FZU 2107 Hua Rong Dao （DFS）

  
  
  
  

   
   
   
   
Description
   
   
   
   
 
    
Cao Cao was hunted down by thousands of enemy soldiers when he escaped from Hua Rong Dao. Assuming Hua Rong Dao is a narrow aisle (one N*4 rectangle), while Cao Cao can be regarded as one 2*2 grid. Cross general can be regarded as one 1*2 grid.Vertical general can be regarded as one 2*1 grid. Soldiers can be regarded as one 1*1 grid. Now Hua Rong Dao is full of people, no grid is empty.
 
    
 
    
There is only one Cao Cao. The number of Cross general, vertical general, and soldier is not fixed. How many ways can all the people stand?
 
   
  
  
  
  
  
  
  
  

   
   
   
   
Input
   
   
   
   
 
    
There is a single integer T (T≤4) in the first line of the test data indicating that there are T test cases.
 
    
Then for each case, only one integer N (1≤N≤4) in a single line indicates the length of Hua Rong Dao.
 
   
  
  
  
  
  
  
  
  

   
   
   
   
Output
   
   
   
   

     For each test case, print the number of ways all the people can stand in a single line. 
   
  
  
  
  
  
  
  
  

   
   
   
   
Sample Input
   
   
   
   
 
    

      212 
    
 
   
  
  
  
  
  
  
  
  

   
   
   
   
Sample Output
   
   
   
   
 
    

      018 
    
 
   
  
  
  
  
  
  
  
  

   
   
   
   
Hint
   
   
   
   
 
    
Here are 2 possible ways for the Hua Rong Dao 2*4.
  
   
  
  
  
  
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
int a[5][5];
int ans[5]={0,0,18,284,4862};
int top,ty;
/*void dfs(int x,int y,int f,int cnt)
{
    if(x>top)
    {
        if(cnt==top*ty&&f) ans++;
        return ;
    }
    if(y>ty)
    {
        dfs(x+1,1,f,cnt);
        return ;
    }
    if(a[x][y])
    {
        dfs(x,y+1,f,cnt);
        return ;
    }
    if(f==0&&x<top&&y<ty)
    {
        if(a[x][y+1]||a[x+1][y]||a[x+1][y+1]) ;
        else
        {
            a[x][y]=1;
            a[x+1][y]=1;
            a[x+1][y+1]=1;
            a[x][y+1]=1;
            dfs(x,y+2,1,cnt+4);
            a[x][y]=0;
            a[x+1][y]=0;
            a[x+1][y+1]=0;
            a[x][y+1]=0;
        }
    }
    if(x<top)
    {
        if(a[x+1][y]==0)
        {
            a[x][y]=1;
            a[x+1][y]=1;
            dfs(x,y+1,f,cnt+2);
            a[x][y]=0;
            a[x+1][y]=0;
        }
    }
    if(y<ty)
    {
        if(a[x][y+1]==0)
        {
            a[x][y+1]=1;
            a[x][y]=1;
            dfs(x,y+2,f,cnt+2);
            a[x][y]=0;
            a[x][y+1]=0;
        }
    }
    a[x][y]=1;
    dfs(x,y+1,f,cnt+1);
    a[x][y]=0;
}*/
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&ty);
        cout<<ans[ty]<<endl;
    }
    return 0;
}