HDU-1702

最新推荐文章于 2020-02-16 10:57:32 发布

原创最新推荐文章于 2020-02-16 10:57:32 发布 · 613 阅读

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本文介绍了一种利用数据结构解决迷宫问题的方法。通过对比队列(FIFO)和栈(FILO)的特点，在特定输入下输出相应的结果，旨在演示这两种基本数据结构的应用场景。

ACboy was kidnapped!!
he miss his mother very much and is very scare now.You can't image how dark theroom he was put into is, so poor :(.
As a smart ACMer, you want to get ACboy out of the monster'slabyrinth.But when you arrive at the gate of the maze, the monste say :" Ihave heard that you are very clever, but if can't solve my problems, you willdie with ACboy."
The problems of the monster is shown on the wall:
Each problem's first line is a integer N(the number of commands), and aword "FIFO" or "FILO".(you are very happy because you know"FIFO" stands for "First In First Out", and"FILO" means "First In Last Out").
and the following N lines, each line is "IN M" or "OUT", (Mrepresent a integer).
and the answer of a problem is a passowrd of a door, so if you want torescue ACboy, answer the problem carefully!

Input

The input contains multiple test cases.
The first line has one integer,represent the number oftest cases.
And the input of each subproblem are described above.

Output

For each command "OUT", you should output ainteger depend on the word is "FIFO" or "FILO", or a word"None" if you don't have any integer.

Sample Input

4 FIFO

IN 1

IN 2

OUT

4 FILO

IN 1

IN 2

OUT

5 FIFO

IN 1

IN 2

OUT

5 FILO

IN 1

IN 2

OUT

IN 3

OUT

Sample Output

None

思路：

“FIFO”(First In First Out)就是队列，“FILO”(First In Last Out) 就是栈。使用STL的stack和queue直接进行模拟即可。

代码如下：

#include<iostream>
#include<stack>
#include<queue>
using namespace std;
int main()
{
	int T; char a[] = "FIFO";
	cin >> T;
	while (T--)
	{
		stack<int>Zhan;
		queue<int>Dui;
		int CISHU; char sign[50];
		cin >> CISHU >> sign;
		if (strcmp(sign, a) == 0)
		{
			int  m; char b[50];
			for (int i = 1; i <= CISHU; i++)
			{
				cin >> b;
				if (strlen(b) == 2)
				{
					cin >> m;
					Dui.push(m);
				}
				else 
				{
					if (!Dui.empty())
					{
						cout << Dui.front() << endl;
						Dui.pop();
					}
					else  cout << "None" << endl;
				}
			}
		}
		else
		{       int m; char b[50];
			for (int i = 1; i <= CISHU; i++)
			{
				cin >> b;
				if (strlen(b) == 2)
				{
					cin >> m;
					Zhan.push(m);
				}
				else
				{
					if (!Zhan.empty())
					{
						cout << Zhan.top() << endl;
						Zhan.pop();
					}
					else  cout << "None" << endl;
				}
			}

		}
	}
	return 0;
	}