首先将坐标系顺时针旋转45度,得到一个新的坐标系,这个坐标系
对应的坐标的manhattan距离就是原图中的距离,然后快排,利用前缀和
数组O(N)求所有的答案,然后找最小值就行了,总时间O(NlogN),今天
体力不足,在此不再赘述。。。
/**************************************************************
Problem:
User: BLADEVIL
Language: Pascal
Result: Accepted
Time: ms
Memory: kb
****************************************************************/
//By BLADEVIL
var
n :int64;
size :array[..,..] of int64;
ans :array[..] of int64;
sum :array[..] of int64;
print :int64;
function min(a,b:int64):int64;
begin
if a>b then min:=b else min:=a;
end;
procedure swap(var a,b:int64);
var
c :int64;
begin
c:=a; a:=b; b:=c;
end;
procedure init;
var
i :longint;
x, y :int64;
begin
read(n);
for i:= to n do
begin
read(x,y);
size[,i]:=x+y;
size[,i]:=y-x;
end;
end;
procedure qs(low,high,s:int64);
var
i, j, xx :int64;
begin
i:=low; j:=high; xx:=size[s,(i+j) div ];
while i<j do
begin
while size[s,i]<xx do inc(i);
while size[s,j]>xx do dec(j);
if i<=j then
begin
swap(size[,i],size[,j]);
swap(size[,i],size[,j]);
swap(ans[i],ans[j]);
inc(i); dec(j);
end;
end;
if i<high then qs(i,high,s);
if j>low then qs(low,j,s);
end;
procedure main;
var
i :longint;
begin
qs(,n,);
for i:= to n do sum[i]:=int64(size[,i]);
for i:= to n do sum[i]:=sum[i]+sum[i-];
for i:= to n do
ans[i]:=((i-)*size[,i]-sum[i-])+((sum[n]-sum[i])-(n-i)*size[,i]);
qs(,n,);
for i:= to n do sum[i]:=int64(size[,i]);
for i:= to n do sum[i]:=sum[i]+sum[i-];
for i:= to n do
ans[i]:=ans[i]+((i-)*size[,i]-sum[i-])+((sum[n]-sum[i])-(n-i)*size[,i]);
print:=maxlongint*maxlongint;
for i:= to n do print:=min(print,ans[i]);
print:=print div ;
writeln(print);
end;
begin
init;
main;
end.