UVA10969计算几何+交叉圆形成的圆弧长

/*UVA 10969计算几何
这道题和LA2572相似，但相对简单些。
思路：求圆间的交点，顺序枚举出圆上的圆弧，中点判断是否被覆盖。
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <ctype.h>
#include <string>
#include <iostream>
#include <sstream>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <set>
#include <algorithm>
#define INF 0x3f3f3f3f
#define eps 1e-7
#define eps2 1e-3
using namespace std;

struct Point
{
    double x,y;
    Point() {}
    Point(double xx,double yy)
    {
        x=xx;
        y=yy;
    }
};
struct Circle
{
    Point O;
    double r;
    Circle() {}
    Circle(Point O1,double r1)
    {
        O=O1;
        r=r1;
    }
    Point point(double a)
    {
        return Point(O.x+cos(a)*r,O.y+sin(a)*r);
    }
};
typedef Point Vector;

bool operator==(Point A,Point B)
{
    if ((fabs(A.x-B.x)<1e-10) && (fabs(A.y-B.y)<1e-10)) return true;
    else return false;
}
Vector operator-(Point A,Point B)//表示A指向B
{
    return Vector(A.x-B.x,A.y-B.y);
}
Vector operator*(Vector A,double k)
{
    return Vector(A.x*k,A.y*k);
}
Vector operator+(Point A,Point B)//表示A指向B
{
    return Vector(B.x+A.x,B.y+A.y);
}
double Dot(Vector A,Vector B)
{
    return A.x*B.x+A.y*B.y;
}
double Length(Vector A)
{
    return sqrt(Dot(A,A));
}
double Cross(Vector A,Vector B)
{
    return A.x*B.y-A.y*B.x;
}
int dcmp(double x)
{
    if(fabs(x)<1e-10) return 0;
    else if(x>0) return 1;
    else return -1;
}
double angle(Vector v)
{
    return atan2(v.y,v.x);
}
double normal(double rad)
{
    return rad-2*M_PI*floor(rad/(2*M_PI));
}

Point GetMid(double a,double b,double r,double x,double y)//圆上两点，获得中点的坐标
{
    double arf=(b+a)/2;
    double xx=r*cos(arf)+x;
    double xy=r*sin(arf)+y;
    Point P=Point(xx,xy);
    return P;
}
bool PInCircle(Point P,Circle C)//判断点在圆内
{
    double dis=Length(P-C.O);
    if (dis-C.r>1e-6) return false;
    else return true;
}
void getcircleinter(Point c1,double r1,Point c2,double r2,vector<double>& rad)
{
    double d=Length(c1-c2);
    if(dcmp(d)==0) return;
    if(dcmp(r1+r2-d)<0) return;
    if(dcmp(fabs(r1-r2)-d)>0) return;
    double a=angle(c2-c1);
    double da=acos((r1*r1+d*d-r2*r2)/(2*r1*d));
    rad.push_back(normal(a+da));
    rad.push_back(normal(a-da));
}

int n;
Circle CC[110];//圆
vector<double> Arf[110];//圆心角

int cas;
int main()
{
    cin>>cas;
    while(cas--)
    {
        double ans=0.0;
        cin>>n;
        for(int i=1; i<=n; i++)//读取信息
        {
            double x,y,r;
            cin>>r>>x>>y;
            CC[i]=Circle(Point(x,y),r);
        }
        for(int i=1; i<=n; i++) {Arf[i].clear();Arf[i].push_back(0);Arf[i].push_back(2*M_PI);}

        for(int i=1; i<=n; i++)//获得圆上的交点的角度
        for(int j=1; j<=n; j++)
            getcircleinter(CC[i].O,CC[i].r,CC[j].O,CC[j].r,Arf[i]);

        for(int i=1; i<=n; i++) //枚举n个圆上的圆弧
        {
            int k=Arf[i].size();
            sort(Arf[i].begin(),Arf[i].end());

            for(int j=0; j<k-1; j++) //枚举每条弧
            {
                Point Mid=GetMid(Arf[i][j],Arf[i][j+1],CC[i].r,CC[i].O.x,CC[i].O.y);

                bool ok=true;//表示这条圆弧可见
                for(int l=i+1; l<=n; l++) //判断中点是否可见
                    if (PInCircle(Mid,CC[l])) ok=false;

                if(ok) ans+=(Arf[i][j+1]-Arf[i][j])*CC[i].r;
            }
        }

        printf("%.3lf
",(ans));//放缩回来
    }
    return 0;
}