Submit][Status][Discuss] Description 给你一棵包含N个节点的树,设每条边一开始的边权为0,现在有两种操作: 1)给出参数U,V,C,表示把U与V之间的路径上的边权变成C(保证C≥0) 2)给出参数U,V,C,表示把U与V之间的路径上的边权加上C.但是如果U至V之间路径某条边的边权加上C小于0,那…

题目链接:http://poj.org/problem?id=3237 You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edges are numbered 1 through N − 1. Each edge is associated with a weight. Then you are to execute a series of instructions on th…

Description You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edges are numbered 1 through N − 1. Each edge is associated with a weight. Then you are to execute a series of instructions on the tree. The instructions…

Do use segment tree Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://www.bnuoj.com/v3/problem_show.php?pid=39566 Description Given a tree with n (1 ≤ n ≤ 200,000) nodes and a list of q (1 ≤ q ≤ 100,000) queries, process the queries in order and out…

To your surprise, Jamie is the final boss! Ehehehe. Jamie has given you a tree with n vertices, numbered from 1 to n. Initially, the root of the tree is the vertex with number 1. Also, each vertex has a value on it. Jamie also gives you three types o…

欢迎访问~原文出处——博客园-zhouzhendong 去博客园看该题解 题目传送门 - POJ3237 题意概括 Description 给你由N个结点组成的树.树的节点被编号为1到N,边被编号为1到N-1.每一条边有一个权值.然后你要在树上执行一系列指令.指令可以是如下三种之一: CHANGE i v:将第i条边的权值改成v. NEGATE a b:将点a到点b路径上所有边的权值变成其相反数. QUERY a b:找出点a到点b路径上各边的最大权值. Input多组数据,数据为T<=20,对…

D. Water Tree time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Mad scientist Mike has constructed a rooted tree, which consists of n vertices. Each vertex is a reservoir which can be either…

[CF725G]Messages on a Tree 题意:给你一棵n+1个节点的树,0号节点是树根,在编号为1到n的节点上各有一只跳蚤,0号节点是跳蚤国王.现在一些跳蚤要给跳蚤国王发信息.具体的信息传输过程如下: 1.信息的发起者把信息上传给他父亲节点处的跳蚤,然后自身进入等待状态.3.跳蚤国王在收到信息时会将信息立刻下传到发来信息的那个儿子,跳蚤国王可以在同一时刻下传多份信息.4.上传:a把信息传给b.如果b正处于等待状态,则b会立刻将a发来的信息下传回去.如果同时有好多个信息传给b,则b会…

Water Tree http://codeforces.com/problemset/problem/343/D time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Mad scientist Mike has constructed a rooted tree, which consists of n vertices. Ea…