#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <map>using namespace std;typedef long long ll;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define read(a) a=getint()#define print(a) printf("%d", a)#define dbg(x) cout << (#x) << " = " << (x) << endl#define error(x) (!(x)?puts("error"):0)#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }const double PI=acos(-1), eps=1e-6;int dcmp(double x) { return abs(x)<eps?0:(x<0?-1:1); }struct ipt { double x, y; ipt(double _x=0, double _y=0) : x(_x), y(_y) {} };struct icir { ipt x; double r; };typedef ipt ivt;ivt operator- (ipt &a, ipt &b) { return ivt(a.x-b.x, a.y-b.y); }double iangle(ivt v) { return atan2(v.y, v.x); }double sqr(double x) { return x*x; }double dis(ipt &a, ipt &b) { return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y)); }bool getCC(icir &a, icir &b, double &ang1, double &ang2) {	static double d, ang;	d=dis(a.x, b.x);	if(dcmp(d)==0) return false;	if(dcmp(a.r+b.r-d)<=0) return false;	if(dcmp(a.r-b.r-d)>=0) return false;	ang=iangle(b.x-a.x);	double da=acos((sqr(d)+sqr(a.r)-sqr(b.r))/a.r/d/2);	ang1=ang-da; //dbg(da);	ang2=ang+da;	return true;}const int N=1005;int n;icir a[N];struct dat { double l, r; }line[N*2];bool cmp(const dat &a, const dat &b) { return a.l==b.l?a.r>b.r:a.l<b.l; }double work(int now) {	int ln=0;	static double ang1, ang2;	for1(i, now+1, n) { double d=dis(a[now].x, a[i].x); if(dcmp(a[i].r-a[now].r-d)>=0) return 0; } //puts("here");	for1(i, now+1, n) {		if(getCC(a[now], a[i], ang1, ang2)==false) continue;		if(ang1<0) ang1+=PI*2;		if(ang2<0) ang2+=PI*2;		if(dcmp(ang1-ang2)>0) {			++ln; line[ln].l=0; line[ln].r=ang2;			++ln; line[ln].l=ang1; line[ln].r=PI*2;		}		else { ++ln; line[ln].l=ang1; line[ln].r=ang2; }	}	sort(line+1, line+1+ln, cmp);	double w=0, mx=0;	for1(i, 1, ln) {		if(dcmp(mx-line[i].r)>=0) continue;		mx=max(mx, line[i].l);		w+=line[i].r-mx;		mx=max(mx, line[i].r);	}	return a[now].r*(PI*2-w);}int main() {	read(n);	for1(i, 1, n) scanf("%lf%lf%lf", &a[i].r, &a[i].x.x, &a[i].x.y);	double ans=0;	for1(i, 1, n) ans+=work(i);	printf("%.3f\n", ans);	return 0;}

有n个圆盘从天而降，后面落下的可以盖住前面的。求最后形成的封闭区域的周长。看下面这副图, 所有的红色线条的总长度即为所求. 

n ri xi y1 ... rn xn yn

最后的周长，保留三位小数

数据规模

n<=1000