POJ 2051 Argus

A data stream is a real-time, continuous, ordered sequence of items. Some examples include sensor data, Internet traffic, financial tickers, on-line auctions, and transaction logs such as Web usage logs and telephone call records. Likewise, queries over streams run continuously over a period of time and incrementally return new results as new data arrives. For example, a temperature detection system of a factory warehouse may run queries like the following.
Query-1: “Every five minutes, retrieve the maximum temperature over the past five minutes.”
Query-2: “Return the average temperature measured on each floor over the past 10 minutes.”

We have developed a Data Stream Management System called Argus, which processes the queries over the data streams. Users can register queries to the Argus. Argus will keep the queries running over the changing data and return the results to the corresponding user with the desired frequency.

For the Argus, we use the following instruction to register a query:
Register Q_num Period

Q_num (0 < Q_num <= 3000) is query ID-number, and Period (0 < Period <= 3000) is the interval between two consecutive returns of the result. After Period seconds of register, the result will be returned for the first time, and after that, the result will be returned every Period seconds.

Here we have several different queries registered in Argus at once. It is confirmed that all the queries have different Q_num. Your task is to tell the first K queries to return the results. If two or more queries are to return the results at the same time, they will return the results one by one in the ascending order of Q_num.

Input
The first part of the input are the register instructions to Argus, one instruction per line. You can assume the number of the instructions will not exceed 1000, and all these instructions are executed at the same time. This part is ended with a line of “#”.

The second part is your task. This part contains only one line, which is one positive integer K (<= 10000).

Output
You should output the Q_num of the first K queries to return the results, one number per line.

Sample Input
Register 2004 200
Register 2005 300
#
5

Sample Output
2004
2005
2004
2004
2005

题意：
给出用户ID和出现的时间间隔，以‘#’结束，输入一个正整数K，输出前K次出现的用户ID
思路：
创建一个user 结构体，存储用户ID，间隔时间和下一次出现的时间并重载运算符’<’，将user对象添加到优先队列中，利用优先队列的自动排序，将最前出现的放在最前面，每次pop（）出的对象后将其删除，重置下一次出现的时间后再添加进队列

#include <iostream>#include <queue>using namespace std;int K;struct User//定义一个结构体，储存用户ID，间隔时间，和下一次出现的时间{    int IdMumber;    int period;    int timeOfNext;    friend bool operator < (User u2,User u1)//重载'<'运算符，利用优先队列的自动排序，将先出现的排在队列最前面    {        if(u1.timeOfNext!=u2.timeOfNext)            return u1.timeOfNext<u2.timeOfNext;        else            return u1.IdMumber<u2.IdMumber;    }};priority_queue<User> Que;//声明优先队列void process(){    User u;    for(int i=0;i<K;i++)    {        u=Que.top();//每次取出队列中第一个，即最先出现的用户        Que.pop();//删除第一个        cout<<u.IdMumber<<endl;        u.timeOfNext+=u.period;//重置下次出现的时间        Que.push(u);//添加进队列，自动排序    }}int main(){    string str;    int id,time;    User u;    while(cin>>str&&str!="#")    {        cin>>id>>time;        u.IdMumber=id;        u.period=time;        u.timeOfNext=time;        Que.push(u);    }    cin>>K;    process();    return 0;}