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!
Source: Zhejiang University Local Contest 2006, Preliminary
写的代码在zoj系统里面提交过了,但在virdual judge里面却没过
题意:有两种操作:put 把信息放入 GET把信息输出
#include<stdio.h> #include<string.h> #include<queue> #include<algorithm> using namespace std; struct node//队列重载 { int priority; char name[1100]; int can; bool operator<(const node a)const { return a.priority<priority; } }; int main() { char com[200]; priority_queue<node>q; while(~scanf("%s",com)) { node t; if(strcmp(com,"PUT")==0)//输入信息 { scanf("%s %d %d",t.name,&t.can,&t.priority); q.push(t); } else//输出信息 { if(q.size()==0) printf("EMPTY QUEUE!\n"); else { printf("%s %d\n",q.top().name,q.top().can); q.pop(); } } } return 0; }