3170: [Tjoi 2013]松鼠聚会

对一个点(x,y)定义x'=x+y,y'=x-y，曼哈顿距离就可以写成max(|x1'-x2'|,|y1'-y2'|)。

回到原题，容易发现，两个点(x1,y1)和(x2,y2)的距离为max(|x1-x2|,|y1-y2|)。

把上面的处理倒过来，就可以回到我们熟悉的曼哈顿距离了。

把点转换后枚举每个点，然后O(1)求出它到所有点的曼哈顿距离和。（排序后前缀和优化）。

RunID	User	Problem	Result	Memory	Time	Language	Code_Length	Submit_Time
410862	lbz007	3170	Accepted	5696 kb	696 ms	Pascal/Edit	1487 B	2013-05-13 07:13:26

var
  p,q,sum,a,x,y:array[0..100000]of double;
  b,rank:Array[1..100000]of longint;
  ans,an,xx,yy:extended;
  n,i,aa,bb:longint;
procedure sort(ll,rr:longint);
  var
    t,i,j:longint;
    xx,tt:double;
  begin
  i:=ll;j:=rr;
  xx:=a[ll+random(rr-ll+1)];
  while i<=j do
    begin
    while a[i]<xx do inc(i);
    while a[j]>xx do dec(j);
    if i<=j then
      begin
      tt:=a[i];a[i]:=a[j];a[j]:=tt;
      t:=b[i];b[i]:=b[j];b[j]:=t;
      inc(i);dec(j);
      end;
    end;
  if ll<j then sort(ll,j);
  if rr>i then sort(i,rr);
  end;
function abs(xx:extended):extended;
  begin
  if xx>0 then exit(xx)
  else exit(-xx);
  end;
begin
  readln(n);
  for i:=1 to n do
    begin
    readln(aa,bb);
    x[i]:=(aa+bb)/2;
    y[i]:=(aa-bb)/2;
    end;
  a:=x;
  for i:=1 to n do b[i]:=i;
  sort(1,n);
  for i:=1 to n do
    sum[i]:=sum[i-1]+a[i];
  for i:=1 to n do
    begin
    p[b[i]]:=a[i]*(i-1)-sum[i-1];
    p[b[i]]:=p[b[i]]+sum[n]-sum[i]-a[i]*(n-i);
    end;
  ans:=9000000000000000;
  a:=y;
  for i:=1 to n do b[i]:=i;
  sort(1,n);
  sum[0]:=0;
  for i:=1 to n do
    sum[i]:=sum[i-1]+a[i];
  for i:=1 to n do
    begin
    q[b[i]]:=a[i]*(i-1)-sum[i-1];
    q[b[i]]:=q[b[i]]+sum[n]-sum[i]-a[i]*(n-i);
    if q[b[i]]+p[b[i]]<ans then ans:=p[b[i]]+q[b[i]];
    end;
  writeln(ans:0:0);
  end.