bzoj1043

每次做计算几何题都要做好久

考虑每个圆对答案的贡献，也就是每个圆被后面圆覆盖还有多少

可以把覆盖当成盖住一段弧度，看最后有多少没被覆盖

这就相当于线段覆盖问题了，

推推公式，算极角然后排序即可

md，pascal算极角就是麻烦

 uses math;
 const pi=3.1415926535897932384626433832795;
       eps=1e-4;
 type node=record
        l,r:double;
      end;

 var q:array[..] of node;
     x,y,r:array[..] of double;
     t,j,i,n:longint;
     ans:double;

 function dis(i,j:longint):double;
   begin
     exit(sqrt(sqr(x[i]-x[j])+sqr(y[i]-y[j])));
   end;

 function have(i,j:longint):boolean;
   begin
     exit(r[j]-r[i]>=dis(i,j));
   end;

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

 function get(x,y:double):double;
   begin
     if x= then
     begin
       if y> then exit(pi/)
       else exit(-pi/);
     end;
     get:=arctan(y/x);
     if (x<) then get:=get+pi;
   end;

 procedure sort(l,r:longint);
   var i,j:longint;
       x:double;
   begin
     i:=l;
     j:=r;
     x:=q[(l+r) shr ].l;
     repeat
       while q[i].l<x do inc(i);
       while x<q[j].l do dec(j);
       if not(i>j) then
       begin
         swap(q[i],q[j]);
         inc(i);
         dec(j);
       end;
     until i>j;
     if l<j then sort(l,j);
     if i<r then sort(i,r);
   end;

 function cal(j:longint):double;
   var i:longint;
       d,now,l,z,an:double;
   begin
     for i:=j+ to n do
       if have(j,i) then exit();
     t:=;
     for i:=j+ to n do
     begin
       d:=dis(i,j);
       if not have(i,j) and (r[j]+r[i]>d) then
       begin
         inc(t);
         an:=get(x[i]-x[j],y[i]-y[j]);
         l:=(sqr(r[j])-sqr(r[i])+sqr(d))/(*d);
         z:=arccos(l/r[j]);
         q[t].l:=an-z;
         q[t].r:=an+z;
      //   writeln(an,' ',z,' ',i,' ',x[i]-x[j],' ',y[i]-y[j]);
      //   readln;
       end;
     end;
     for i:= to t do
     begin
       if q[i].l>*pi then q[i].l:=q[i].l-*pi;
       if q[i].r>*pi then q[i].r:=q[i].r-*pi;
       if q[i].l< then q[i].l:=q[i].l+*pi;
       if q[i].r< then q[i].r:=q[i].r+*pi;
       if q[i].l>q[i].r then
       begin
         inc(t);
         q[t].l:=;
         q[t].r:=q[i].r;
         q[i].r:=*pi;
       end;
     end;
     sort(,t);
     cal:=;
     now:=;
     for i:= to t do
       if q[i].l>now then
       begin
         cal:=cal+(q[i].l-now);
         now:=q[i].r;
       end
       else if now<q[i].r then now:=q[i].r;

     cal:=cal+*pi-now;
     if cal< then cal:=;
     cal:=cal*r[j];
   end;

 begin
   readln(n);
   for i:= to n do
     readln(r[i],x[i],y[i]);

   for i:= to n do
     ans:=ans+cal(i);

   writeln(ans::);
 end.