HDU 2442 bricks

Hdu_2442

If recursive DP is used, you need to record the state of the two lines of contour lines, and then discuss whether each graph can be embedded into the current state one by one based on the State-by-grid DP.

Because of the write rub, it is time to directly submit the TLE, so we simply pre-process all States first, and then output the correct results every time O (1.

# Include <stdio. h> # Include < String . H> # Include <Algorithm> # Define Maxn 110 # Define Maxm 6 # Define St 4096 Int N, m, d, f [maxn] [maxm] [st], DP [maxn] [maxm];  Void Prepare ( Int N, Int  M ){  Int  I, J, K;  For (I = 0 , D = 1 ; I <m; I ++) d <= 2  ;  For (I = 0 ; I <= N; I ++)  For (J = 0 ; J <m; j ++ )  For (K = 0 ; K <D; k ++) f [I] [J] [k] = 0  ; F [I] [J] [  0 ] = 0  ;  For (I = 0 ; I <n; I ++ ){ For (J = 0 ; J <m; j ++ )  For (K = 0 ; K <D; k ++ ){  Int  St; ST = ~ ( 3 < 2 * J) & K | ( 1 < 2 * J & K) < 1 | 1 < 2 * J;  If (J = m- 1 ) F [I + 1 ] [ 0 ] [St] = STD: max (F [I + 1 ] [ 0  ] [St], F [I] [J] [k]);  Else F [I] [J + 1 ] [St] = STD: max (F [I] [J + 1 ] [St], F [I] [J] [k]);  If (I> = 2 & J <m- 1 & (K> 2 * J & 7 ) = 7  ) {St = K &~ ( 7 < 2 * J); F [I] [J + 1 ] [St] = STD: max (F [I] [J +1 ] [St], F [I] [J] [k] + 4  );}  If (I> = 2 & J> = 1 & J <m- 1 & (K> 2 * J- 1 & 15 ) = 15  ) {St = K &~ (15 < 2 * J- 1  ); F [I] [J + 1 ] [St] = STD: max (F [I] [J + 1 ] [St], F [I] [J] [k] + 5  );}  If (I> = 1 & J> = 1 & J <m- 1 & (K &1 < 2 * J- 1 ) & (K & 1 < 2 * J) & (K & 1 < 2 * J + 2  ) {St = K &~ ( 1 < 2 * J + 1 ) ^1 < 2 * J- 1 ^ 1 < 2 * J ^ 1 < 2 * J + 2  ; F [I] [J + 1 ] [St] = STD: max (F [I] [J + 1 ] [St], F [I] [J] [k] + 4  );} If (I> = 1 & J <m- 2 & (K & 1 < 2 * J) & (K & 1 < 2 * J + 2 ) & (K & 1 < 2 * J + 4  ) {St = K &~ (1 < 2 * J + 1 ) ^ 1 < 2 * J ^ 1 < 2 * J + 2 ^ 1 < 2 * J + 4  ; F [I] [J + 1 ] [St] = STD: max (F [I] [J +1 ] [St], F [I] [J] [k] + 4  );}  If (I> = 2 & J <m- 1 & (K & 1 < 2 * J) & (K & 1 < 2 * J + 1 ) & (K & 1 <2 * J + 3  ) {St = K ^ 1 < 2 * J ^ 1 < 2 * J + 1 ^ 1 < 2 * J + 3  ; F [I] [J + 1 ] [St] = STD: max (F [I] [J + 1 ] [St], F [I] [J] [k] + 4  );}}  For (K = 0 , DP [I] [m- 1 ] = 0 ; K <D; k ++) DP [I] [m- 1 ] = STD: max (DP [I] [m- 1 ], F [I + 1 ] [ 0  ] [K]) ;}}  Int Main () {memset (DP,  0 , Sizeof  (DP ));  For ( Int I = 2 ; I <= 6 ; I ++) Prepare ( 100  , I );  While (Scanf ( "  % D  " , & N, & M) =2  ) {Printf (  "  % D \ n  " , DP [n- 1 ] [M- 1  ]);}  Return   0  ;}