H - Windows Message Queue

最新推荐文章于 2019-07-20 19:09:15 发布

原创最新推荐文章于 2019-07-20 19:09:15 发布 · 199 阅读

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本文介绍了一个基于优先级的消息队列模拟程序，该程序能够处理多达60,000条命令，包括添加（PUT）和获取（GET）消息操作。文章提供了完整的C++代码实现，并解释了如何通过重载比较运算符来实现优先级队列的功能。

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Description

Message queue is the basic fundamental of windows system. For each process, the system maintains a message queue. If something happens to this process, such as mouse click, text change, the system will add a message to the queue. Meanwhile, the process will do a loop for getting message from the queue according to the priority value if it is not empty. Note that the less priority value means the higher priority. In this problem, you are asked to simulate the message queue for putting messages to and getting message from the message queue.

Input

There's only one test case in the input. Each line is a command, "GET" or "PUT", which means getting message or putting message. If the command is "PUT", there're one string means the message name and two integer means the parameter and priority followed by. There will be at most 60000 command. Note that one message can appear twice or more and if two messages have the same priority, the one comes first will be processed first.(i.e., FIFO for the same priority.) Process to the end-of-file.

Output

For each "GET" command, output the command getting from the message queue with the name and parameter in one line. If there's no message in the queue, output "EMPTY QUEUE!". There's no output for "PUT" command.

Sample Input

GET
PUT msg1 10 5
PUT msg2 10 4
GET
GET
GET

Sample Output

EMPTY QUEUE!
msg2 10
msg1 10
EMPTY QUEUE!
#include <iostream>
#include <queue>
#include <cstdio>
using namespace std;

struct cmp
{
    char s[30];
    int n1,n2;
    bool operator <(const cmp&n)const
    {
        return n.n2<n2;
    }
};
int main()
{
    priority_queue<cmp>pq;
    char commed[20];
    cmp message;
  while(scanf("%s",commed)!=EOF)
    {

        if(commed[0]=='G')
        {
            if(pq.empty())
                printf("EMPTY QUEUE!\n");
            else
            {
                printf("%s %d\n",pq.top().s,pq.top().n1);
                pq.pop();
            }
        }
        else if(commed[0]=='P')
        {
            scanf("%s%d%d",message.s,&message.n1,&message.n2);
            pq.push(message);
        }

    }
    return 0;
}
此题一看便知是优先队列 重载一下<运算符便可 把原来的元素删掉 ，要定义一个新的结构来承接要添进去的数字 然后把它加进去。