ACboy needs your help again!（杭电1702）（栈和队列的基础应用）

最新推荐文章于 2021-03-12 04:36:50 发布

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在这篇博客中，我们将探讨如何通过解决一系列复杂的迷宫难题来成功救援被绑架的AC男孩。每一步都需要仔细思考，运用先进队列和栈操作技巧。从输入的数据开始，我们一步步解析迷宫指令，利用'FIFO'（先进先出）或'FILO'（先进后出）原则，找出正确的路径，最终将AC男孩从困境中解救出来。

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ACboy needs your help again!

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4832 Accepted Submission(s): 2473

Problem Description

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

提交用C++AC，用G++就WA了，基本的队列和栈的操作，基础题。
#include<stdio.h>
#include<stack>
#include<queue> 
using namespace std;
int main()
{
	char s[4];
	int test,n,k;
	scanf("%d",&test);
	while(test--)
	{
		scanf("%d %s",&n,s);
		if(s[2]=='F')
		{
			queue<int> b;
			while(n--)
			{
				scanf("%s",s);
				if(s[0]=='I')
				{
					scanf("%d",&k);
					b.push(k);
				}
				else if(b.empty())
					printf("None\n");
			    else{   
					printf("%d\n",b.front());
					b.pop();
				}
		    }
		}
		else
		{
			stack<int> a;
			while(n--)
			{
				scanf("%s",s);
				if(s[0]=='I')
				{
					scanf("%d",&k);
					a.push(k);
				}
				else if(a.empty())
				    printf("None\n");
				else{
					printf("%d\n",a.top());
					a.pop();
				} 
			}
		}
	}
}