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.