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本身就依据key的值排了序   相应2。3  我们仅仅须要输出最大或最小即可了并从map中删除该键值

#include<cstdio>
#include<map>
using namespace std;
map<int, int> a;
int main()
{
map<int, int>::iterator it;
int n,k,p;
while (scanf ("%d", &n), n)
{
if (n == 1)
{
scanf ("%d%d", &k, &p);
a[p] = k;
}
else
{
if (a.empty()) printf ("0\n");
else
{
if(n==2) it = a.end(),--it;
else it=a.begin();
printf ("%d\n", it->second);
a.erase (it->first);
}
}
}
return 0;
}
Double Queue

Description

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 K and, 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
K P 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

POJ 3481 Double Queue(STL)的更多相关文章

  1. 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 ...

  2. POJ 3481 Double Queue(Treap模板题)

    Double Queue Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15786   Accepted: 6998 Des ...

  3. POJ 3481 Double Queue(set实现)

    Double Queue The new founded Balkan Investment Group Bank (BIG-Bank) opened a new office in Buchares ...

  4. 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 ...

  5. POJ 3481 Double Queue (treap模板)

    Description The new founded Balkan Investment Group Bank (BIG-Bank) opened a new office in Bucharest ...

  6. poj 3841 Double Queue (AVL树入门)

    /****************************************************************** 题目: Double Queue(poj 3481) 链接: h ...

  7. hdu 1908 Double Queue

    题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1908 Double Queue Description The new founded Balkan ...

  8. 【Map】Double Queue

    Double Queue Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13258   Accepted: 5974 Des ...

  9. queue STL

    //queue STL //queue is just a container adaptor, which is a class that use other container. //just l ...

随机推荐

  1. 关于linux上cron服务的python封装工具

    关于cron:定时任务服务,一般linux自带且已启动.(pgrep cron查看cron服务是否启动了.) 关于plan:一个通过python来定制cron服务的工具.其官网:http://plan ...

  2. Leetcode:Longest Substring Without Repeating Characters 解题报告

    Longest Substring Without Repeating Characters Given a string, find the length of the longest substr ...

  3. Java SWT编程基础

    SWT常用组件列表及使用 https://blog.csdn.net/u013310025/article/details/52939452 SWT编程基础-控件和图形资源 https://blog. ...

  4. Adventures in Functions

    速度还行,两天看完一章,就是有细节没去扣.书上的大部分知识点和代码都看了,这个还是可以的. 今天继续来学习函数的高级特性,要涉及到以下的主题. 内联函数(inline function) 引用变量(r ...

  5. Emacs代码折叠

    进入HideShow mode: M-x hs-minor-mode(幸亏有tab键..要不这么长的命令=.=) 主要的功能: * C-c @ C-M-s 显示所有的代码 * C-c @ C-M-h ...

  6. CAS (11) —— CAS TicketRegistry使用Ehcache的集群方案

    CAS (11) -- CAS TicketRegistry使用Ehcache的集群方案 摘要 CAS TicketRegistry使用Ehcache的集群方案 版本 tomcat版本: tomcat ...

  7. JAVA环境变量配置详解(Windows)

    JAVA环境变量配置详解(Windows)   JAVA环境变量JAVA_HOME.CLASSPATH.PATH设置详解  Windows下JAVA用到的环境变量主要有3个,JAVA_HOME.CLA ...

  8. sql 表连接基本的语法

    SQL连接能够分为内连接.外连接.交叉连接. 1.内连接:内连接查询操作列出与连接条件匹配的数据行,它使用比較运算符比較被连接列的列值. 1.1 select * from Table1 as a, ...

  9. 【异常】IOException parsing XML document from class path resource [xxx.xml]

    1.IDEA导入项目运行出现异常 org.springframework.beans.factory.BeanDefinitionStoreException: IOException parsing ...

  10. android 避免线程的重复创建(HandlerThread、线程池)

    最近在android开发中,用到都是new Thread(){...}.start()这种方式.本来这样是可以,但是最近突然爆出Performing stop of activity that is ...