Obtain a rectangle that covers more than K times with the image watermark (11983) and

K is very small, and each storage can overwrite 0 ~ K-times interval and K-times or more are all K-times. You can create a template.

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# Include <cstdio> # Include <Cstring> # Include <Algorithm> Using   Namespace  STD; typedef  Long   Long  LLD;  # Define Lson L, M, RT <1 # Define Rson m + 1, R, RT <1 | 1 Const  Int Maxn = 60000  ;  Int Sum [maxn < 2 ] [ 15  ];  Int Cov [maxn < 2  ];  Int  X [maxn];  Int  N, m;  Struct  SEG { Int  L, R, h;  Int  Flag; seg () {} seg (  Int _ L, Int _ R, Int _ H, Int  _ Flag): L (_ L), R (_ r), H (_ H), flag (_ flag ){}  Bool   Operator <( Const SEG & CMP) Const  { Return H < CMP. h ;}} ss [maxn];  Void Pushup ( Int RT, Int L, Int  R ){  If (COV [RT]> = M | L = R) {memset (sum [RT],  0 , Sizeof  (Sum [RT]);  Int T = cov [RT] <m?Cov [RT]: m; sum [RT] [T] = X [R + 1 ]- X [l];  Return  ;}  Int  I;  For (I = 0 ; I <cov [RT]; I ++) sum [RT] [I] = 0  ;  For (I = cov [RT]; I <m; I ++ ) Sum [RT] [I] = Sum [RT < 1 ] [I-cov [RT] + sum [RT < 1 | 1 ] [I- Cov [RT]; sum [RT] [m] = 0  ;  For (I = m-cov [RT]; I <= m; I ++ ) Sum [RT] [m] + = Sum [RT < 1 ] [I] + sum [RT < 1 | 1  ] [I];}  Void Build ( Int L,Int R, Int  RT) {cov [RT] = 0 ; Sum [RT] [ 0 ] = X [R + 1 ]- X [l];  For ( Int I = 1 ; I <= m; I ++) sum [RT] [I] = 0  ;  If (L = r) Return ;  Int M = (L + r)> 1  ; Build (lson); Build (rson );}  Void Update ( Int L, Int R, Int Val, Int L, Int R, Int  RT ){  If (L <= L & R <= R) {cov [RT] + = Val; pushup (RT, L, R );  Return  ;}  Int M = (L + r)> 1  ;  If (L <= M) Update (L, R, Val, lson );  If (R> M) Update (L, R, Val, rson); pushup (RT, L, R );}  Int  Main (){  Int T, CA = 1 , I, J, K, x1, x2, Y1, Y2; scanf (  "  % D  " ,& T );  While (T -- ) {Scanf (  "  % D  " , & N ,& M );  Int TOT = 0  ;  For (I =1 ; I <= N; I ++ ) {Scanf (  "  % D  " , & X1, & Y1, & X2, & Y2); x2 ++; Y2 ++ ; X [tot] = X1; SS [tot ++] = Seg (x1, x2, Y1, 1  ); X [tot] = X2; SS [tot ++] = Seg (x1, x2, Y2 ,- 1  );} Sort (x, x +TOT); sort (SS, SS + TOT );  Int Nx = unique (x, x + ToT )- X; build (  0 , NX- 1 , 1 ); //  Printf ("% d \ n", sum [1] [0]); LLD ans = 0  ;  For (I = 0 ; I <tot- 1 ; I ++ ){  Int Left = lower_bound (x, x + NX, ss [I]. L )- X;  Int Right = lower_bound (x, x + NX, ss [I]. R)-X- 1  ; Update (left, right, ss [I]. Flag,  0 , NX- 1 , 1  ); Ans + = (LLD) sum [ 1 ] [M] * (LLD) (ss [I + 1 ]. H-Ss [I]. H);} printf (  "  Case % d: % LLD \ n  " , CA ++ , ANS );}  Return   0  ;}