问题 K: 【USACO2012Feb】植草 {Bronze题2}

按着矩形周长的思路，到当前边的时候，前一层的覆盖数乘以高度加入 ans 就行，然而真正的算法可能并不是这个。。只能想到这个了

 const maxe=;
 type
   node=record
     l,r,mid,sum,delta:longint;
   end;
   arr=array[..] of longint;
 var i,j,n,bs,x1,y1,x2,y2,b1,b2,l,r,ans,_sum,v:longint;
 bx:array[..maxe] of arr;
 tree:array[..*maxe] of node;
 procedure qsx(t,w:longint);
 var mid1,mid2,l,r:longint;
 tem:arr;
 begin
   l:=t; r:=w; mid1:=bx[(l+r) shr ,]; mid2:=bx[(l+r) shr ,];
   //writeln(l,' ',r);
   repeat
   begin
     while (bx[l,]<mid1) or ((bx[l,]=mid1) and (bx[l,]>mid2)) do inc(l);
     while (bx[r,]>mid1) or ((bx[r,]=mid1) and (bx[r,]<mid2)) do dec(r);
     if l<=r then
     begin
       tem:=bx[l];
       bx[l]:=bx[r];
       bx[r]:=tem;
       inc(l); dec(r);
     end;
   end;
   until l>r;
   if t<r then qsx(t,r);
   if l<w then qsx(l,w);
 end;
 procedure build(o,l,r:longint);
 begin
   tree[o].l:=l; tree[o].r:=r; tree[o].mid:=(l+r) shr ;
   tree[o].delta:=; tree[o].sum:=;
   if l<>r then
   begin
     build(o*,l,tree[o].mid);
     build(o*+,tree[o].mid+,r);
   end;
 end;
 procedure maintain(o:longint);
 begin
   tree[o].sum:=;
   if tree[o].l<tree[o].r then tree[o].sum:=tree[*o].sum+tree[*o+].sum;
   if tree[o].delta> then tree[o].sum:=tree[o].r-tree[o].l+;
 end;
 procedure update(o:longint);
 begin
   if (l<=tree[o].l) and (tree[o].r<=r) then inc(tree[o].delta,v)
   else begin
     if l<=tree[o].mid then update(*o);
     if r>tree[o].mid then update(*o+);
   end;
   maintain(o);
 end;
 begin
   readln(n); bs:=*n;
   for i:= to n do
   begin
     readln(x1,y2,x2,y1);
     inc(x1,); inc(x2,);
     inc(y1,); inc(y2,);
     b1:=*i-; b2:=*i;  //writeln(y1,' ',y2);
     bx[b1,]:=; bx[b1,]:=x1; bx[b1,]:=x2; bx[b1,]:=y1;
     bx[b2,]:=-;bx[b2,]:=x1; bx[b2,]:=x2; bx[b2,]:=y2;
   end;
   qsx(,bs); ans:=;
   build(,,);
   for i:= to bs do
   begin
     l:=bx[i,]; r:=bx[i,]-; v:=bx[i,];
     update();
     inc(ans,_sum*(bx[i,]-bx[i-,]));
     _sum:=tree[].sum;
   end;
   writeln(ans);
 end.

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