题目代号：UVA 540
题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=481
题目描述：
Team Queue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2645 Accepted Submission(s): 910
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
题目大意：给你有n个队伍。 对于每个ENQUEUE x 命令。 如果x所在的队伍已经在队列中， 则x排在队列中它的队伍的尾巴， 否则排在队列的末尾。 可以理解为队列中的队列的味道。对于DEQUEUE命令输出总队列中第一个小组的第一个元素。对于STOP命令，代表停止命令的输入。
解题思路：代码中的注释有详细描述。
AC代码：
# include <stdio.h> # include <string.h> # include <stdlib.h> # include <iostream> # include <fstream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <math.h> # include <algorithm> using namespace std; # define pi acos(-1.0) # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define For(i,n,a) for(int i=n; i>=a; --i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define Fo(i,n,a) for(int i=n; i>a ;--i) typedef long long LL; typedef unsigned long long ULL; char str[10]; int team[1000005];//最大编号不会超过999999所以用这个来表示编号所属的小组号 int main() { int n,t,num,cases=1; while(scanf("%d",&t),t) { mem(team,0); queue<int>Q[1005]; queue<int>que;//这样定义在while内就不需要通过循环清空队列，循环一次队列就初始化一次 for(int i=1; i<=t; i++) { scanf("%d",&n); for(int j=1; j<=n; j++) { scanf("%d",&num); team[num]=i;//初始化该编号成员所属小组 } } printf("Scenario #%d\n",cases++); do { scanf("%s",str); if(!strcmp(str,"ENQUEUE")) { scanf("%d",&num); if(Q[team[num]].empty())que.push(team[num]);//如果该编号成员所属小组的队列没有人，则在总队列尾部添加该小组 Q[team[num]].push(num);//小组所属的队列尾部添加该成员 } else if(!strcmp(str,"DEQUEUE")) { printf("%d\n",Q[que.front()].front());//输出总队列中首部的小组所对应的第一个成员 Q[que.front()].pop();//删除已经输出编号的成员 if(Q[que.front()].empty())//如果该小组人已经没人了，则在总队列中删除这个小组，之后的循环是对下一个小组进行操作 que.pop(); } } while(strcmp(str,"STOP")); cout<<endl; } return 0; }