Poj 1753 flip game

Poj_1753

After several days, we have already written this question 3rd Times. After being familiar with bit operations, we have solved this problem more efficiently.

At the same time, this question also has a little skill in enumeration. If we use brute force enumeration, once the checkerboard reaches 16*16, it will obviously not be enough (an Eastern European semi-finals question). In fact, we can only enumerate the first row of operations. Afterwards, if we want to turn all the pawns into a color, the operations in the second row will be fixed (because the status of the pawns in the first row restricts the flip of the pawns in the second row, if you want to turn the pawns in the first line into white, the position at the bottom of the black pawns in the second line must be flipped, and vice versa ), the operations on the third and fourth rows are obviously fixed.

In this way, even the 16*16 board has only 15*16*2 ^ 16 complexity, which is acceptable.

# Include <stdio. h>
# Include < String . H>
 Int Fresh, steps, DX [] = { 0 ,-1 , 1 , 0 , 0 }, Dy [] = { 0 , 0 , 0 ,- 1 , 1 };
 Void Init ()
{
 Char Ch;
Fresh = 0 ;
For ( Int I = 0 ; I < 4 ; I ++)
{
 For ( Int J = 0 ; J < 4 ; J ++)
Fresh = (fresh < 1 ) + (CH = getchar () = '  B  ' ?1 : 0 );
Getchar ();
}
}
 Void Flip ( Int X, Int Y, Int & St)
{
 If (X> = 0 & X < 4 & Y> = 0 & Y <4 )
St ^ = 1 <(X * 4 + Y );
}
 Void DFS ( Int Cur, Int Num, Int St, Int Flag)
{
 If (Cur = 4 )
{
If (St = 0 xFFFF | ST = 0 )
Steps = num <steps? Num: steps;
 Return ;
}
 Int X, Y;
 For ( Int I = cur- 1 , J = 0 ; J < 4 ; ++ J)
 If (St &( 1 <(I * 4 + J)> (I * 4 + J) ^ flag)
{
 For ( Int K = 0 ; K < 5 ; ++ K)
{
X = cur + dx [k];
Y = J + dy [k];
Flip (X, Y, St );
}
++ Num;
}
DFS (cur + 1 , Num, St, flag );
}
 Int Solve ()
{
Steps = 0x7fffffff ;
 For ( Int I = 0 ; I < 16 ; ++ I)
{
 Int Num = 0 , St = fresh, X, Y;
For ( Int J = 0 ; J < 4 ; ++ J)
{
 If (I &( 1 <J ))
{
 For ( Int K = 0 ; K < 5 ; K ++)
{
X =0 + Dx [k];
Y = J + dy [k];
Flip (X, Y, St );
}
Num ++;
}
}
DFS ( 1 , Num, St, 0 );
DFS ( 1 , Num, St, 1 );
}
 Return Steps = 0x7fffffff ? - 1 : Steps;
}
 Int Main ()
{
Init ();
 If (Solve ()! =- 1 )
Printf ( "  % D \ n  " , Steps );
 Else 
Printf ( "  Impossible \ n  " );
 Return   0 ;
}