Double Queue

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 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

题意就是规定了四种操作：
1表示输入优先级为P的数；
2表示输出优先级最高的并删除；
3表示输出优先级最低的并删除；
0表示操作结束。

代码：

#include
#include
#include
#include
using namespace std;
struct node {
	int id,p;
	bool operator <(const node &t ) const {	//结构体内嵌比较函数，括号中的const表示参数a对象不会被修改，最后的const表明调用函数对象不会被修改!
		return p<t.p; //从小到大排序
	}
};
int main() {
	set s; //定义一个set集合
	set ::iterator it; //遍历元素
	int op;
	while(~scanf("%d",&op)&&op) {
		node e;
		switch(op) {
			case 1: {
				cin>>e.id>>e.p;
				s.insert(e); //插入
				break;
			}
			case 2: {
				if(s.empty()) {
					puts(“0”);
					break;
				}
				it=–s.end(); //我不知道这是什么意思，为什么要–
				printf("%d\n",it->id);
				s.erase(it); //删除
				break;
			}
			case 3: {
				if(s.empty()) {
					puts(“0”);
					break;
				}
				it=s.begin();
				printf("%d\n",it->id);
				s.erase(it);
				break;
			}
		}
	}
}