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Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
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.
Finally, a list of commands follows. There are three different kinds of commands:
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.
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
Scenario #1 101 102 103 201 202 203 Scenario #2 259001 259002 259003 259004 259005 260001
t个团队的人在排一个长队,新来一个人时,如果有他的队友在排队,那么新来的人就会插队到最后一个队友的身后,否则排在长队的队尾。
有三种指令:
ENQUEUE x:编号为x的人进入长队。
DEQUEUE:长队的队首出列。
STOP:停止模拟。
思路:用两个queue来记录长队的排列情况。
#include<iostream> #include<cstdio> #include<queue> #include<string> #include<map> using namespace std; int main() { int cas=0; int t; while (scanf("%d", &t) != EOF&&t) { cout << "Scenario #" << ++cas << endl; int n, x; map<int, int>team; //记录所有人的团队编号,team[x]表示编号为x的人所在的团队编号。 for (int i = 0; i < t; i++) { cin >> n; while (n--) { cin >> x; team[x] = i; } } string op; int num; queue<int>teamate[1010]; //各个团队的成员的队列 queue<int>teamm; //团队的队列 for (;;) { cin >> op; if (op[0] == 'S') break; if (op[0] == 'E') { cin >> num; if (teamate[team[num]].empty()) teamm.push(team[num]); //新团队入队 teamate[team[num]].push(num); } if (op[0] == 'D') { int temp = teamm.front(); cout << teamate[temp].front() << endl; teamate[temp].pop(); if (teamate[temp].empty()) teamm.pop(); //长队中已经没有该队的人 } } cout << endl; } }
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原文地址:http://blog.csdn.net/qq_18738333/article/details/45006089