题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=481
Time limit 1000 ms Memory limit 32768 kB

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 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

Problem solving report:

Description: 给出n个队列，当出现ENQUEUE指令的时候，就将ENQUEUE后面的那个数据加入队伍当中，加入队伍的规则是：如果有同属一个队列的数据已经在队伍里，那么就直接加到这个已经在队伍里的那个数据的右边，如果一个都没有，那么就将这个数据加入到队伍的末尾。
Problem solving: 以每只队伍建立队列，再以队伍序号建立一个队列，将每个元素入到它所属的队伍队列中，若序号队列没有此队伍的序号，则将序号入队，当某个队伍队列为空时，将序号队列中的此序号出队。

Accepted Code:

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1005;
int main() {
    char str[10];
    int n, m, s, kase = 0;
    while (scanf("%d", &n), n) {
        map <int, int> spt;
        queue <int> Q, Q1[MAXN];
        for (int i = 0; i < n; i++) {
            scanf("%d", &m);
            while (m--) {
                scanf("%d", &s);
                spt[s] = i;
            }
        }
        printf("Scenario #%d\n", ++kase);
        while (scanf("%s", str)) {
            if (!(str[0] - 'S'))
                break;
               if (str[0] != 'D') {
                   scanf("%d", &m);
                   if (Q1[spt[m]].empty())
                       Q.push(spt[m]);
                   Q1[spt[m]].push(m);
            }
            else {
                m = Q.front();
                printf("%d\n", Q1[m].front());
                Q1[m].pop();
                if (Q1[m].empty())
                    Q.pop();
            }
        }
        printf("\n");
    }
    return 0;
}