POJ 3481 Double Queue(set实现)
Double Queue
The new founded Balkan Investment Group Bank (BIG-Bank) opened a new office in Bucharest, equipped with a modern computing environment provided by IBM Romania, and using modern information technologies. As usual, each client of the bank is identified by a positive integer Kand, upon arriving to the bank for some services, he or she receives a positive integer priority P. One of the inventions of the young managers of the bank shocked the software engineer of the serving system. They proposed to break the tradition by sometimes calling the serving desk with the lowest priority instead of that with the highest priority. Thus, the system will receive the following types of request:
0 | The system needs to stop serving |
1 KP | Add client K to the waiting list with priority P |
2 | Serve the client with the highest priority and drop him or her from the waiting list |
3 | Serve the client with the lowest priority and drop him or her from the waiting list |
Your task is to help the software engineer of the bank by writing a program to implement the requested serving policy.
Input
Each line of the input contains one of the possible requests; only the last line contains the stop-request (code 0). You may assume that when there is a request to include a new client in the list (code 1), there is no other request in the list of the same client or with the same priority. An identifier K is always less than 106, and a priority P is less than 107. The client may arrive for being served multiple times, and each time may obtain a different priority.
Output
For each request with code 2 or 3, the program has to print, in a separate line of the standard output, the identifier of the served client. If the request arrives when the waiting list is empty, then the program prints zero (0) to the output.
Sample Input
2
1 20 14
1 30 3
2
1 10 99
3
2
2
0
Sample Output
0
20
30
10
0
解题思路:
本题要求处理银行排队问题,根据每行的输入处理数据;若输入为1则继续输入两个整数,以第一个数为权值,第二个数为优先级(数值越大优先级越高)加入排队系统中;若输入为2 则处理当前优先级最高的用户,并输出用户的权值;若输入为3 则处理当前优先级最低的用户,并输出用户权值;若输入为0结束运行。若当没有排队人数则无论输入2、3都输出0。
利用set自带红黑树排序的性质,可以在集合中直接获取最高与最低优先级,用pair记录权值和优先级,将pair传入集合。用迭代器可以获得集合首位(最大优先级)和集合末尾(最小优先级),数据处理完毕后可以通过erase()删除当前用户以便后续操作。
样例分析:
Sample Input
2 //处理高优先级用户,当前无人排队输出0
1 20 14 //优先级为14的用户20开始排队
1 30 3 //优先级为3的用户30开始排队
2 //处理高优先级用户,当前队列中有 20 14 30 3 最高优先级14 输出20
1 10 99 //优先级为99的用户10开始排队
3 //处理低优先级用户,当前队列中有 10 99 30 3 最低优先级3 输出30
2 //处理高优先级用户,当前队列中有 10 99 最高优先级99 输出10
2 //处理高优先级用户, 当前无人排队输出0
0 //输入为0结束
#include <cstdio>
#include <iterator>
#include <set>
//bits/stdc++.h编译错误
using namespace std;
//利用set的排序性质
struct cmp{ //set比较方法以pair的第二个成员为基准,数值越大的在前
bool operator() (const pair<int, int> &a, const pair<int, int> &b)
{
return a.second > b.second;
}
};
set<pair<int, int>, cmp> waitingList; //用set记录当前队列
int main()
{
int n;
while(scanf("%d", &n) != EOF){ //输入数据
if(n == ) //输入为0结束运行
break;
if(n == ){ //输入为1
pair<int, int> temp;
scanf("%d%d", &temp.first, &temp.second); //记录用户权值与优先级
waitingList.insert(temp); //用户开始排队
}
if(n == ){ //输入为2
if(!waitingList.empty()){
set<pair<int, int>, cmp>::iterator it = waitingList.begin();
//获取优先级最高的用户
printf("%d\n", (*it).first); //输出用户权值
waitingList.erase(it); //删除用户
}else{
printf("0\n"); //队中无人
}
}
if(n == ){
if(!waitingList.empty()){
set<pair<int, int>, cmp>::iterator it = waitingList.end();
it--;
//获取优先级最小的用户
printf("%d\n", (*it).first); //输出用户权值
waitingList.erase(it); //删除用户
}else{
printf("0\n");//队中无人
}
}
}
return ;
}
POJ 3481 Double Queue(set实现)的更多相关文章
- POJ 3481 Double Queue STLmap和set新学到的一点用法
2013-08-08 POJ 3481 Double Queue 这个题应该是STL里较简单的吧,用平衡二叉树也可以做,但是自己掌握不够- -,开始想用两个优先队列,一个从大到小,一个从小到大,可是 ...
