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HDU1509-Windows Message Queue

2014年09月05日 ⁄ 综合 ⁄ 共 2077字 ⁄ 字号 评论关闭

Windows Message Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1837    Accepted Submission(s): 713

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
 
题意:windows 有很多操作,每个操作都有一个优先级和操作序号,优先权小的优先级较高,优先权相同时序号越小越提前。
用优先队列存储信息!
 

#include<stdio.h>
#include<iostream>
#include<queue>
using namespace std;
struct node
{
    friend bool operator < (node n1,node n2)
    {
        if(n1.priority==n2.priority)
            return n1.id>n2.id;
        else return
            n1.priority>n2.priority;
    } 
    int id;
    int value;
    int priority;
    char s[50];
};
int main()
{
   
    int k=1;
    char s[5];
    priority_queue<node>q;
    struct node p;
    while(scanf("%s",s)!=EOF)
    {
        if(s[0]=='P')
        {
            scanf("%s %d %d",p.s,&p.value,&p.priority);
            p.id=k++;
            q.push(p);
        }
        else
        {
            if(!q.empty())
            {
                p=q.top();
                printf("%s %d\n",p.s,p.value);
                q.pop();
            }
            else printf("EMPTY QUEUE!\n");
        }
    }
    return 0;
}

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