Windows Message Queue 优先队列

题目

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!

题意

给你最多60000条指令，"GET"表示获取指令，如果队列为空，输出"EMPTY QUEUE!"。"PUT"表示放置指令，每条指令都有一个参数和一个表示优先级的数，且如果优先级数越大，优先级越小，如果优先级相同，则先出现的消息优先被处理。

分析

优先队列，注意优先级相同时，先出现的消息优先被处理，所以记录一下每条消息出现的时间，在结构体内重载一下'<'就可以了（写这个题就是为了记个模板）。

代码

#include
#include
#include
#include
using namespace std;
#define inf 0x3f3f3f
const int N=6e4+5;
struct node{
	char s[1005];
	int can,id,time;
	bool operator < (const node &a) const//重载'<' 
	{
		if(id==a.id)
			return time>a.time;
		return id>a.id;
	}
}num;
int main()
{
	priority_queue q;
	char t[5];
	int cas=1;
	while(~scanf("%s",t))
	{
		if(t[0]=='G')
		{
			if(q.empty())
				printf("EMPTY QUEUE!\n");
			else
			{
				node tmp=q.top();
				q.pop();
				printf("%s %d\n",tmp.s,tmp.can);
			}
		}
		else if(t[0]=='P')
		{
			scanf("%s",num.s);
			scanf("%d%d",&num.can,&num.id);
			num.time=cas++;
			q.push(num);
		}
	}
	return 0;
}