题意  公司中有n个员工  除了boss  每个员工都有自己的上司  自己下属的下属也是自己的下属  当给一个员工分配任务时  这个员工会把任务也分配到自己的所有下属   每个员工都只做最后一个被分配的任务  对于每个C x  输出员工x正在做的任务  没有就输出-1

把员工的关系数建成类似并查集的结构  把每个直接分配任务的员工的任务和任务分配时间保存起来  查询时只要找这个员工所有父节点中最晚分配的任务

#include <bits/stdc++.h>
using namespace std;
const int N = 50005;
struct employee{int task, t;
} e[N];
int par[N];int main()
{int cas, a, b, n, m;char op[5];scanf("%d", &cas);for(int k = 1; k <= cas; ++k){printf("Case #%d:\n", k);memset(par, -1, sizeof(par));scanf("%d", &n);for(int i = 1; i < n; ++i){e[i].t = e[i].task = 0;scanf("%d%d", &a, &b), par[a] = b;}e[n].t = e[n].task = 0;scanf("%d", &m);int t = 0, last, ans;while(m--){scanf("%s%d", op, &a);if(op[0] == 'C'){last = 0;//所有祖先节点最晚任务的时间while(a != -1){if(e[a].t > last)last = e[a].t, ans = e[a].task;a = par[a];}printf("%d\n", last ? ans : -1);}else{scanf("%d", &b);e[a].task = b, e[a].t = ++t;}}}return 0;
}
//Last modified :   2015-04-22 20:48

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 subordinate, and all his subordinates are your subordinates as well. If you are nobody's boss, then you have no subordinates,the employee who has no immediate boss is the leader of whole company.So it means the N employees form a tree.

The company usually assigns some tasks to some employees to finish.When a task is assigned to someone,He/She will assigned it to all his/her subordinates.In other words,the person and all his/her subordinates received a task in the same time. Furthermore,whenever a employee received a task,he/she will stop the current task(if he/she has) and start the new one.

Write a program that will help in figuring out some employee’s current task after the company assign some tasks to some employee.

Input
The first line contains a single positive integer T( T <= 10 ), indicates the number of test cases.

For each test case:

The first line contains an integer N (N ≤ 50,000) , which is the number of the employees.

The following N - 1 lines each contain two integers u and v, which means the employee v is the immediate boss of employee u(1<=u,v<=N).

The next line contains an integer M (M ≤ 50,000).

The following M lines each contain a message which is either

"C x" which means an inquiry for the current task of employee x

or

"T x y"which means the company assign task y to employee x.

(1<=x<=N,0<=y<=10^9)

Output
For each test case, print the test case number (beginning with 1) in the first line and then for every inquiry, output the correspond answer per line.
Sample Input
1 5 4 3 3 2 1 3 5 2 5 C 3 T 2 1C 3 T 3 2 C 3
Sample Output
Case #1: -1 1 2
Source
2011 Multi-University Training Contest 14 - Host by FZU

HDU 3974 Assign the task(树 并查集)相关推荐

  1. HDU - 3974 Assign the task (线段树 + dfs序)

    HDU - 3974 题意:有个公司有一些关系,每个人(除了boss)都有且仅有一个上司,这就是一棵树的关系,然后会有一些操作,C i,询问第i个人现在的任务,T x y,把y任务给x, 给x相当于给 ...

  2. HDU 3974 Assign the task(DFS序+线段树单点查询,区间修改)

    描述 There is a company that has N employees(numbered from 1 to N),every employee in the company has a ...

  3. HDU 3974 Assign the task(dfs时间戳+线段树成段更新)

    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 ...

  4. HDU - 3974 Assign the task (DFS建树+区间覆盖+单点查询)

    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 ...

  5. HDU 1512 Monkey King 左偏树 + 并查集

    题目:http://acm.hdu.edu.cn/showproblem.php?pid=1512 题意:有n个猴子,一开始每个猴子只认识自己.每个猴子有一个力量值,力量值越大表示这个猴子打架越厉害. ...

  6. 可持久化线段树【主席树】可持久化并查集【主席树+并查集】

    笼统的主席树原理 众所周知, 主席树是可以持久化的, 换言之你能知道你所维护信息的所有历史状态. 主席树是这样做的: 1. 首先建一颗朴素的线段树,代表初始状态 (下图黑色) , 也就是第0次操作后的 ...

  7. HDU 1512 Monkey King(左偏树+并查集)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1512 题       意: 有n个猴子,一开始每个猴子只认识自己.每个猴子有一个力量值,力量值越大表示 ...

  8. HDU 1272 - 小希的迷宫(并查集判树)

    目录 Problem Description C++代码 Problem Description 上次Gardon的迷宫城堡小希玩了很久(见Problem B),现在她也想设计一个迷宫让Gardon来 ...

  9. HDU 3047 Zjnu Stadium (带权并查集)

    链接: http://acm.hdu.edu.cn/showproblem.php?pid=3047 题目: Problem Description In 12th Zhejiang College ...

最新文章

  1. csu 1554: SG Value 思维题
  2. html简单网页代码表白_表白网页代码,不会代码也可以操作,告别单身
  3. SAP:HANA为高性能数据分析保驾护航
  4. IAAS、PAAS与SAAS
  5. ORA-01925:maximum of 80 enabled roles exceeded
  6. Matplotlib从文件绘图时Y轴坐标不正确
  7. 计算1!+2!+...+n!
  8. python 把函数作为参数 ---高阶函数
  9. 实现第一个JDBC程序(详细)
  10. 将科学计数法的数值转化为字符
  11. CSS美化超链接样式
  12. 注册界面模板HTML+CSS
  13. RTKLIB专题学习(十二)—支持的信号ID/观测类型及读取优先级
  14. 射频中经常是用50欧姆作为阻抗匹配的标准的原因
  15. 虚拟机无法连接虚拟设备sata0:1
  16. OCR财务报表识别方案应用
  17. python知到答案_Python语言应用_知到网课答案
  18. SSH框架相关准备与入门学习
  19. android文字闪动效果
  20. 教你正确使用文字识别软件ABBYY的转换格式功能

热门文章

  1. Java中类方法可以继承吗_在Java语言中,类方法(静态方法)可以被继承,但是不能被( )。_学小易找答案...
  2. 大唐仙侠传java,大唐仙侠传官网版,大唐仙侠传手游官网版预约 v1.0-手游汇
  3. 炒菜机器人康名牌_机器人帮炒菜,九阳、美的产品各有何本领?
  4. “智慧交通”转型升级+创新发展策略
  5. telnet命令及测试网络端口的几种方法
  6. linux 单引号和双引号区别
  7. 优盘数据恢复怎么做?3个方法分享!
  8. Java--The import *** cannot be resolved问题解决方案
  9. CANoe-发送CAN消息的几种方式
  10. mac睡眠状态下远程唤醒_如何控制Mac自动进入睡眠状态的时间