Team Queue. id:540

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Queues and Priority Queues are data structures which are known to most computer scientists. TheTeam 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 itsteammates (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 teamst ( $1 \le t \le 1000$). 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", wherek 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 

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

Sample Output 

Scenario #1101102103201202203Scenario #2259001259002259003259004259005260001


分析:

用到两个队列,一个是每个团队自己是一个队列,而每个团队整体之间又构成一个队列。

#include <iostream>#include <queue>#include <cstdio>#include <cstring>#include <map>#define maxn 1010using namespace std;int main(){    int kase = 0, t;    while(scanf("%d", &t) == 1 && t) {        printf("Scenario #%d\n", ++kase);        queue<int> q;                       //表示由团队组成的队列        queue<int> team_q[maxn];    //team_q[i]表示编号为i的团队的队列         map<int, int> team;             //team[i]表示在队列中编号为i的成员所在的团队编号        for(int i = 0; i < t; i++) {            //之所以用map容器而不直接用数组来映射的原因是用int数组可能会导致内存开销过大            int n, x;                                   //提交到oj上会产生runtime error            scanf("%d", &n);            while(n--) {                scanf("%d", &x);    //在团队中的编号为x的成员                team[x] = i;            //将其所属团队的编号设置为i            }        }        while(1) {            char cmd[10];            scanf("%s", cmd);            if(cmd[0] == 'S')   break;            else if(cmd[0] == 'E') {                int x;                scanf("%d", &x);                int i = team[x];    //找到编号为x的队列成员所属的团队编号                if(team_q[i].empty())   q.push(i);                team_q[i].push(x);            } else if(cmd[0] == 'D') {                int x = q.front();      //找到队列中最前列即将出列的成员所属团队的编号                printf("%d\n", team_q[x].front());                team_q[x].pop();                if(team_q[x].empty())   q.pop();    //如果出列的是该团队的最后一名成员,则在q中将整个团队pop()            }        }        printf("\n");    }    return 0;}


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