• F


    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.

    he 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

    题目意思:

    以一些数为一个团队,在排队时,如果有团队的成员,就会排到自己团队的成员后面,否则排到队伍的后面

    ENQUEUE XXXX

    使XXXX进入队列,

    DEQUEUE

    使当前队列的第一个出队;

    STOP

    结束

    解法:

    首先用map给每个团队编号;

    在执行命令的时候,用一个队列储存团队编号的先后,用一个队列数组来储存每个团队成员

     

     1 #include <iostream>
     2 #include <queue>
     3 #include <string>
     4 #include <string.h>
     5 #include <stdio.h>
     6 #include <map>
     7 
     8 using namespace std;
     9 
    10 const int MAX = 1000 + 500;
    11 int N;
    12 int N0;
    13 
    14 
    15 int main()
    16 {
    17     N0 = 1;
    18    while(scanf("%d",&N) == 1&&N)
    19    {
    20        cout<<"Scenario #"<<N0++<<endl;
    21 
    22        map< int ,int > a;
    23 
    24        for(int i = 0;i < N;i++)
    25        {
    26              int n,temp;
    27              cin>>n;
    28              while(n--)
    29            {
    30                    cin>>temp;
    31                    a[temp] = i;
    32            }
    33        }
    34 
    35        queue< int > P1,P2[MAX];
    36        for(;;)
    37        {
    38            char str[10];
    39            scanf("%s",str);
    40            if( str[0] == 'S') break;
    41             else if( str[0] == 'E')
    42            {
    43                int temp,t;
    44                 cin >> temp;
    45                 t = a[temp];
    46                 if( P2[t].empty() ) P1.push(t);
    47                 P2[t].push(temp);
    48 
    49            }
    50             else  if( str[0] == 'D')
    51            {
    52                 int t;
    53                 t = P1.front();
    54                 cout<<P2[t].front()<<endl;
    55                 P2[t].pop();
    56                 if(P2[t].empty()) P1.pop();
    57             }
    58 
    59        }
    60        cout<<endl;
    61 
    62    }
    63     return 0;
    64 }

     

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  • 原文地址:https://www.cnblogs.com/a2985812043/p/7211910.html
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