A - ACboy needs your help again!

最新推荐文章于 2021-08-05 17:56:09 发布

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最新推荐文章于 2021-08-05 17:56:09 发布

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文章标签：栈 stack queue 队列

本文链接：https://blog.youkuaiyun.com/with_wine/article/details/118692543

这篇博客介绍了一种使用栈和队列解决怪物迷宫问题的方法。通过理解‘FIFO’（先进先出）和‘FILO’（先进后出）的概念，博主分别用队列和栈处理了输入和输出命令。当遇到'IN'命令时，将数字压入对应的数据结构，遇到'OUT'命令时，从数据结构中弹出并输出数字。如果数据结构为空，则输出'None'。示例输入和输出展示了算法的正确运行。

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ACboy was kidnapped!!
he miss his mother very much and is very scare now.You can't image how dark the room he was put into is, so poor :(.
As a smart ACMer, you want to get ACboy out of the monster's labyrinth.But when you arrive at the gate of the maze, the monste say :" I have heard that you are very clever, but if can't solve my problems, you will die 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 a word "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", (M represent a integer).
and the answer of a problem is a passowrd of a door, so if you want to rescue 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 a integer depend on the word is "FIFO" or "FILO", or a word "None" if you don't have any integer.

Sample Input
4
4 FIFO
IN 1
IN 2
OUT
OUT
4 FILO
IN 1
IN 2
OUT
OUT
5 FIFO
IN 1
IN 2
OUT
OUT
OUT
5 FILO
IN 1
IN 2
OUT
IN 3
OUT
Sample Output
1
2
2
1
1
2
None
2
3

思路：stack和queue的基本应用

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n;
	cin>>n;
	while(n--)
	{
		int a,d;
		char b[10],c[10];
		cin>>a>>b;
		if(b[2]=='F')//先进先出 
		{
			queue<int>q;
			while(a--)
			{
				cin>>c;
				if(c[0]=='I')
				{
					cin>>d; 
					q.push(d);
				}
				else
				{
					if(!q.empty())
					{
						cout<<q.front()<<endl;
						q.pop();
					}
					else
					{
						cout<<"None"<<endl;
					}
				}
			}
		}
		else//先进后出 
		{
			stack<int>s;
			while(a--)
			{
				cin>>c;
				if(c[0]=='I')
				{
					cin>>d;
					s.push(d);
				}
				else
				{
					if(!s.empty())
					{
						cout<<s.top()<<endl;
						s.pop();
					}
					else
					{
						cout<<"None"<<endl;
					}
				}
			}
		}
	}
	return 0;
}