bzoj 3132 二维树状数组

题意：一个n*m初始为空的矩阵，资磁两种操作：（1）某个子矩阵的值都增加c （2）输出某个子矩阵的权值和

用a[i,j]表示（i,j）~（n,m）的增量，则对于（1,1）~（x,y）的权值和ans

ans=sigma（a[i,j]*(x-i+1）*（y-j+1））(1<=i<=x,1<=j<=y)

      =sigma(a[i,j]*(x+1)*(y+1)-j*(x+1)*a[i,j]-i*(y+1)*a[i,j]+i*j*a[i,j])  (1<=i<=x,1<=j<=y)

      =(x+1)*(y+1)*sigma(a[i,j])-(x+1)*sigma(j*a[i,j])-(y+1)*sigma(i*a[i,j])+sigma(i*j*a[i,j])  (1<=i<=x,1<=j<=y)

维护四个树状数组分别为 

1：sigma(a[i,j])

2：sigma(j*a[i,j])

3：sigma(i*a[i,j])

4：sigma(i*j*a[i,j])

{$Q-}var        n,m,x1,x2,y1,y2,c       :longint;        ch                      :char;        t                       :array[0..4,0..2105,0..2105] of longint;procedure _add(z,x,y,v:longint);var        tt:longint;begin   while x<=n do   begin      tt:=y;      while tt<=m do      begin         inc(t[z,x,tt],v);         inc(tt,tt and (-tt));      end;      inc(x,x and (-x));   end;end;procedure change(x1,y1,x2,y2,c:longint);begin   _add(1,x1,y1,c);   _add(1,x2+1,y2+1,c);   _add(1,x1,y2+1,-c);   _add(1,x2+1,y1,-c);   //   _add(2,x1,y1,y1*c); _add(3,x1,y1,x1*c);   _add(2,x2+1,y2+1,(y2+1)*c); _add(3,x2+1,y2+1,(x2+1)*c);   _add(2,x1,y2+1,-(y2+1)*c); _add(3,x1,y2+1,-x1*c);   _add(2,x2+1,y1,-y1*c); _add(3,x2+1,y1,-(x2+1)*c);   //   _add(4,x1,y1,x1*y1*c);   _add(4,x2+1,y2+1,(x2+1)*(y2+1)*c);   _add(4,x1,y2+1,-x1*(y2+1)*c);   _add(4,x2+1,y1,-(x2+1)*y1*c);end;function find(z,x,y:longint):longint;var        tt,ans:longint;begin   ans:=0;   while x>0 do   begin      tt:=y;      while tt>0 do      begin         inc(ans,t[z,x,tt]);         dec(tt,tt and (-tt));      end;      dec(x,x and (-x));   end;   exit(ans);end;function sum(x,y:longint):longint;begin   exit(find(1,x,y)*(x+1)*(y+1)-(x+1)*find(2,x,y)-(y+1)*find(3,x,y)+find(4,x,y));end;begin   read(ch);   readln(n,m);   while not eof do   begin      read(ch);      if ch='L' then      begin         readln(x1,y1,x2,y2,c);         change(x1,y1,x2,y2,c);      end else      begin         readln(x1,y1,x2,y2);         writeln(sum(x2,y2)-sum(x1-1,y2)-sum(x2,y1-1)+sum(x1-1,y1-1));      end;   end;end.

——by Eirlys