每个圆盘只会受到后边的圆盘的影响
所以算一下每个圆盘和后边的圆盘相交的圆心角，然后求个并即可
可以用余弦定理
复杂度n^2 log n
注意特判没有交-_-
#include<iostream>
#include<cstring>
#include<ctime>
#include<cmath>
#include<algorithm>
#include<iomanip>
#include<cstdlib>
#include<cstdio>
#include<map>
#include<bitset>
#include<set>
#include<stack>
#include<vector>
#include<queue>
using namespace std;
#define MAXN 1010
#define MAXM 1010
#define ll long long
#define eps 1e-8
#define MOD 1000000007
#define INF 1000000000
const double pi=acos(-1);
struct cir{
double x;
double y;
double r;
};
struct itv{
double l;
double r;
itv(){
}
itv(double _l,double _r){
l=_l;
r=_r;
}
friend bool operator <(itv x,itv y){
return x.l<y.l;
}
};
itv a[MAXN*2];
cir c[MAXN];
double ans;
int tot;
int n;
double sq(double x){
return x*x;
}
double dis(cir x,cir y){
return sqrt(sq(x.x-y.x)+sq(x.y-y.y));
}
double cal(){
int i;
double re=0;
sort(a+1,a+tot+1);
double r=-pi;
for(i=1;i<=tot;i++){
if(a[i].l>r){
re+=a[i].l-r;
}
r=max(r,a[i].r);
}
re+=pi-r;
return re;
}
int main(){
int i,j;
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%lf%lf%lf",&c[i].r,&c[i].x,&c[i].y);
}
for(i=n;i;i--){
bool flag=0;
tot=0;
for(j=i+1;j<=n;j++){
if(dis(c[i],c[j])<c[i].r+c[j].r){
if(dis(c[i],c[j])+c[i].r<=c[j].r){
flag=1;
break;
}else if(dis(c[i],c[j])+c[j].r<=c[i].r){
}else{
double angf=acos((sq(c[i].r)+sq(dis(c[i],c[j]))-sq(c[j].r))/(2*c[i].r*dis(c[i],c[j])));
double ang=atan2(c[j].y-c[i].y,c[j].x-c[i].x);
double l=ang-angf,r=ang+angf;
if(l<-pi){
a[++tot]=itv(l+2*pi,pi);
a[++tot]=itv(-pi,r);
}else if(r>pi){
a[++tot]=itv(l,pi);
a[++tot]=itv(-pi,r-2*pi);
}else{
a[++tot]=itv(l,r);
}
}
}
}
if(flag){
continue ;
}
ans+=cal()*c[i].r;
}
printf("%.3lf\n",ans);
return 0;
}
/*
*/