bzoj1132

每次都选最左边的点，然后以这个点为原点

统计和这个点构成的三角形面积和

不难想到极角排序然后由叉积很容易求出

const oo=1 shl 30;
      eps=1e-8;
var i,j,k,m,n:longint;
    x,y:array[0..6010] of longint;
    z:array[0..6010] of double;
    ans,xx,yy:int64;

procedure swap(var a,b:longint);
  var c:longint;
  begin
    c:=a;
    a:=b;
    b:=c;
  end;

procedure sort(l,r:longint);
  var i,j:longint;
      p,q:double;
  begin
    i:=l; j:=r;
    p:=z[(l+r) shr 1];
    repeat
      while z[i]<p-eps do inc(i);
      while z[j]>p+eps do dec(j);
      if i<=j then
      begin
        swap(x[i],x[j]);
        swap(y[i],y[j]);
        q:=z[i]; z[i]:=z[j]; z[j]:=q;
        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);
  for i:=1 to n do
    readln(x[i],y[i]);
  for i:=1 to n-2 do
  begin
    k:=i;
    for j:=i to n do
      if x[j]<x[k] then k:=j;
    swap(x[i],x[k]);
    swap(y[i],y[k]);
    for j:=i+1 to n do
      if x[j]=x[i] then
        if y[j]>y[i] then z[j]:=oo
        else z[j]:=-oo
      else z[j]:=(y[j]-y[i])/(x[j]-x[i]);
    sort(i+1,n);
    xx:=0; yy:=0;
    for j:=i+1 to n do
    begin
      ans:=ans+(x[j]-x[i])*yy-(y[j]-y[i])*xx;
      xx:=xx+x[j]-x[i]; yy:=yy+y[j]-y[i];
    end;
  end;
  writeln(abs(ans)/2:0:1);
end.