題目描述：

F - Team Queue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2645 Accepted Submission(s): 910

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

這個題目用到了map函數和隊列的問題。

map: 就是通過（key）來找（value），就是對於每個不同的鑰匙都有相對應的值。就是數學中的函數。map函數的頭文件#include<map> 定義方式：map<鑰匙類型，值的類型>名稱

接下來就是隊列，隊列不同與棧隊列是先進先出更加符合我們的要求。隊列的頭文件：#include <queue>.隊列的定義當式也和棧差不多：queue<類型>名稱 隊列裏面的函數和也是差不多的，除了取首元素，棧用的是top()而隊列是front()。

我們這題用map函數來對各團隊進行分類，用編號作為鑰匙，所在的團隊作為value值。而定義一個隊列來儲存團隊的排序，一個數組隊列來表示團隊隊員的排序。剩下的就不細講了。看看代碼吧。

AC代碼：

 1 #include <iostream>
 2 #include <map>
 3 #include <queue>
 4 #include <string.h>
 5 const int Max=1010;
 6 using namespace std;
 7 int main()
 8 {
 9     int n,Scenario=1;
10     while(cin>>n&&n)
11     {
12         long long x;
13         map <int ,int > team;
14         for(int i=0;i<n;i++)
15         {
16             int t;
17             cin>>t;
18             for(int k=0;k<t;k++)
19                 {
20                     cin>>x;
21                     team[x]=i;
22                 }
23         }
24         cout<<"Scenario #"<<Scenario<<endl;
25         Scenario++;
26        queue <int> p,q[Max];
27        while(1)
28        {
29            string a;
30            cin>>a;
31            if(a[0]==‘E‘)
32            {
33                int t;
34               cin>>x;
35                t=team[x];
36                if(q[t].empty()) p.push(t);
37                q[t].push(x);
38            }
39            else if(a[0]==‘D‘)
40            {
41                int t=p.front();
42                cout<<q[t].front()<<endl;
43                q[t].pop();
44                if(q[t].empty()) p.pop();
45            }
46            else if(a[0]==‘S‘) break;
47        }
48        cout<<endl;
49     }
50     return 0;
51 }

Week 1 # F Team Queue