BZOJ1914

对于每一个点,它与原点的连线将平面划分为两个半平面,我们统计连线的逆时针方向的半平面中的点的个数N,
这N个点与这个点构成的N(N-1)/2个三角形都是不包含原点的.最后只要用总个数减去这些三角形的和就是答案.
统计过程极角排序后可以做到O(N).

 

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/**************************************************************

    Problem: 1914

    User: exponent

    Language: Pascal

    Result: Accepted

    Time:260 ms

    Memory:1008 kb

****************************************************************/

   

type  point=record

      x,y:longint;

end;

const maxn=100001;

var   p:array[0..maxn] of point;

      ans,tmp,count:int64;

      n,i,j:longint;

   

   

      function cross(a,b:point):int64;

      begin

            exit(int64(a.x)*int64(b.y)-int64(b.x)*int64(a.y));

      end;

   

      function cmp(a,b:point):boolean;   //return angle(a)<angle(b)

      begin

            if (a.y>0)and(b.y<=0) then exit(true);

            if (b.y>0)and(a.y<=0) then exit(false);

            if cross(a,b)>0 then exit(true);

            exit(false);

      end;

   

      procedure sort(l,r:longint);

      var   i,j:longint;

            mid,temp:point;

      begin

            i:=l; j:=r;

            mid:=p[(l+r)>>1];

            while i<=j do

            begin

                  while cmp(p[i],mid) do inc(i);

                  while cmp(mid,p[j]) do dec(j);

                  if i<=j then

                  begin

                        temp:=p[i];

                        p[i]:=p[j];

                        p[j]:=temp;

                        inc(i); dec(j);

                  end;

            end;

            if i<r then sort(i,r);

            if j>l then sort(l,j);

      end;

   

   

begin

      readln(n);

      for i:=1 to n do readln(p[i].x,p[i].y);

   

      sort(1,n);

   

      j:=1;

      count:=0;

      for i:=1 to n do

      begin

            while cross(p[i],p[j mod n+1])>=0 do

            begin

                  if j mod n+1=i then break;

                  j:=j mod n+1;

                  inc(count);

            end;

            inc(tmp,count*(count-1)>>1);

            dec(count);

      end;

      ans:=int64(n)*int64(n-1) div 2*int64(n-2) div 3;

      writeln(ans-tmp);

end.