HDU 3624 get the Treasury

Hdu_3624

This question requires that the volume of the part that covers three or more times be obtained.

If we discretize the zcoordinate, if a part of each layer Z is overwritten by three or more times, it is equivalent to that the projection of this layer Z on the XY plane is covered by three or more times, therefore, for each layer of Z, We can first find the area covered by the projection three or more times, then multiply the height of the layer Z, that is, the layer Z is covered by three or more times.

# Include <stdio. h> # Include < String . H> # Include <Stdlib. h> # Define Maxd 2010 # Define K 3 Int N, m, Z, S, Ty [maxd], TZ [maxd], cover [4 * Maxd] [ 4 ], CNT [ 4 * Maxd];  Struct  REC {  Int  X1, Y1, Z1, X2, Y2, Z2;} rec [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 Cmpseg ( 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 (cover [cur],  0 , Sizeof  (Cover [cur]); cover [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; scanf (  "  % D  " ,& N );  For (I = 0 ; I <n; I ++) {Scanf (  "  % D  " , & Rec [I]. x1, & rec [I]. y1, & rec [I]. z1, & rec [I]. x2, & rec [I]. y2 ,& REC [I]. Z2); TZ [I < 1 ] = Rec [I]. Z1, TZ [(I < 1 ) | 1 ] = REC [I]. Z2; ty [I < 1 ] = Rec [I]. Y1, Ty [(I < 1 ) | 1 ] = REC [I]. Y2;} qsort (tz, n < 1 , Sizeof (Tz [ 0  ]), Cmpint); Z =- 1  ;  For (I = 0 ; I <(n < 1 ); I ++ )  If (I = 0 | TZ [I]! = TZ [I-1  ]) TZ [ ++ Z] = TZ [I]; 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 = (MAX + min)> 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 (cover [cur],  0 ,Sizeof  (Cover [cur]);  If (CNT [cur]> = K) cover [cur] [k] = Ty [Y + 1 ]- Ty [x];  Else   If (X = Y) cover [cur] [CNT [cur] = Ty [Y + 1 ]- Ty [x];  Else  {  Int I;  For (I = CNT [cur]; I <= K; I ++ ) Cover [cur] [I] = Cover [ls] [I-CNT [cur] + cover [RS] [I- CNT [cur];  For (I = k-CNT [cur] + 1 ; I <= K; I ++ ) Cover [cur] [k] + = Cover [ls] [I] + Cover [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, L, R;  Long   Long   Int Ans = 0  , Temp;  For (I = 0 ; I <z; I ++ ) {S = 0  ;  For (J = 0 ; J <n; j ++ )  If (REC [J]. Z1 <= TZ [I] & rec [J]. Z2> = TZ [I + 1  ]) {Seg [s]. x = Rec [J]. x1, SEG [s]. y1 = rec [J]. y1, SEG [s]. y2 = rec [J]. y2, SEG [s]. col = 1  ; ++ S; seg [s]. x = Rec [J]. x2, SEG [s]. y1 = rec [J]. y1, SEG [s]. y2 = rec [J]. y2, SEG [s]. col =- 1  ; ++ S;} qsort (SEG, S,  Sizeof (SEG [ 0 ]), Cmpseg); seg [s]. x = Seg [s- 1  ]. X; temp = 0  ;  For (J = 0 ; J <s; j ++ ) {L = BS (SEG [J]. Y1), r = BS (SEG [J]. Y2); refresh (  1 , 0 , M- 1 , L, R- 1 , SEG [J]. col); temp + = ( Long   Long   Int ) Cover [ 1 ] [ 3 ] * (SEG [J + 1 ]. X- SEG [J]. X);} ans + = Temp * (tz [I + 1 ]- TZ [I]);} printf (  "  % I64d \ 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  ;}