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= shl ;
       eps=1e-8;
 var i,j,k,m,n:longint;
     x,y:array[..] of longint;
     z:array[..] 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 ];
     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:= to n do
     readln(x[i],y[i]);
   for i:= to n- 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+ 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+,n);
     xx:=; yy:=;
     for j:=i+ 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)/::);
 end.