几何题。。。
先把所有圆储存起来，然后对于每个圆我们求得之后放下的圆挡住了的部分，求个并集，并把没被挡到的周长加进答案。
#include <cstdlib>
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <cctype>
#include <algorithm>
#define rep(i, l, r) for(int i=l; i<=r; i++)
#define pi acos(-1)
#define maxn 1234
typedef long long ll;
using namespace std;
inline int read()
{
int x=0, f=1; char ch=getchar();
while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while (isdigit(ch)) x=x*10+ch-'0', ch=getchar();
return x*f;
}
struct line{double l, r;} q[maxn*2];
bool cmp(line a, line b){return a.l<b.l || (a.l==b.l && a.r<b.r);}
double r[maxn], x[maxn], y[maxn];
double dis(int a, int b){return sqrt((x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b]));}
int main()
{
int n=read(); double ans=0;
rep(i, 1, n) scanf("%lf%lf%lf", &r[i], &x[i], &y[i]);
rep(a, 1, n)
{
int top=0;
bool end=0; rep(b, a+1, n) if (r[a]+dis(a, b)<=r[b]) end=1; if (end) continue;
rep(b, a+1, n) if (r[b]+dis(a, b)>r[a] && r[a]+r[b]>dis(a, b))
{
double r1=r[a], r2=r[b], d=dis(a, b), t, v;
t=acos((r1*r1-r2*r2+d*d)/(2*d)/r1);
v=atan2((x[a]-x[b]),(y[a]-y[b]));
q[++top]=(line){v-t, v+t};
}
rep(i, 1, top)
{
if (q[i].l<0) q[i].l+=2*pi;
if (q[i].l>2*pi) q[i].l-=2*pi;
if (q[i].r<0) q[i].r+=2*pi;
if (q[i].r>2*pi) q[i].r-=2*pi;
if (q[i].l>q[i].r)
q[++top]=(line){0,q[i].r}, q[i].r=2*pi;
}
sort(q+1, q+top+1, cmp);
double now=0, tmp=0;
rep(i, 1, top)
{
if (q[i].l>now) tmp+=q[i].l-now;
now=max(now, q[i].r);
}
tmp+=2*pi-now;
ans+=tmp*r[a];
}
printf("%.3lf
", ans);
return 0;
}