做题笔记9

做题笔记9

文章目录

• 一、题目
• 二、代码
• 总结

---

一、题目

二、代码

#include 
#include

using namespace std;

int whi(int s, int** a, int m, int n)
{
	int w = -1;
	for (int i = 0; i < m; i++)
	{
		for (int j = 0; j < n; j++)
		{
			if (s == a[i][j])
			{
				w = i;
				break;
			}
		}
	}
	return w;
}

int main()
{
	int t;
	cin >> t;
	int n;
	int** have = new int* [t];
	for (int m = 0; m < t; m++)
	{
		cin >> n;
		int temp;
		have[m] = new int[n];
		for (int i = 0; i < n; i++)
		{
			cin >> temp;
			have[m][i] = temp;
		}
	}
	queue* answer = new queue[t];
	int num = 0;
	int m = 0;
	int p = 0;
	while (1)
	{
		string want;
		cin >> want;

		if (want == "ENQUEUE")
		{
			int temp;
			int lo;
			cin >> temp;
			if (num == 0)
			{
				answer[m].push(temp);
				num++;
			}
			else
			{
				int w = whi(temp, have, t, n);
				int bo = -1;
				for (int i = 0; i <= m; i++)
				{
					lo = answer[i].front();
					if (whi(lo, have, t, n) == w)
					{
						bo = i;
						break;
					}
				}
				if (bo == -1)
				{
					answer[m + 1].push(temp);
					m = m + 1;
				}
				else
				{
					answer[bo].push(temp);
				}
			}
		}
		else if (want == "DEQUEUE")
		{
			for (int i = 0; i < t;)
			{
				if (answer[i].empty())
				{
					i++;
				}
				else
				{
					if (p == 0)
					{
						cout << answer[i].front();
						answer[i].pop();
						p++;
						break;
					}
					else
					{
						cout << " " << answer[i].front();
						answer[i].pop();
						break;
					}
				}
			}
		}
		else if (want == "STOP")
		{
			break;
		}
	}

	return 0;
}

总结

在该题中，运用队列的仅有后面的插入与输出，前面的部分可以用二维数组表示。问题主要在插入的数属于队列组中的哪一个队列。判断时只需要与每个队列的第一个或最后一个元素比较即可，队列不适用遍历且不需要遍历。