PKU 2007 Scrambled Polygon 极角排序

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输入一个凸包且没有三点共线，输入的第一个点始终是（0，0），从（0，0）开始逆时针输出所有点。

因为只有三个象限有点，所以分象限极角排序输出即可。

var  u,d1,d2,d3,d4,i,x,y:longint;  s1,s2,s3,s4:array[0..50,1..2] of longint;procedure work1;var  i,j:longint;begin  for i:=1 to d1-1 do    for j:=i+1 to d1 do      if s1[i,1]*s1[j,2]-s1[i,2]*s1[j,1]<0 then      begin        s1[0]:=s1[i];s1[i]:=s1[j];s1[j]:=s1[0];      end;  for i:=1 to d1 do    writeln('(',s1[i,1],',',s1[i,2],')');end;procedure work2;var  i,j:longint;begin  for i:=1 to d2-1 do    for j:=i+1 to d2 do      if s2[i,1]*s2[j,2]-s2[i,2]*s2[j,1]<0 then      begin        s2[0]:=s2[i];s2[i]:=s2[j];s2[j]:=s2[0];      end;  for i:=1 to d2 do    writeln('(',s2[i,1],',',s2[i,2],')');end;procedure work3;var  i,j:longint;begin  for i:=1 to d3-1 do    for j:=i+1 to d3 do      if s3[i,1]*s3[j,2]-s3[i,2]*s3[j,1]<0 then      begin        s3[0]:=s3[i];s3[i]:=s3[j];s3[j]:=s3[0];      end;  for i:=1 to d3 do    writeln('(',s3[i,1],',',s3[i,2],')');end;procedure work4;var  i,j:longint;begin  for i:=1 to d4-1 do    for j:=i+1 to d4 do      if s4[i,1]*s4[j,2]-s4[i,2]*s4[j,1]<0 then      begin        s4[0]:=s4[i];s4[i]:=s4[j];s4[j]:=s4[0];      end;  for i:=1 to d4 do    writeln('(',s4[i,1],',',s4[i,2],')');end;begin  while not eof do   begin    readln(x,y);    if (x<>0)and(y<>0) then      if (x>0)and(y>0)        then begin               inc(d1);               s1[d1,1]:=x;               s1[d1,2]:=y;             end        else if (x<0)and(y>0)               then begin                      inc(d2);                      s2[d2,1]:=x;                      s2[d2,2]:=y;                    end               else if (x<0)and(y<0)                      then begin                             inc(d3);                             s3[d3,1]:=x;                             s3[d3,2]:=y;                           end                      else begin                             inc(d4);                             s4[d4,1]:=x;                             s4[d4,2]:=y;                           end;  end;  if d1=0    then u:=2    else if d2=0           then u:=3           else if d3=0                  then u:=4                  else u:=1;  writeln('(0,0)');  for i:=1 to 3 do  begin    case u of      1:work1;      2:work2;      3:work3;      4:work4;    end;    u:=u mod 4+1;  end;end.

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