UVA540 TeamQueue【map+queue】

题面：

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 file 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. 
Warning: A test case may contain up to 200000 (two hundred thousand) commands, so the implementation 
of the team queue should be efficient: both enqueing and dequeuing of an element should 
only take constant time. 
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

题目大意：

现在需要构建一个特殊的队列，这个队列的特点是当一个元素入队时，先在队列里看是否有这个元素的队友，如果有就将这个元素放在最后一个队友的后面，如果没有就放在队尾。要求有较低的时间复杂度。

大致思路：

可以利用map将一个队中的所有元素标记。并且每一队都有一个专门的队列，用来放入队的元素。还有一个主队列，里面放的都是不同队的元素。 
当入队时，先根据对应的map值看其专门队里是否为空，如果不为空就入其专门的队列，如果为空不光入专门的队，还要入主队。 
出队时，从专门的队中出元素，直到专门的队空时，才从主队中将这个元素剔除。

代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    //freopen("in.txt","r",stdin);
    char str[50];
    int t,n,x,cnt=1;
    while(cin>>t&&t)
    {
        map<int,int> mp;
        queue<int> ma,team[1010];//主队和专门的队
        for(int i=0;i<t;++i){
            cin>>n;
            for(int j=0;j<n;++j){
                cin>>x;
                mp[x]=i;
            }
        }
        cout<<"Scenario #"<<cnt<<endl;
        while(1)
        {
            cin>>str;
            if(str[0]=='S')
                break;
            if(str[0]=='E'){
                cin>>x;
                if(team[mp[x]].empty())//看专门的队是否为空
                    ma.push(x);
                team[mp[x]].push(x);
            }else{
                x=ma.front();
                cout<<team[mp[x]].front()<<endl;
                team[mp[x]].pop();
                if(team[mp[x]].empty())//专门队为空再出主队
                    ma.pop();
            }
        }
        cnt++;
        cout<<endl;
    }
    return 0;
}