【Map】Double Queue
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 13258 | Accepted: 5974 |
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 |
1 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
题目大意:
有三个操作,1 K P:设置点K的优先级为P;
2 :输出优先级最高的点,并且删除该点
3 :输出优先级最低的点,并且删除该点
0:退出
用map处理,插入和查找的复杂度都为log n;
#include <iostream>
#include <algorithm>
#include <string>
#include <queue>
#include <stdio.h>
#include <map>
#include <string.h>
#include <stdlib.h>
using namespace std;
int main()
{
int T,a,P,K;
map<int,int>ID;
map<int,int>::iterator it;
while(scanf("%d",&a)!=EOF){
switch(a){
case :
scanf("%d%d",&P,&K);
ID[K]=P;break;
case :
if(ID.empty())printf("0\n");
else {
it=ID.end();it--;
printf("%d\n",it->second);
ID.erase(it);
}
break;
case :
if(ID.empty())printf("0\n");
else {
it=ID.begin();
printf("%d\n",it->second);
ID.erase(it);
}
break;
case : return ;
}
} return ;
}
最新文章
- GDB 完全教程
- XMPP环境搭建
- CSS3 里添加自定义字体
- springMVC返回json
- Dynamic CRM 2013学习笔记(十三)附件上传 / 上传附件
- blur效果,模糊效果css
- poj2104 线段树 划分树
- [JS进阶] 编写可维护性代码 (1)
- 李洪强iOS开发之OC[013] -类的创建的练习
- xss(跨站脚本攻击),crsf(跨站请求伪造),xssf
- Dreamweaver安装jQuery插件jQuery_API.mxp
- Keil C51中变量的使用
- BZOJ1782: [Usaco2010 Feb]slowdown 慢慢游
- SKPhysicsJointSpring类
- Node.js之操作文件系统(二)
- NOIP 2018 大翻车记
- python之路--装饰器
- 201621123002《JAVA程序设计》第五周学习总结
- BZOJ1202 [HNOI2005]狡猾的商人 spfa
- C++学习(十五)(C语言部分)之 数组二