- POJ 3481 Double Queue(Treap模板题)
Double Queue Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 15786 Accepted: 6998 Des ...
- POJ 3481 Double Queue(STL)
题意 模拟银行的排队系统 有三种操作 1-加入优先级为p 编号为k的人到队列 2-服务当前优先级最大的 3-服务当前优先级最小的 0-退出系统 能够用stl中的map 由于map本身 ...
- POJ 3481 Double Queue
平衡树.. 熟悉些fhq-Treap,为啥我在poj读入优化不能用啊 #include <iostream> #include <cstdio> #include <ct ...
- POJ 3481 Double Queue (treap模板)
Description The new founded Balkan Investment Group Bank (BIG-Bank) opened a new office in Bucharest ...
- poj 3841 Double Queue (AVL树入门)
/****************************************************************** 题目: Double Queue(poj 3481) 链接: h ...
- hdu 1908 Double Queue
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1908 Double Queue Description The new founded Balkan ...
- 【Map】Double Queue
Double Queue Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 13258 Accepted: 5974 Des ...
- poj 2259 Team Queue
Team Queue Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 2977 Accepted: 1092 Descri ...
随机推荐
- [Erlang24]使用zotonic搭建网站记录
zotonic的搭建网站(blog)记录: zotonic:用Erlang做的一个web 框架: 和wordpress 类似,但是官网称比PHP CMS要快10倍以上 先看看我的成果:正弦 ...
- Windows服务器管理与优化
一.服务器自动重启windows服务器运行时间长了,内存会爆满,比如数据库会缓存大量的数量,IIS进程也会缓存数据而没有及时释放.这样需要定时重启服务器来释放内存. 创建任务计划,如在 每周一/周三/ ...
- Katalon Studio简单使用(二)
距离上一篇更新katalon学习部分已有两个月的时间 ,我的博文的访问量为400多,(*^__^*) 嘻嘻…… 说明还是很多同学在学习这个小tools的.所以再记录下 近两个月来对katalon的体验 ...
- deepin获取root权限
ctrk+alt+t 打开终端 输入 sudo passwd root mywork@mywork-PC:~$ sudo passwd root[sudo] mywork 的密码: [sudo] 输入 ...
- djang系列5.5-- 图书管理系统实例
一.表格设计 E-R图 分析图 models.py from django.db import models # Create your models here. class Author(model ...
- Syncthing源码解析 - 启动过程
我相信很多朋友会认为启动就是双击一下Syncthing程序图标,随后就启动完毕了!如果这样认为,对,也不对!对,是因为的确是这样操作,启动了Syncthing:不对是因为在调试Syncthing启动过 ...
- 深入了解java虚拟机(JVM) 第十一章 类的加载
一.类加载机制概述 虚拟机把描述类的数据从class文件加载到内存并对数据进行效验,解析和初始化,最终形成可以被虚拟机直接使用的java类型,这就是虚拟机的类加载机制. 二.类加载的机制 类加载的过程 ...
- 【Oracle 12c】CUUG OCP认证071考试原题解析(35)
35.choose the best answer View the Exhibit and examine the description of the EMPLOYEES table. Evalu ...
- 【OCP认证12c题库】CUUG 071题库考试原题及答案(26)
26.choose two Examine the structure of the PRODUCTS table. Which two statements are true? A) EXPIRY_ ...
- AI 的下一个重大挑战:理解语言的细微差别
简评:人类语言非常博大精妙,同一句话在不同的语境下,就有不同的含义.连人类有时候都不能辨别其中细微的差别,机器能吗?这就是人工智能的下一个巨大挑战:理解语言的细微差别.本文原作者是 Salesforc ...