Team Queue（多队列技巧处理）

Team Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2009    Accepted Submission(s): 696

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

 题解：题意就是一个队列相同组的可以插队，插到相同队员的后面，思路是弄多个队列，用一个队列来确定队编号的序列，一个数组判断当前队是否还有队员，没有队员标记为空；

代码：

import java.util.HashMap;
import java.util.LinkedList;
import java.util.Scanner;


public class hdu1387 {
    static Scanner cin = new Scanner(System.in);
    private static HashMap<Integer, Integer>mp = new HashMap<Integer, Integer>();
    private static LinkedList<Integer>q[] = new LinkedList[1010];
    private static int vis[] = new int[1010];
    static{
        for(int i = 0; i < q.length; i++){
            q[i] = new LinkedList<Integer>();
        }
    }
    public static void main(String[] args) {
        int t, n, e, i, kase = 1;
        while((t = cin.nextInt()) != 0){
            init(t);
            for(i = 0; i < t; i++){
                n = cin.nextInt();
                while(n-- > 0){
                    e = cin.nextInt();
                    mp.put(e, i);
                }
            }
            String command;
            System.out.println("Scenario #" + kase++);
            while(!(command = cin.next()).equals("STOP")){
                if("ENQUEUE".equals(command)){
                    e = cin.nextInt();
                    i = mp.get(e);
                    q[i].add(e);
                    if(vis[i] == 0){
                        q[t].add(i);
                        vis[i] = 1;
                    }
                    //0 1 0 0 1 0 1 1
                }
                else{
                    i = q[t].getFirst();
                    while(q[i].isEmpty()){
                        q[t].poll();
                        vis[i] = 0;
                        i = q[t].getFirst();
                    }
                    e = q[i].getFirst();
                    q[i].poll();
                    System.out.println(e);
                }
            }
            System.out.println();
        }
        
    }
    private static void init(int t) {
        for(int i = 0; i <= t; i++){
            q[i].clear();
        }
        for(int i = 0; i < 1010; i++){
            vis[i] = 0;
        }
        mp.clear();
    }
}