7.18

Team Queue

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

23 101 102 1033 201 202 203ENQUEUE 101ENQUEUE 201ENQUEUE 102ENQUEUE 202ENQUEUE 103ENQUEUE 203DEQUEUEDEQUEUEDEQUEUEDEQUEUEDEQUEUEDEQUEUESTOP25 259001 259002 259003 259004 2590056 260001 260002 260003 260004 260005 260006ENQUEUE 259001ENQUEUE 260001ENQUEUE 259002ENQUEUE 259003ENQUEUE 259004ENQUEUE 259005DEQUEUEDEQUEUEENQUEUE 260002ENQUEUE 260003DEQUEUEDEQUEUEDEQUEUEDEQUEUESTOP0

样例输出:

Scenario #1101102103201202203Scenario #2259001259002259003259004259005260001

解题思路： 首先需要明确的一点是：根据题意，可得 哪一支队伍先进来，那么在执行"D"操作时，首先出来的就是这支队伍的第一个队员，所以"先进先出"要用到队列；

又因为有多支队伍，每支队伍又有很多队员，所以要用到 一个队列的数组（用来每一组的队员），一个队列（用来存组号）；

因为又分了组，所以组员怎么确定？！   用map，

用到的数据结构有：队列(queue) ，map

代码如下：

#include <map>#include <queue>#include <iostream>#include<cstdio>using namespace std;const int Max =1005;int main(){    int t;    int kase=0;    while (~scanf("%d",&t) && t)    {        map<int,int> team;        int p,q,n,x;        for (int i=1;i<=t;i++)        {            scanf("%d",&n);            for (int j=0;j<n;j++)            {                scanf("%d",&x);                team[x]=i;            }        }        kase++;   printf("Scenario #%d\n",kase);   queue<int> q1;//用来存放组号   queue<int> q2[Max];//队列的一个数组，用来存放每一组的成员   for (;;)//题目说可能有20000组数据   {      char c[10];      scanf("%s",c);      if (c[0]=='S')        break;      else if (c[0]=='E')      {          int T;          int p;          scanf("%d",&p);          T=team[p];          if (q2[T].empty())//第几组的判断            q1.push(T);          q2[T].push(p);      }      else if (c[0]=='D')      {            int T;            T=q1.front();            printf("%d\n",q2[T].front());            q2[T].pop();            if (q2[T].empty())                q1.pop();      }  }  printf("\n");}  return 0;}