hdoj 1702 ACboy needs your help again!【数组模拟+STL实现】

ACboy needs your help again!

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4241    Accepted Submission(s): 2164

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

 

#include<stdio.h>
#include<string.h>
int main()
{
	int n,m,j,i,sum,t,k,top;
	char a[5]={"FIFO"};
	char b[5]={"FILO"};
	char d[5]={"IN"};
	char e[5]={"OUT"};
	char s[5];
	char c[5];
	int zhan[1100];
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		scanf("%s",s);
		top=0;k=0;
		for(i=0;i<n;i++)
		{
		    scanf("%s",c);
			if(strcmp(c,d)==0)  //如果输入的字符串为IN 
			{
				scanf("%d",&m);  //则输入其后的数字并存入栈中 
				zhan[top++]=m;
			}
			else if(strcmp(c,e)==0)  //如果输入字符串为OUT 
			{
				if(strcmp(s,b)==0)   //则判断字符串s是 FIFO还是FILO即是先进先出还是先进后出 
				{                      //此处为先进后出 
					if(top>0)       //如果栈不为空 
					{
						printf("%d\n",zhan[top-1]);//则输出栈尾元素 
					    top--;  //并删除此元素 
					}
					else if(top<=0)
					printf("None\n");//如果栈为空则输出None 
				}
				else if(strcmp(s,a)==0)
				{
					if(k<top)  //如果k<top即栈不为空 
					{
						printf("%d\n",zhan[k++]);//输出栈首元素并删除 
					}
					else if(k>=top)  //如果栈为空 
					printf("None\n");//则输出None 
				}
			}
		}
	}
	return 0;
}

　　STL：

#include<stdio.h>
#include<string.h>
#include<stack>
#include<queue>
#define MAX 1100
using namespace std;
char s1[MAX];
char str1[5]={"IN"};
char str4[5]={"FILO"};
int main()
{
	int n,m,j,i,t,k,l;
	int a,b;
	char str[MAX];
	stack<int>s;
	queue<int>q;
	scanf("%d",&t);
	getchar();
	while(t--)
	{
		scanf("%d%s",&n,str);
		a=0;
		if(strcmp(str,str4)==0)
		{
			while(n--)
			{
				scanf("%s",s1);
				//scanf("%s %d",s1,&m);
				if(strcmp(s1,str1)==0)
				{
				    scanf("%d",&m);
					s.push(m);
				}
				else
				{
					if(s.empty())
					{
						printf("None\n");
						continue;
					}
					a=s.top();
					printf("%d\n",a);
					s.pop();
				}
			}
			while(!s.empty())
			{
				s.pop();
			}
		}
		else
		{
			while(n--)
			{
				scanf("%s",s1);
				//scanf("%s %d",s1,&m);
				if(strcmp(s1,str1)==0)
				{
					scanf("%d",&m);
					q.push(m);
				}
				else
				{
					if(q.empty())
					{
						printf("None\n");
						continue;
					}
					a=q.front();
					printf("%d\n",a);
					q.pop();
				}
			}
			while(!q.empty())
			q.pop();
		}
	}
	return 0;
}