hdu-1702 ACboy needs your help again!(栈和队列的结合)

最新推荐文章于 2019-11-25 17:13:58 发布

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这是一个编程问题，玩家需要通过解决怪物提出的基于FIFO和FILO的数据结构问题来营救被囚禁的角色ACboy。输入包括多个测试案例，每个案例包含一组命令，玩家必须正确输出每个OUT命令的结果。

ACboy needs your help again!

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4421 Accepted Submission(s): 2256

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

# include<stdio.h>
# include<string.h>
# include<queue>
# include<stack>
using namespace std;

void f(int m)
{
   int t,i,a,b;
   char str[10];
   queue<int>q;//stl中queue的引用。
   while(m--)
    {
    	scanf("%s",str);
    	if(!strcmp(str,"IN"))
    	 {
    	 	scanf("%d",&t);
    	 	q.push(t);
    	 }
    	 else 
    	  {
    	  	if(q.empty()) printf("None\n");
    	  	else
			{
			  a=q.front();
    	  	  printf("%d\n",a);
    	  	  q.pop();
			}
    	  }
    }
	while(!q.empty())
	  {
	  	q.pop();
	  }	
}

void g(int m)
{
   int t,i,a,b;
   char str[10];
   stack<int>q;
   while(m--)
    {
    	scanf("%s",str);
    	if(!strcmp(str,"IN"))
    	 {
    	 	scanf("%d",&t);
    	 	q.push(t);
    	 }
    	 else 
    	  {
    	  	if(q.empty()) printf("None\n");
    	  	else
			{
			  b=q.top();
    	  	  printf("%d\n",b);
    	  	  q.pop();
			}
    	  }
}
 while(!q.empty())
	  {
	  	q.pop();
	  }	//注意最后要将栈释放，与动态数组相似。
}

int main()
{
	int z;
	scanf("%d",&z);
	while(z--)
	 {
	 	int m,n;
	 	char str[10];
	 	scanf("%d %s",&m,str);
	 	if(!strcmp(str,"FIFO")) f(m); //如果字符串为FIFO则调用函数f进行先进先出的运算。
	 	else g(m);//先进后出。
	 }
	return 0;
}