实现功能——对于一个N×M的方格,1:输入一个区域,将此区域全部值作加法;2:输入一个区域,求此区域全部值的和
其实和一维线段树同理,只是不知道为什么速度比想象的慢那么多,求解释。。。@acphile
(还有代码略恶心,求原谅。。。^_^)
1 const tvp=8000000; 2 var 3 i,j,k,l,m,n,a1,a2,a3,a4,a5:longint; 4 a,b:array[0..tvp] of longint; 5 c1,c2:char; 6 function max(x,y:longint):longint;inline; 7 begin 8 if x>y then max:=x else max:=y; 9 end; 10 function min(x,y:longint):longint;inline; 11 begin 12 if x<y then min:=x else min:=y; 13 end; 14 function op(z,x1,y1,x2,y2,lx,ly,rx,ry,nu,d:longint):longint;inline; 15 var 16 a1,a2,a3,a4,a5:longint; 17 begin 18 if (lx>rx) or (ly>ry) then exit(0); 19 if (x1=lx) and (y1=ly) and (x2=rx) and (y2=ry) then 20 begin 21 b[z]:=b[z]+nu; 22 exit(nu*(rx-lx+1)*(ry-ly+1)); 23 end; 24 a2:=op(z*4-2,x1,y1,(x1+x2) div 2,(y1+y2) div 2,lx,ly,min(rx,(x1+x2) div 2),min(ry,(y1+y2) div 2),nu,d); 25 a3:=op(z*4-1,x1,(y1+y2) div 2+1,(x1+x2) div 2,y2,lx,max(ly,(y1+y2) div 2+1),min(rx,(x1+x2) div 2),ry,nu,d); 26 a4:=op(z*4,(x1+x2) div 2+1,y1,x2,(y1+y2) div 2,max(lx,(x1+x2) div 2+1),ly,rx,min(ry,(y1+y2) div 2),nu,d); 27 a5:=op(z*4+1,(x1+x2) div 2+1,(y1+y2) div 2+1,x2,y2,max(lx,(x1+x2) div 2+1),max(ly,(y1+y2) div 2+1),rx,ry,nu,d); 28 a[z]:=a[z]+a2+a3+a4+a5; 29 exit(a2+a3+a4+a5); 30 end; 31 function cal(z,x1,y1,x2,y2,lx,ly,rx,ry,d:longint):longint;inline; 32 var a1,a2,a3,a4,a5:longint; 33 begin 34 if (lx>rx) or (ly>ry) then exit(0); 35 d:=d+b[z]; 36 if (x1=lx) and (y1=ly) and (x2=rx) and (y2=ry) then exit(a[z]+d*(rx-lx+1)*(ry-ly+1)); 37 a2:=cal(z*4-2,x1,y1,(x1+x2) div 2,(y1+y2) div 2,lx,ly,min(rx,(x1+x2) div 2),min(ry,(y1+y2) div 2),d); 38 a3:=cal(z*4-1,x1,(y1+y2) div 2+1,(x1+x2) div 2,y2,lx,max(ly,(y1+y2) div 2+1),min(rx,(x1+x2) div 2),ry,d); 39 a4:=cal(z*4,(x1+x2) div 2+1,y1,x2,(y1+y2) div 2,max(lx,(x1+x2) div 2+1),ly,rx,min(ry,(y1+y2) div 2),d); 40 a5:=cal(z*4+1,(x1+x2) div 2+1,(y1+y2) div 2+1,x2,y2,max(lx,(x1+x2) div 2+1),max(ly,(y1+y2) div 2+1),rx,ry,d); 41 exit(a2+a3+a4+a5); 42 end; 43 begin 44 readln(c1,n,m); 45 fillchar(a,sizeof(a),0); 46 fillchar(b,sizeof(b),0); 47 while not(eof) do 48 begin 49 read(c1,a1,a2,a3,a4); 50 case c1 of 51 'L':begin 52 readln(a5); 53 op(1,1,1,n,m,a1,a2,a3,a4,a5,0); 54 end; 55 'k':begin 56 readln; 57 writeln(cal(1,1,1,n,m,a1,a2,a3,a4,0)); 58 end; 59 end; 60 end; 61 end.