hdu 4667（几何）

解题思路：求出所有可能在外围的点。也就是将两圆两两求外公切线，得到所有的切点；将圆和三角形的三个点依次做切线，

也得到切点，再加上所有三角形的三个点。最后对这些点求凸包，对于凸包上的每条边，如果它们在同一个圆上，用相应的圆弧替代。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<math.h>
#include<bitset>
#include<string>
#include<queue>
#include<deque>
#include<ctype.h>
#include<vector>
#include<set>
#include<map>
#include<list>
#include<stack>
#include<functional>
#include<algorithm>

using namespace std;

#define N 100010

const double eps=1e-12;
const double pi=acos(-1.0);

struct Point{
    double x,y;
    int id;
    Point(double a=0,double b=0):x(a),y(b){}
};
typedef Point Vector;

Point operator-(Point a,Point b){ return Point(a.x-b.x,a.y-b.y); }
Point operator+(Point a,Point b){ return Point(a.x+b.x,a.y+b.y); }
double operator*(Vector a,Vector b){ return a.x*b.x+a.y*b.y; }
int operator==(Point a,Point b){ return a.x==b.x&&a.y==b.y; }
double Len(Vector a){  return sqrt(a.x*a.x+a.y*a.y); }

Vector Rotate(Vector a,double rad){ //向量逆时针旋转rad弧度
    return Vector(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}

int dcmp(double x){
    if(x>-eps&&x<eps) return 0;
    if(x>eps) return 1;
    return -1;
}
double cross(Point a,Point b){ //叉积
    return a.x*b.y-a.y*b.x;
}
double angle(Vector a,Vector b){ //两个向量的夹角
    return acos((a*b)/(Len(a)*Len(b)));
}
int online(Point a,Point b,Point c){ //判断点c是否在线段ab上
    return  (a==c)||(b==c)||(cross(a-c,b-c)==0&&((a-c)*(b-c)<0));
}

int segmentIntersection(Point a1,Point a2,Point b1,Point b2) //判断线段a1a2和b1b2相交（不包含端点）
{
    double c1=cross((a2-a1),(b1-a1)),c2=cross((a2-a1),(b2-a1));
    double c3=cross((b2-b1),(a1-b1)),c4=cross((b2-b1),(a2-b1));
    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}

int Polygon_contains(Point point,Point *points,int n) {  //判断一个点在多边形内
    int i, j, status=0;
    for (i=0,j=n-1;i<n;j=i++){
        if(online(points[i],points[j],point)) return 0;
        if((((points[i].y<=point.y)&&(point.y<points[j].y))
        ||((points[j].y<=point.y)&&(point.y<points[i].y)))
        &&(point.x <(points[j].x - points[i].x)*(point.y - points[i].y)
        /(points[j].y - points[i].y) + points[i].x))
            status = !status;
    }
    return status;
}

Point P[N];
bool cmp(Point a,Point b){   //极坐标排序
    if(cross(a-P[0],b-P[0])==0)
        return dcmp(Len(a-P[0])-Len(b-P[0]))<=0;
    else
        return cross(a-P[0],b-P[0])>0;
}

void Graham(int n,vector<Point> &sta){  //求解二维凸包
    for(int i=1;i<n;i++){
        if(dcmp(P[i].y-P[0].y)<0||(P[i].y==P[0].y&&P[i].x<P[0].x))
            swap(P[i],P[0]);
    }
    sort(P+1,P+n,cmp);
    sta.push_back(P[0]);
    sta.push_back(P[1]);
    for(int i=2;i<n;i++){
        while(sta.size()>1&&dcmp(cross(sta[sta.size()-1]-sta[sta.size()-2],P[i]-sta[sta.size()-1]))<=0)
            sta.pop_back();
        sta.push_back(P[i]);
    }
}
struct circle{
    double x,y,r;
    Point make_Point(double a){
        return Point(x+cos(a)*r,y+sin(a)*r);
    }
}C[60];
int getTangents(Point p,circle cc,Vector *t){ //求一个点p到圆cc的切线向量
    Vector u=Point(cc.x,cc.y)-p;
    double dist=Len(u);
    if(dcmp(dist-cc.r)<0) return 0;
    else if(dcmp(dist-cc.r)==0){
        t[0]=Rotate(u,pi/2);
        return 1;
    }else {
        double ang=asin(cc.r/dist);
        t[0]=Rotate(u,-ang);
        t[1]=Rotate(u,+ang);
        return 2;
    }
}
void meet(Point p,circle cc,Point *t){ //求一个点p到圆cc的切点
    double dis,l;
    Vector u=Point(cc.x,cc.y)-p;
    getTangents(p,cc,t);l=Len(t[0]);
    dis=sqrt(Len(u)*Len(u)-cc.r*cc.r);
    t[0].x=t[0].x*dis/l;t[0].y=t[0].y*dis/l;
    t[1].x=t[1].x*dis/l;t[1].y=t[1].y*dis/l;
    t[0]=t[0]+p;
    t[1]=t[1]+p;
}

