map常见用法——POJ 3481

对应POJ题目：点击打开链接

Double Queue

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10251 Accepted: 4658

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 priorityP. 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 thelowest priority instead of that with the highest priority. Thus, the system will receive the following types of request:

0The system needs to stop serving1 K PAdd client K to the waiting list with priority P2Serve the client with the highest priority and drop him or her from the waiting list3Serve 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 priorityP 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

21 20 141 30 321 10 993220

Sample Output

02030100

#include<cstdio>#include<cstdlib>#include<cmath>#include<map>#include<queue>#include<stack>#include<vector>#include<algorithm>#include<cstring>#include<string>#include<map>#include<iostream>using namespace std;const int MAXN=100050+10;const int INF=1<<25;int main(){//freopen("in.txt","r",stdin);int op;map<int,int>m;map<int,int>::iterator it;map<int,int>::reverse_iterator ii;while(scanf("%d", &op), op){int k,p;if(op==1){scanf("%d%d", &k,&p);m[p]=k;continue;}if(op==2){if(m.size()){ii=m.rbegin();cout<<ii->second<<endl;m.erase(m.rbegin()->first);}else cout<<0<<endl;continue;}if(op==3){if(m.size()){it=m.begin();cout<<it->second<<endl;m.erase(m.begin());}else cout<<0<<endl;}}return 0;}

count()            返回指定key出现的次数,只能是0,和1，so~其作用应该是判断是否存在此key

find()                查找一个key，返回该迭代器，如果没有，则返回尾部迭代器end();

insert ()           插入元素

max_size()     返回可以容纳的最大元素个数

swap()            交换两个map,   如：swap(m1,m2)    ||    m1.swap(m2);

upper_bound()      返回的是第一个大于这个key的iterator