Double Queue

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 116    Accepted Submission(s): 68

Problem 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:

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 10 6, and a priority P is less than 10 7. 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
Source
 
 
解题：
比较长的题目，惹自己少了耐心去看，其实后来才发现题目意思并不难。当code为1的时候，则K在排队，他的优先权是P。
当code为2的时候，从优先权高的先行出列处理。而code为3的时候，则从优先权低的先行出列。这题可以用STL的map做，自动
排序，但是时间会比较慢一点；也可以用二分法去做。
#pragma warning(disable:4786) #include <iostream> #include <map> using namespace std; int main() { int n,k,p; map<int ,int > aQueue; while (scanf("%d",&n)!=EOF && n!=0) { if (n==1) { cin>>k>>p; aQueue.insert(map<int ,int>::value_type(p,k)); } else if (n==3) { if (!aQueue.empty()) { cout<<aQueue.begin()->second<<endl; aQueue.erase(aQueue.begin()); } else { cout<<0<<endl; } } else { if (!aQueue.empty()) { cout<<aQueue.rbegin()->second<<endl; aQueue.erase(aQueue.find(aQueue.rbegin()->first)); } else { cout<<0<<endl; } } } return 0; }