# HDU 1702 ACboy needs your help again! (栈与行列)

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HDU 1702 ACboy needs your help again! (栈与队列)

# ACboy needs your help again!

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3701    Accepted Submission(s): 1885

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```

Source
2007省赛集训队练习赛（1）

```#include <stdio.h>
#include <string.h>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
int main()
{
int t;
char s[5];
scanf("%d",&t);
while(t--)
{
int n,m;
int i;
int l=0;
int a[222];
char s1[5];
queue<int>q;  //队列
stack<int>Q;   //栈
int flag=0;
scanf("%d%s",&n,s);
for(i=1;i<=n;i++)
{
if(strcmp(s,"FIFO")==0)
{
flag=1;
scanf("%s",s1);
if(s1[0]=='O' &&s1[1]=='U' &&s1[2]=='T' &&q.empty())
a[l++]=-444;   //  队列为空的话赋值为负数，遇到这个数输出none。
if(s1[0]=='I' &&s1[1]=='N')
{
scanf("%d",&m);
q.push(m);//   入队
}
else
{
if(!q.empty())
{
a[l++]=q.front();//  遇到OUT用a数组存栈的首元素
q.pop();//  弹栈
}
}
}
else
{
flag=2;
scanf("%s",s1);
if(s1[0]=='O' &&s1[1]=='U' &&s1[2]=='T' &&Q.empty())
a[l++]=-444;
if(s1[0]=='I' &&s1[1]=='N')
{
scanf("%d",&m);
Q.push(m);
}
else
{
if(!Q.empty())
{
a[l++]=Q.top();
Q.pop();
}
}
}
}//   第二种栈的情况和队列一样，把小q改成大Q就好了
if(flag==1)// 第一种情况
{
for(i=0;i<l;i++)
{
if(a[i]!=-444)
printf("%d\n",a[i]);
else
printf("None\n");
}
}
else if(flag==2)//第二种情况
{
for(i=0;i<l;i++)
{
if(a[i]!=-444)
printf("%d\n",a[i]);
else
printf("None\n");
}
}
}
return 0;
}```