POJ 2259 - Team Queue - [队列的邻接表]

题目链接：http://poj.org/problem?id=2259

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.
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

题意：

给出 $t$ 个团队，每个团队一行输入，给出正整数 $n$ 以及团队中 $n$ 个人各自的编号。

又给出一系列命令，分别是入队编号为 $x$ 的人和 出队编号为 $x$ 的人。

入队一个人，他会寻找队列中自己的队友，并排在他们的后面；若没有找到队友，则直接排在队伍最末尾。

出队一个人，则直接出队队首。要求给出每次出队的人的编号。

要求每次出入队都是常数时间复杂度。

题解：

用一种类似于邻接表的形式存储：开一个大队列，用于存储在队伍中的团队，再开若干小队列，用于存储每个团队内部的成员。

AC代码：

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxid=1e6+;
const int maxt=1e3+;

int t;
int id2team[maxid];
queue<int> T;
queue<int> Q[maxt];
bool vis[maxt];

int main()
{
    int kase=;
    while(scanf("%d",&t) && t)
    {
        memset(id2team,,sizeof(id2team));
        for(int team=,n;team<=t;team++)
        {
            scanf("%d",&n);
            for(int i=,id;i<=n;i++)
            {
                scanf("%d",&id);
                id2team[id]=team;
            }
        }
        while(!T.empty()) T.pop();
        for(int team=;team<=t;team++) while(!Q[team].empty()) Q[team].pop();
        memset(vis,,sizeof(vis));
        char op[];
        printf("%sScenario #%d\n",kase>?"\n":"",++kase);
        while(scanf("%s",op) && op[]!='S')
        {
            if(op[]=='E')
            {
                int id,team;
                scanf("%d",&id);
                team=id2team[id];
                if(vis[team]) Q[team].push(id);
                else Q[team].push(id), T.push(team), vis[team]=;
            }
            if(op[]=='D')
            {
                int team=T.front();
                printf("%d\n",Q[team].front());
                if(Q[team].size()>) Q[team].pop();
                else Q[team].pop(), T.pop(), vis[team]=;
            }
        }
    }
}