Poj 2411 Mondriaan's dream solution report [Dynamic Planning] (Code Part)

# Include <iostream>

# Include <cstdio>

# Include <cstring>

Using namespace STD;

 

Long long DP [12] [3000];

// 0 indicates the upper part of the horizontal or vertical column. 1 indicates the lower part of the vertical column. Each row is a binary number (<2048)

Int main ()

{

Long long H, W, ANS, I, J, K, tmp1, tmp2, L, C, Cs;

Scanf ("% LLD", & CS );

While (Cs --)

{

Scanf ("% LLD", & H, & W );

If (H * W & 1) {printf ("0 \ n"); continue ;}

If (W> H) {tmp1 = H; H = W; W = tmp1;} // H * 2 ^ W, So let W be the smaller one.

Memset (DP, 0, sizeof (DP ));

Tmp1 = W; L = 1; while (tmp1 --) L * = 2;

For (I = 0; I <L; I ++)

{// Obtain the possible conditions of the first line and assign a value of 1

Tmp1 = I; C = W;

While (c)

{

If (tmp1 & 1) tmp1> = 1, C --;

Else if (tmp1 & 3) break;

Else if (C <2) break;

Else tmp1> = 2, c-= 2;

}

If (! C) DP [0] [I] = 1;

}

For (I = 1; I

{// Recursive Addition

For (j = 0; j <L; j ++)

{// Adds the compatibility status of the underlying layer to each status of each row

For (k = 0; k <L; k ++)

{// Search for the status of integration and Increment

Tmp1 = J; tmp2 = K; C = W;

While (c)

{

If (tmp2 & 1)

{

If (tmp1 & 1) break;

Else tmp1> = 1, tmp2> = 1, C --;

}

Else

{

If (tmp1 & 1) tmp1> = 1, tmp2> = 1, C --;

Else if (C <2 | (tmp1 & 3) | (tmp2 & 3) break;

Else tmp1> = 2, tmp2> = 2, c-= 2;

}

}

If (! C) DP [I] [J] + = DP [I-1] [k];

}

If (I = h-1) break; // The top layer must all be 0, and the result is found to be 0.

}

}

Printf ("% LLD \ n", DP [h-1] [0]);

}

Return 0;

}