Codeforces863B-Kayaking

标签：do begin people ans1 kayaks Kayaking Codeforces863B instability

Kayaking

Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.

Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·n people in the group (including Vadim), and they have exactly n - 1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. i-th person's weight is w_i, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.

Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.

Help the party to determine minimum possible total instability!

Input

The first line contains one number n (2 ≤ n ≤ 50).

The second line contains 2·n integer numbers w₁, w₂, ..., w_2n, where w_i is weight of person i (1 ≤ w_i ≤ 1000).

Output

Print minimum possible total instability.

Example

Input

2 1 2 3 4

Output

1

Input

4 1 3 4 6 3 4 100 200

Output

5

题解：我们发现n的范围比较小，所以可以直接暴力贪心求解。

Code：

var
  n,ans,i,j,cnt,k,ans1:longint;
  a,b:array[1..105] of longint;
procedure sort(l,r:longint);
var i,j,x,y:longint;
begin
  i:=l;j:=r;x:=a[(l+r) div 2];
  repeat
    while a[i]<x do inc(i);
    while a[j]>x do dec(j);
    if not(i>j) then
    begin
      y:=a[i];a[i]:=a[j];a[j]:=y;
      inc(i);dec(j);
    end;
  until i>j;
  if l<j then sort(l,j);
  if i<r then sort(i,r);
end;
begin
  readln(n);
  n:=n*2;ans:=maxlongint;
  for i:=1 to n do read(a[i]);
  sort(1,n);
  for i:=1 to n-1 do
    for j:=i+1 to n do
    begin
      ans1:=0;cnt:=0;
      for k:=1 to n do
      if (k<>i)and(k<>j) then
      begin
        inc(cnt);
        b[cnt]:=a[k];
      end;
      for k:=1 to n div 2-1 do
        ans1:=ans1+abs(b[k*2]-b[k*2-1]);
      if ans1<ans then ans:=ans1;
    end;
  writeln(ans);
end.

标签：do,begin,people,ans1,kayaks,Kayaking,Codeforces863B,instability
From： https://blog.51cto.com/u_15888102/5878354