There is a company that has N employees(numbered from 1 to N),every employee in the company has a immediate boss (except for the leader of whole company).If you are the immediate boss of someone,that person is your subordinate, and all his subordinat…

题意:给定一棵树的公司职员管理图,有两种操作, 第一种是 T x y,把 x 及员工都变成 y, 第二种是 C x 询问 x 当前的数. 析:先把该树用dfs遍历,形成一个序列,然后再用线段树进行维护,很简单的线段树. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib>…

题目链接:https://vjudge.net/problem/HDU-3974 There is a company that has N employees(numbered from 1 to N),every employee in the company has a immediate boss (except for the leader of whole company).If you are the immediate boss of someone,that person is…

There is a company that has N employees(numbered from 1 to N),every employee in the company has a immediate boss (except for the leader of whole company).If you are the immediate boss of someone,that person is your subordinate, and all his subordinat…

题目链接: 题目 Assign the task Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) 问题描述 There is a company that has N employees(numbered from 1 to N),every employee in the company has a immediate boss (except for the leader of…

https://cn.vjudge.net/problem/HDU-3974 题意 有一棵树,给一个结点分配任务时,其子树的所有结点都能接受到此任务.有两个操作,C x表示查询x结点此时任务编号,T x y表示给x结点分配编号为y的任务. 分析 题目读起来就很有区间修改的味道,将一个区间变为一个值.问题在于怎么把这棵树对应到区间上. 对于一个结点,其控制的范围是它的子树,对应区间范围可以看作是以dfs序表示的区间.好像有点绕..就是给每个结点再对应一个dfs序,然后在dfs时把这个点控制的子树看…

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=3974 题意:给定一棵树,50000个节点,50000个操作,C x表示查询x节点的值,T x y表示更新x节点及其子节点的值为y 大致把边存一下那一棵树来举例子 2 3 5 4 1 例如像这样的一棵树,可以将2->1,3->2,4->3,1->4,5->5按照dfs序来编号,然后用线段树进行区间修改,稍微想一想 应该都会了. #include <iostream> #…

Description There is a company that has N employees(numbered from 1 to N),every employee in the company has a immediate boss (except for the leader of whole company).If you are the immediate boss of someone,that person is your subordinate, and all hi…

1.HDU 5877  Weak Pair 2.总结:有多种做法,这里写了dfs+线段树(或+树状树组),还可用主席树或平衡树,但还不会这两个 3.思路:利用dfs遍历子节点,同时对于每个子节点au,查询它有多少个祖先av满足av<=k/au. (1)dfs+线段树 #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm>…

Codeforces1110F dfs + 线段树 + 询问离线 F. Nearest Leaf Description: Let's define the Eulerian traversal of a tree (a connected undirected graph without cycles) as follows: consider a depth-first search algorithm which traverses vertices of the tree and enu…