double D(Point a,Point b,int id){ //求a点到b点在的圆上的圆弧长度
    double ang1,ang2;
    Vector v1,v2;
    v1=a-Point(C[id].x,C[id].y);
    v2=b-Point(C[id].x,C[id].y);
    ang1=atan2(v1.y,v1.x);
    ang2=atan2(v2.y,v2.x);
    if(ang2<ang1) ang2+=2*pi;
    return C[id].r*(ang2-ang1);
}

int getgqx(circle A,circle B,Point* sa,Point* sb) //求圆A和B的公切线
{
    int cnt=0;
    Point *a=sa,*b=sb;
    if(A.r<B.r){ swap(A,B);swap(a,b);}
    double d2=(A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y);
    double rdiff=A.r-B.r;
    double rsum=A.r+B.r;
    if(dcmp(d2-rdiff*rdiff)<0) return 0;    //内含

    double base=atan2(B.y-A.y,B.x-A.x);
    if(dcmp(d2)==0&&dcmp(A.r-B.r)==0) return -1; //无限多条切线
    if(dcmp(d2-rdiff*rdiff)==0)
    {
        a[cnt]=A.make_Point(base);b[cnt]=B.make_Point(base);cnt++;
        return 1;
    }
    //有外共切线
    double ang=acos((A.r-B.r)/sqrt(d2));
    a[cnt]=A.make_Point(base+ang);b[cnt]=B.make_Point(base+ang);cnt++;
    a[cnt]=A.make_Point(base-ang);b[cnt]=B.make_Point(base-ang);cnt++;
    if(dcmp(d2-rsum*rsum)==0)//一条内共切线
    {
        a[cnt]=A.make_Point(base);b[cnt]=B.make_Point(pi+base);cnt++;
    }
    else if(dcmp(d2-rsum*rsum)>0)
    {
        double ang=acos((A.r+B.r)/sqrt(d2));
        a[cnt]=A.make_Point(base+ang);b[cnt]=B.make_Point(pi+base+ang);cnt++;
        a[cnt]=A.make_Point(base-ang);b[cnt]=B.make_Point(pi+base-ang);cnt++;
    }
    return cnt;
}

int main(){
    int n,m,i,M,j;
    while(scanf("%d%d",&n,&m)!=EOF){
        for(i=0;i<n;i++){
            scanf("%lf%lf%lf",&C[i].x,&C[i].y,&C[i].r);
        }
        for(i=0;i<m;i++){
            scanf("%lf%lf",&P[3*i].x,&P[3*i].y);
            scanf("%lf%lf",&P[3*i+1].x,&P[3*i+1].y);
            scanf("%lf%lf",&P[3*i+2].x,&P[3*i+2].y);
        }
        if(n==1&&m==0){
            printf("%lf\n",pi*2*C[0].r);
            continue;
        }
        M=3*m;m*=3;
        for(i=0;i<M;i++) P[i].id=i-M-M;
        double ans=0;
        vector<Point> sta;
        if(M>0){
            Graham(M,sta);
            for(i=0;i<sta.size();i++) P[i]=sta[i];
            M=sta.size();m=M;
        }
        Point t[4],T[4];
        for(i=0;i<n;i++){
            for(j=i+1;j<n;j++){
                getgqx(C[i],C[j],t,T);
                t[0].id=t[1].id=i;
                T[0].id=T[1].id=j;
                P[m++]=t[0];P[m++]=t[1];
                P[m++]=T[0];P[m++]=T[1];
            }
        }
        for(i=0;i<M;i++){
            for(j=0;j<n;j++){
                meet(P[i],C[j],t);
                t[0].id=t[1].id=j;
                P[m++]=t[0];P[m++]=t[1];
            }
        }
        M=m;
        sta.clear();
        Graham(M,sta);
        if(sta[0].id==sta[sta.size()-1].id)
            ans+=D(sta[sta.size()-1],sta[0],sta[0].id);
        else
            ans+=Len(sta[0]-sta[sta.size()-1]);
        for(i=1;i<int(sta.size());i++){
            if(sta[i].id==sta[i-1].id)
                ans+=D(sta[i-1],sta[i],sta[i].id);
            else
                ans+=Len(sta[i]-sta[i-1]);
        }
        printf("%.8lf\n",ans);
    }
    return 0;
}