hdu 1387（Team Queue） STL

Posted 2020-08-25 heaventouch

Team Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2051    Accepted Submission(s): 713

Problem Description

Queues and Priority Queues are data structures which are known to most computer scientists. The Team Queue, however, is not so well known, though it occurs often in everyday life. At lunch time the queue in front of the Mensa is a team queue, for example. 
In a team queue each element belongs to a team. If an element enters the queue, it first searches the queue from head to tail to check if some of its teammates (elements of the same team) are already in the queue. If yes, it enters the queue right behind them. If not, it enters the queue at the tail and becomes the new last element (bad luck). Dequeuing is done like in normal queues: elements are processed from head to tail in the order they appear in the team queue. 

Your task is to write a program that simulates such a team queue. 

 

 

Input

The input will contain one or more test cases. Each test case begins with the number of teams t (1<=t<=1000). Then t team descriptions follow, each one consisting of the number of elements belonging to the team and the elements themselves. Elements are integers in the range 0 - 999999. A team may consist of up to 1000 elements. 

Finally, a list of commands follows. There are three different kinds of commands: 

ENQUEUE x - enter element x into the team queue 
DEQUEUE - process the first element and remove it from the queue 
STOP - end of test case 
The input will be terminated by a value of 0 for t. 

 

 

Output

For each test case, first print a line saying "Scenario #k", where k is the number of the test case. Then, for each DEQUEUE command, print the element which is dequeued on a single line. Print a blank line after each test case, even after the last one. 

 

 

Sample Input

2 3 101 102 103 3 201 202 203 ENQUEUE 101 ENQUEUE 201 ENQUEUE 102 ENQUEUE 202 ENQUEUE 103 ENQUEUE 203 DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 2 5 259001 259002 259003 259004 259005 6 260001 260002 260003 260004 260005 260006 ENQUEUE 259001 ENQUEUE 260001 ENQUEUE 259002 ENQUEUE 259003 ENQUEUE 259004 ENQUEUE 259005 DEQUEUE DEQUEUE ENQUEUE 260002 ENQUEUE 260003 DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 0

 

 

Sample Output

Scenario #1 101 102 103 201 202 203 Scenario #2 259001 259002 259003 259004 259005 260001

 

 

Source

University of Ulm Local Contest 1998

 

 

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排队问题，一个term为一个单位，前面如果有你的term你可以直接进term的队尾，没有你就在整个队最后，自己新建自己的term并排在队尾。相当于双重队列

开始直接用两个队列模拟过程 TLE 了，换一个做法，有两个操作：
1、进队：
　　用map匹配先队列中有没有你的term，有的话在你排在 term最后，没有的话新建term，并排在整体最后；

2、出队

　　直接取整体第一个term，并取出第一个人，取完后判断term是否为空，为空解散；

 

 1 //608MS    4544K    1051B    G++
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<string.h>
 5 #include<queue>
 6 #include<map>
 7 #include<vector>
 8 using namespace std;
 9 const int N = 1005;
10 typedef queue<int> INTQ;
11 
12 int main()
13 {
14     int n,m,a;
15     char op[10];
16     int cas=1;
17     while(cin>>n && n){
18         map<int ,int>TERM;
19         for(int i=1;i<=n;i++){
20             cin>>m;
21             for(int j=0;j<m;j++){
22                 cin>>a;
23                 TERM[a]=i;
24             }
25         }
26         
27         INTQ V[N];
28         map<int, int>M2V;
29         int index=1;
30         int head_index=index;
31         cout<<"Scenario #"<<cas++<<endl;
32         while(cin>>op){
33             if(strcmp(op, "STOP")==0){
34                 break;
35             }
36             else if(strcmp(op, "ENQUEUE")==0){
37                 cin>>a;
38                 if(M2V[TERM[a]]==0){
39                     INTQ tQ;
40                     tQ.push(a);
41                     V[index] =  tQ;
42                     M2V[TERM[a]] = index;
43                     index = (index+1)%N;
44                 }else{
45                     V[M2V[TERM[a]]].push(a);
46                 }
47             
48             }else{
49                 int a=V[head_index].front();
50                 V[head_index].pop();
51                 cout<<a<<endl;
52                 if(V[head_index].empty()){
53                     M2V[TERM[a]] = 0;
54                     head_index = (head_index+1)%N;
55                 }
56             }    
57 
58         }
59         cout<<endl;
60     }
61     return 0;    
62 }

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