POJ 3481 & HDU 1908 Double Queue (map运用)

题目链接：

PKU:http://poj.org/problem?id=3481

HDU:http://acm.hdu.edu.cn/showproblem.php?pid=1908

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

Source

Southeastern Europe 2007

题意：

三个操作：

1、插入两个数，第一个数是客户的名字，第二个数是优先级！

2、输出当前客户中优先级最高的名字，并删除！

3、输出当前客户中优先级最低的名字，并删除！

代码如下：

#include <cstdio>
#include <map>
#include <cstring>
#include <algorithm>
using namespace std;
#include <string>
map<int,int> m;
int main()
{
    int op;
    int name, p;
    while(~scanf("%d",&op))
    {
        if(op == 0)
            break;
        if(op == 2 && m.size() == 0)
        {
            printf("0\n");
            continue;
        }
        if(op == 3 && m.size() == 0)
        {
            printf("0\n");
            continue;
        }
        if(op == 1)
        {
            scanf("%d%d",&name,&p);
            m[p] = name;
            continue;
        }
        if(op == 2)
        {
            printf("%d\n",(*m.rbegin()).second);
            m.erase(m.find((*m.rbegin()).first));
            //printf("%d\n",(*m.end()).second);
            //m.erase(m.end());
            //注意此处不能用*m.end(),因为*m.end()返回的是数组的最后一个单元+1的指针
        }
        else
        {
            //printf("%d\n",(*m.rend()).second);//同样
            //m.erase(m.find((*m.rend()).first));
            printf("%d\n",(*m.begin()).second);
            m.erase(m.begin());
        }
    }
    return 0;
}