題干：

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.

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Input

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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.

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Output

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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.

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Sample Input

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

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Sample Output

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EMPTY QUEUE! msg2 10 msg1 10 EMPTY QUEUE!

解題報告：

? ?優先隊列直接解。剛開始數組開太大了瘋狂超內存。我還在node上加了個[MAX] ?出現了這個報錯 Segmentation Fault

AC代碼：

#include<bits/stdc++.h>using namespace std; const int MAX = 6e4 + 5; struct Node {int id;char name[400];int k;//參數int vip;//Node(int id,char name[MAX],int k,int vip):id(id),name(name),k(k),vip(vip){}bool operator<(const Node b) const {if(vip!=b.vip) return vip>b.vip;return id>b.id;} } node; priority_queue <Node> pq; char tmpname[MAX]; int main() {char op[10];int cnt = 0;int k,v;while(~scanf("%s",op) ) {if(op[0] == 'G') {if(!pq.empty() ) {Node cur = pq.top();pq.pop();printf("%s %d\n",cur.name,cur.k);}else printf("EMPTY QUEUE!\n");}else {cnt++;scanf("%s%d%d",tmpname,&k,&v);Node tmp;tmp.id = cnt;tmp.k = k;tmp.vip = v;strcpy(tmp.name,tmpname);pq.push(tmp);}}return 0 ; }

總結

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