【杭电oj】1509 - Windows Message Queue（优先队列）

Windows Message Queue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4979 Accepted Submission(s): 1993

Problem 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!

Author

ZHOU, Ran

Source

Zhejiang University Local Contest 2006, Preliminary

优先队列的题，这道题在杭电上需要用一个变量记录输入的时间，在ZOJ则不需要。

代码如下：

#include <cstdio>
#include <algorithm>
#include <queue>
#include <stack>
#include <cstring>
using namespace std;
struct node
{
	char name[22];
	int v,k,t;
	bool friend operator<(node a,node b)
	{
		if (a.v==b.v)
			return a.t>b.t;
		else
			return a.v>b.v;
	}
}a;
int main()
{
	priority_queue<node>q;
	char op[4];
	int num=1;		//记录输入时间 
	while (~scanf ("%s",op))
	{
		if (strcmp(op,"PUT")==0)
		{
			scanf ("%s %d %d",a.name,&a.k,&a.v);
			a.t=num++;
			q.push(a);
		}
		else
		{
			if (q.empty())
				printf ("EMPTY QUEUE!\n");
			else
			{
				a=q.top();
				printf ("%s %d\n",a.name,a.k);
				q.pop();
			}
		}
	}
	return 0;
}