这道题题目很短，看一下数据范围n<1000，直接枚举圆盘。
对于圆盘i，枚举它上面的圆盘，计算圆盘i被它们覆盖的弧的起始位置与终止位置，然后线段覆盖可得圆盘i被覆盖的周长，就可得答案。
注意圆外离与内含的情况。
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define pi 3.1415926535897932384626433832795
using namespace std;
struct circle{double r,x,y;}a[10000];
struct arc{double a,b;}q[10000];
int n,le;double Ans;
double di(circle a,circle b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double ang(double R,double d,double r)
{
return acos((R*R+d*d-r*r)/(2*R*d));
}
bool cmp(arc a,arc b)
{
if (a.a==b.a)return a.b<b.b;
else return a.a<b.a;
}
void add(double l,double r)
{
q[++le].a=l,q[le].b=r;
}
int main()
{
freopen("disc.in","r",stdin);
freopen("disc.out","w",stdout);
scanf("%d",&n);
for (int i=1;i<=n;i++)
scanf("%lf%lf%lf",&a[i].r,&a[i].x,&a[i].y);
for (int i=1;i<=n;i++)
{
bool f=0;le=0;
for (int j=i+1;j<=n;j++)
{
double dis=di(a[i],a[j]);
if (dis<=a[j].r-a[i].r) {f=1;break;}
if (dis>=a[i].r+a[j].r||dis<=a[i].r-a[j].r)
continue;
double x=a[j].x-a[i].x,y=a[j].y-a[i].y;
double a1=atan2(x,y)+pi;
double a2=ang(a[i].r,dis,a[j].r);
if (a1-a2<0) add(0,a1+a2), add(a1-a2+2*pi,2*pi);else
if (a1+a2>2*pi) add(a1-a2,pi*2),add(0,a1+a2-2*pi);
else add(a1-a2,a1+a2);
}
if (f==1)continue;
sort(q+1,q+le+1,cmp);
double l=-1,r=-1,sum=0;
for (int j=1;j<=le;j++)
if (q[j].a<=r)
r=max(r,q[j].b);
else
{ sum+=r-l; l=q[j].a; r=q[j].b; }
sum+=r-l;
Ans+=(2*pi-sum)*a[i].r;
}
printf("%.3f\n",Ans);
return 0;
}