Ural 1034 queens in peaceful positions

Ural_1034

It is very similar to the classic n queen problem. Although N is relatively large, it is possible to reschedule only three queens in the case of C (n, 3, each queen can be placed in at most three positions, so the maximum is C (50, 3) * 3 ^.

For more information about the n queen of bitwise computing, see the matrix67 blog: http://www.matrix67.com/blog/archives/266.

# Include <stdio. h> # Include < String . H> # Define Maxd 60 Int  N, CNT, visr [maxd], VISC [maxd], vis [maxd] [maxd];  Struct  Point { Int  X, Y;} p [maxd];  Void  Init (){  Int  I; memset (VIS,  0 , Sizeof  (VIS ));  For (I = 0 ; I <n; I ++ ) {Scanf (  "  % D  " , & P [I]. X ,&P [I]. Y); vis [p [I]. x] [p [I]. Y] = 1  ;}}  Int Precheck ( Int X, Int Y, Int  Z ){  Int  I, J, K; memset (visr,  0 , Sizeof  (Visr); memset (VISC,  0 ,Sizeof  (VISC ));  For (I = 0 ; I <n; I ++ )  If (I! = X & I! = Y & I! = Z ){  If  (Visr [p [I]. x])  Return   0  ; Visr [p [I]. x] = P [I]. Y;  If (VISC [p [I]. Y])  Return   0  ; VISC [p [I]. Y] = P [I]. X ;}}  Long   Long  Cstate (){  Int  I;  Long   Long Cs = 0  ;  For (I =1 ; I <= N; I ++ )  If  (VISC [I]) CS | = 1ll < I;  Return  CS ;}  Void DFS ( Int Cur, Long   Long CS, Long   Long Ls, Long   Long RS ){  Int  Y;  Long   Long  St;  If (Cur> N ){ ++ CNT;  Return  ;}  If (Y = Visr [cur]) {St = Ls | RS;  If (St & (1ll < Y ))  Return  ; DFS (cur + 1 , Cs, (LS | (1ll <y) < 1 , (RS | (1ll <y)> 1  );}  Else  {St = Cs | ls | RS;  For (Y = 1 ; Y <= N; y ++)  If (St & (1ll <y) = 0 &&! Vis [cur] [Y]) DFS (cur + 1 , CS | (1ll <Y), (LS | (1ll <y) < 1 , (RS | (1ll <y)> 1  );}}  Void  Solve (){  Int  I, J, K, P; CNT = 0 ;  For (I = 0 ; I <n; I ++ )  For (J = I + 1 ; J <n; j ++ )  For (K = J + 1 ; K <n; k ++ )  If  (Precheck (I, j, k) DFS (  1 , Cstate (),0 , 0  ); Printf (  "  % D \ n  "  , CNT );}  Int  Main (){  While (Scanf ( "  % D  " , & N) = 1  ) {Init ();  If (N <4  ) Printf (  "  0 \ n  "  );  Else  Solve ();}  Return   0  ;}