UVA540-队列

题意:

每一个数字有自己所属的团队,如果所属的团队已经有人在队列里,放在团队的最后一个,要不然放队列里最后一个

注意:一个团队里的最多1000个元素,但是入队,出队的操作会达到200000次

解法:循环队列,声明一个n长的数组,数组的每一个元素都是一个循环队列

没有判断队列是否满了,只简单的判断是否为空

AC时间:570ms

主要耗时在查找元素是那个团队时循环了所有元素,可以简单粗暴使用STL,用了就没意义了,o(︶︿︶)o 唉！！！

#include<iostream>
#include <stdio.h>
#include <memory.h>
using namespace std;

struct Node
{
	int num;
	int team;
	Node()
	{
		num = -1;
		team = -1;
	}
};
struct Queue
{
	int ql;
	Node node[1003];
	int position;
	int limit;
	Queue()
	{
		ql = 1003;
		position = 0;
		limit = 0;
	}
	void push(Node n)
	{
		node[position]=n;
		position=(position+1)%ql;
	}
	Node offer()
	{
		Node n= node[limit];
		limit=(limit+1)%ql;
		return n;
	}
	bool empty()
	{
		return limit == position;
	}
};

int findTeam(int number, Node* node, int nl)
{
	for(int i = 0; i < nl; i++)
	{
		if(node[i].num == number)
		{
			return node[i].team;
		}
	}
	return -1;
}
int findQueue(int team, Queue*head, int tl)
{
	for(int i = 0; i < tl; i++)
	{
		if(head[i].node[0].team == team)
		{
			return i;
		}
	}
	return tl;
}
int main()
{
	//freopen("d:\\1.txt", "r", stdin);
	int n;
	string eq = "ENQUEUE";
	string de = "DEQUEUE";
	string stop = "STOP";
	int t = 0;
	while (cin >> n)
	{
		if(n == 0)
			return 0;
		t++;
		cout << "Scenario #" << t << endl;
		Node node[n * 1003];
		Queue head[n];
		int tn = 0;
		for(int i = 0; i < n; i++)
		{
			Queue h;
			head[i] = h;
			int m;
			cin >> m;
			for(int j = 0; j < m; j++)
			{
				Node node2;
				cin >> node2.num;
				node2.team = i;
				node[tn++] = node2;
			}
		}
		string str;
		int number;
		int tl = 0;
		while (true)
		{
			cin >> str;
			if(str == stop)
				break;
			else if(str == de)
			{
				//de
				cout << head[0].offer().num << endl;
				if(head[0].empty())
				{
					int i = 1;
					for(i = 0; i < tl-1; i++)
					{
						head[i] = head[i+1];
					}
					head[tl-1].limit = 0;
					head[tl-1].position = 0;
					tl--;
				}
			}
			else
			{
				//en
				cin >> number;
				int team = findTeam(number, node, tn);
				int i = findQueue(team,head, tl);
				Node nn;
				nn.num = number;
				nn.team = team;
				if(head[i].position == 0)
					tl++;
				head[i].push(nn);
			}
		}
		cout << endl;
	}
	return 0;
}