Ultraviolet A 11983 weird Advertisement

Uva_11983

This question can be converted into finding the area of the rectangle that is covered K times. Simply add X2 and Y2 to the area of 1, which is equivalent to seeing every square unit as a point.

# Include <stdio. h> # Include < String . H> # Include <Stdlib. h> # Define Maxk 11 # Define Maxd 60010 Int N, K, M, Len [ 4 * Maxd] [maxk], CNT [ 4 *Maxd], Ty [maxd];  Struct  SEG {  Int  X, Y1, Y2, Col;} seg [maxd];  Int Cmpint ( Const   Void * _ P, Const   Void * _ Q ){  Int * P = ( Int *) _ P, * q = ( Int *) _ Q;  Return * P <* Q? - 1 : 1  ;}  Int CMPS ( Const   Void * _ P, Const   Void * _ Q) {seg * P = (SEG *) _ p, * q = (SEG * ) _ Q;  Return P-> x <q-> X? - 1 :1  ;}  Void Build ( Int Cur, Int X, Int  Y ){  Int Mid = (x + y)> 1 , Ls = cur < 1 , RS = (cur < 1 ) | 1  ; Memset (LEN [cur],  0 ,Sizeof  (LEN [cur]); Len [cur] [  0 ] = Ty [Y + 1 ]- Ty [X]; CNT [cur] = 0  ;  If (X = Y)  Return  ; Build (LS, X, mid); Build (RS, mid + 1  , Y );}  Void Init (){  Int  I, J, K, X1, Y1, X2, Y2; scanf (  "  % D  " , & N ,& K );  For (I = 0 ; I <n; I ++ ) {J = I < 1 , K = (I < 1 ) | 1  ; Scanf ( "  % D  " , & X1, & Y1, & X2 ,& Y2 ); ++ X2, ++ Y2; seg [J]. x = X1, SEG [K]. x = X2; seg [J]. Y1 = Seg [K]. Y1 = Y1, SEG [J]. y2 = seg [K]. y2 = Y2; seg [J]. Col = 1 , SEG [K]. Col =- 1  ; Ty [J] = Y1, Ty [k] = Y2;} qsort (TY, n < 1 , Sizeof (TY [ 0  ]), Cmpint); m =- 1  ;  For (I = 0 ; I <(n < 1 ); I ++ )  If (I = 0 | Ty [I]! = Ty [I- 1  ]) Ty [ ++ M] =Ty [I]; build (  1 , 0 , M- 1  );}  Int BS ( Int  X ){  Int Mid, min = 0 , Max = m + 1  ;  For  (;) {Mid = (Min + max)> 1 ;  If (Mid = Min)  Break  ;  If (TY [Mid] <= X) min = Mid;  Else  Max = Mid ;}  Return  Mid ;}  Void Update ( Int Cur, Int X, Int  Y ){  Int Ls = cur < 1 , RS = (cur < 1 ) | 1  ; Memset (LEN [cur],  0 , Sizeof  (LEN [cur]);  If (CNT [cur]> = K) Len [cur] [k] = Ty [Y +1 ]- Ty [x];  Else   If (X = Y) Len [cur] [CNT [cur] = Ty [Y + 1 ]- Ty [x];  Else  {  Int  I;  For (I = CNT [cur]; I <= K; I ++ ) Len [cur] [I] + = Len [ls] [I-CNT [cur] + Len [RS] [I-CNT [cur];  For (I = k-CNT [cur] + 1 ; I <= K; I ++ ) Len [cur] [k] + = Len [ls] [I] + Len [RS] [I] ;}}  Void Refresh ( Int Cur, Int X, Int Y, Int S, Int T, Int  C ){ Int Mid = (x + y)> 1 , Ls = cur < 1 , RS = (cur < 1 ) | 1  ;  If (X> = S & Y <= T) {CNT [cur] + = C; Update (cur, x, y );  Return  ;}  If (Mid> =S) Refresh (LS, X, mid, S, T, C );  If (Mid + 1 <= T) Refresh (RS, mid + 1  , Y, S, T, C); Update (cur, x, y );}  Void  Solve (){  Int  I, J, K;  Long   Long   Int Ans = 0 ; Qsort (SEG, n < 1 , Sizeof (SEG [ 0  ]), CMPs); seg [n < 1 ]. X = seg [(n < 1 )- 1  ]. X;  For (I = 0 ; I <(n < 1 ); I ++ ) {J = BS (SEG [I]. Y1), k =BS (SEG [I]. Y2); refresh (  1 , 0 , M- 1 , J, k- 1  , SEG [I]. col); ans + = ( Long   Long   Int ) Len [ 1 ] [K] * (SEG [I + 1 ]. X- SEG [I]. X);} printf (  "  % LLD \ n "  , ANS );}  Int  Main (){  Int  T, TT; scanf (  "  % D  " ,& T );  For (Tt = 0 ; TT <t; TT ++ ) {Init (); printf (  "  Case % d:  " , Tt + 1  ); Solve ();}  Return   0  ;}