HDU 3974 Assign the task (DFS+线段树)】的更多相关文章

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: 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…
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> #…
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…
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…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3974 给你T组数据,n个节点,n-1对关系,右边的是左边的父节点,所有的值初始化为-1,然后给你q个操作: 有两种操作: 操作一:T X Y ,将以X为根的子树上的所有节点都变成Y. 操作二:C X,查询第X号点是多少? 没想到是线段树做,就算想到了也想不到用dfs序做... 例子中给你了这样的树: 2 /     \ 3       5 /    \ 4      1 DFS一遍转化成DFS序:2…
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 whole company).If you are the immediate boss of someone,that person is your subordinate, and all his subordin…
题目大意:公司里有一些员工及对应的上级,给出一些员工的关系,分配给某员工任务后,其和其所有下属都会进行这项任务.输入T表示分配新的任务, 输入C表示查询某员工的任务.本题的难度在于建树,一开始百思不得其解,后来看了lx大大的博客后才明白,用递归建立了各个员工之间的关系,Start[x] 表示x员工为Boss的起点,End[x]表示x员工为Boss的终点.之后对这样的整体线段进行赋值即可. #include <stdio.h> #include <algorithm> #includ…
Problem 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, an…
Assign the task Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3974 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 l…
Assign the task Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 636    Accepted Submission(s): 322 Problem Description There is a company that has N employees(numbered from 1 to N),every employ…
题意:给出一棵树,改变树的一个节点的值,那么该节点及所有子节点都变为这个值.给出m个询问. 思路:DFS序,将树改为线性结构,用线段树维护.start[ ]记录每个节点的编号,End[ ]为该节点的最小子节点的编号,维护线段树时,即是维护start[x] 到End[x]. 代码: #include<queue> #include<cstring> #include<set> #include<map> #include<stack> #inclu…
题意:一共有n名员工, n-1条关系, 每次给一个人分配任务的时候,(如果他有)给他的所有下属也分配这个任务, 下属的下属也算自己的下属, 每次查询的时候都输出这个人最新的任务(如果他有), 没有就输出-1. 题解:需要用DFS建树来确立关系, 然后用线段树进行区间覆盖. DFS建树: 从Boss 开始dfs,通过dfs递归时编号出现的先后顺序来确定某个员工对应的起点与终点. 样例的关系图是这样的 当dfs建树跑完了之后各个节点对应的位置是这样的 其中Start表示这个节点本身的新编号和这个节点…
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…
Assign the task Problem 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…
题意:给定点的上下级关系,规定假设给i分配任务a.那么他的全部下属.都停下手上的工作,開始做a. 操作 T x y 分配x任务y,C x询问x的当前任务: Sample Input 1 5 4 3 3 2 1 3 5 2 5 C 3 T 2 1 C 3 T 3 2 C 3   Sample Output Case #1: -1 1 2 思路: 利用dfs深度优先遍历又一次编号.使一个结点的儿子连续. 然后成段更新. watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5…
http://acm.hdu.edu.cn/showproblem.php?pid=3974 题目大意: 一个公司有N个员工,对于每个员工,如果他们有下属,那么他们下属的下属也是他的下属. 公司会给员工安排任务,分配给一个员工后,他也会把这个任务分配给下属.被分配到任务的人立刻停止 当前在做的工作,接受新的任务. 对于给定的M个操作 C x  输出编号为x的任务 T x y  分配任务y给x 思路: 并查集的实现,分配我们只记录在上司结点里,只不过查询的时候要把它的所有上司全部找一遍. #inc…
根据Rex 的思路才知道可以这么写. 题目意思还是很好理解的,就是找到当前雇员最近的任务. 做法是,可以开辟一个 tim 变量,每次有雇员得到昕任务时候 ++tim 然后取寻找最近的任务的时候写一个搜索就可以 核心代码: while(num != -1){ num = a[num].leader; if(ttime < a[num].time){ ans = a[num].work; ttime = a[num].time; } } Source code: //#pragma comment(…
http://acm.hdu.edu.cn/showproblem.php?pid=3974 Assign the task Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7144    Accepted Submission(s): 2708 Problem Description There is a company that h…
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>…
Dylans loves tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1444    Accepted Submission(s): 329 Problem Description Dylans is given a tree with N nodes. All nodes have a value A[i].Nodes…
Multiply game Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3224    Accepted Submission(s): 1173 Problem Description Tired of playing computer games, alpc23 is planning to play a game on numbe…
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…
HDU 1394 Minimum Inversion Number(线段树求最小逆序数对) ACM 题目地址:HDU 1394 Minimum Inversion Number 题意:  给一个序列由[1,N]构成.能够通过旋转把第一个移动到最后一个.  问旋转后最小的逆序数对. 分析:  注意,序列是由[1,N]构成的,我们模拟下旋转,总的逆序数对会有规律的变化.  求出初始的逆序数对再循环一遍即可了. 至于求逆序数对,我曾经用归并排序解过这道题:点这里.  只是因为数据范围是5000.所以全…
zhrt的数据结构课 这个题目我觉得是一个有一点点思维的dfs+线段树 虽然说看起来可以用树链剖分写,但是这个题目时间卡了树剖 因为之前用树剖一直在写这个,所以一直想的是区间更新,想dfs+线段树,有点点没想明白 后来才知道可以把这个区间更新转化成单点更新,就是查一个结点的子树,如果子树有可以到根节点的,那么这个结点肯定也可以到根节点. #include <cstdio> #include <cstring> #include <algorithm> #include…
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: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…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3974 题意就是:公司有n个员工,关系有n-1个,T x y 代表把工作y交给员工x: 员工可以把工作交给下属: 本来是线段树的题,但是我真心看不懂,所以就找了个能看懂的方法了   #include<stdio.h> #include<iostream> #include<string.h> #include<algorithm> using namespace…
题目http://acm.hdu.edu.cn/showproblem.php?pid=5692 题目说每个点至多经过一次,那么就是只能一条路线走到底的意思,看到这题的格式, 多个询问多个更新, 自然而然的就会想到线段树或者树状数组,在建树前先做处理, 用DFS将从起点0出发到任一点的距离求出, 然后将这些节点按照一条一条完整的路线的顺序建到树中, 比如样例是1---2---3 | 6 ---4----5 所以建树的其中一种顺序是1 4 5 6 2 3 .当查询的时候的区间应该是从现在这个点开始…
http://acm.hdu.edu.cn/showproblem.php?pid=3974 Problem 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…