UVa 10652 Board Wrapping 凸包

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题目地址:pdf
就是求一个凸包吧，在求四个点的坐标的时候，有两种方法，比较欣赏的还是旋转，因为没有求反三角，就是注释中的那一种
代码：
#include<iostream>#include<cmath>#include<cstdio>#include<algorithm>#include<vector>const  double eps=1e-10;const double PI=acos(-1.0);using namespace std;struct Point{    double x;    double y;    Point(double x=0,double y=0):x(x),y(y){}    void operator<<(Point &A) {cout<<A.x<<' '<<A.y<<endl;}};int dcmp(double x)  {return (x>eps)-(x<-eps); }int sgn(double x)  {return (x>eps)-(x<-eps); }typedef  Point  Vector;Vector  operator +(Vector A,Vector B) { return Vector(A.x+B.x,A.y+B.y);}Vector  operator -(Vector A,Vector B) { return Vector(A.x-B.x,A.y-B.y); }Vector  operator *(Vector A,double p) { return Vector(A.x*p,A.y*p);  }Vector  operator /(Vector A,double p) {return Vector(A.x/p,A.y/p);}ostream &operator<<(ostream & out,Point & P) { out<<P.x<<' '<<P.y<<endl; return out;}//bool  operator< (const Point &A,const Point &B) { return dcmp(A.x-B.x)<0||(dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)<0); }bool  operator== ( const Point &A,const Point &B) { return dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)==0;}double  Dot(Vector A,Vector B) {return A.x*B.x+A.y*B.y;}double  Cross(Vector A,Vector B)  {return A.x*B.y-B.x*A.y; }double  Length(Vector A)  { return sqrt(Dot(A, A));}double  Angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B));}double  Area2(Point A,Point B,Point C ) {return Cross(B-A, C-A);}Vector Rotate(Vector A,double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}Vector Normal(Vector A) {double L=Length(A);return Vector(-A.y/L,A.x/L);}Point GetLineIntersection(Point P,Vector v,Point Q,Vector w){    Vector u=P-Q;    double t=Cross(w, u)/Cross(v,w);    return P+v*t;    }double DistanceToLine(Point P,Point A,Point B){    Vector v1=P-A; Vector v2=B-A;    return fabs(Cross(v1,v2))/Length(v2);    }double DistanceToSegment(Point P,Point A,Point B){    if(A==B)  return Length(P-A);        Vector v1=B-A;    Vector v2=P-A;    Vector v3=P-B;        if(dcmp(Dot(v1,v2))==-1)    return  Length(v2);    else if(Dot(v1,v3)>0)    return Length(v3);        else return DistanceToLine(P, A, B);    }Point GetLineProjection(Point P,Point A,Point B){    Vector v=B-A;    Vector v1=P-A;    double t=Dot(v,v1)/Dot(v,v);        return  A+v*t;}bool  SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){    double c1=Cross(b1-a1, a2-a1);    double c2=Cross(b2-a1, a2-a1);    double c3=Cross(a1-b1, b2-b1);    double c4=Cross(a2-b1, b2-b1);        return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0 ;    }bool  OnSegment(Point P,Point A,Point B){    return dcmp(Cross(P-A, P-B))==0&&dcmp(Dot(P-A,P-B))<=0;}double PolygonArea(Point *p,int n){    double area=0;        for(int i=1;i<n-1;i++)    {        area+=Cross(p[i]-p[0], p[i+1]-p[0]);            }    return area/2;    }Point  read_point(){    Point P;    scanf("%lf%lf",&P.x,&P.y);    return  P;}// ---------------与圆有关的--------struct Circle{    Point c;    double r;        Circle(Point c=Point(0,0),double r=0):c(c),r(r) {}        Point point(double a)    {        return Point(c.x+r*cos(a),c.y+r*sin(a));    }        };struct  Line{    Point p;    Vector v;    Line(Point p=Point(0,0),Vector v=Vector(0,1)):p(p),v(v) {}        Point point(double t)    {        return Point(p+v*t);    }    };int getLineCircleIntersection(Line L,Circle C,double &t1,double &t2,vector<Point> &sol){    double a=L.v.x;    double b=L.p.x-C.c.x;    double c=L.v.y;    double d=L.p.y-C.c.y;        double e=a*a+c*c;    double f=2*(a*b+c*d);    double g=b*b+d*d-C.r*C.r;        double delta=f*f-4*e*g;        if(dcmp(delta)<0) return 0;        if(dcmp(delta)==0)    {        t1=t2=-f/(2*e);        sol.push_back(L.point(t1));        return 1;    }        else    {        t1=(-f-sqrt(delta))/(2*e);        t2=(-f+sqrt(delta))/(2*e);                sol.push_back(L.point(t1));        sol.push_back(L.point(t2));                return 2;    }    }// 向量极角公式double angle(Vector v)  {return atan2(v.y,v.x);}int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point> &sol){    double d=Length(C1.c-C2.c);        if(dcmp(d)==0)    {        if(dcmp(C1.r-C2.r)==0)  return -1;  // 重合        else return 0;    //  内含  0 个公共点    }        if(dcmp(C1.r+C2.r-d)<0)  return 0;  // 外离    if(dcmp(fabs(C1.r-C2.r)-d)>0)  return 0;  // 内含        double a=angle(C2.c-C1.c);    double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*C1.r*d));        Point p1=C1.point(a-da);    Point p2=C1.point(a+da);        sol.push_back(p1);        if(p1==p2)  return 1; // 相切    else    {        sol.push_back(p2);        return 2;    }}//  求点到圆的切线int getTangents(Point p,Circle C,Vector *v){    Vector u=C.c-p;        double dist=Length(u);        if(dcmp(dist-C.r)<0)  return 0;        else if(dcmp(dist-C.r)==0)    {        v[0]=Rotate(u,PI/2);        return 1;    }        else    {                double ang=asin(C.r/dist);        v[0]=Rotate(u,-ang);        v[1]=Rotate(u,+ang);        return 2;    }    }//  求两圆公切线int getTangents(Circle A,Circle B,Point *a,Point *b){    int cnt=0;        if(A.r<B.r)    {        swap(A,B); swap(a, b);  //  有时需标记    }        double d=Length(A.c-B.c);        double rdiff=A.r-B.r;    double rsum=A.r+B.r;        if(dcmp(d-rdiff)<0)  return 0;   // 内含        double base=angle(B.c-A.c);        if(dcmp(d)==0&&dcmp(rdiff)==0)   return -1 ;  // 重合 无穷多条切线        if(dcmp(d-rdiff)==0)             // 内切   外公切线    {        a[cnt]=A.point(base);        b[cnt]=B.point(base);        cnt++;        return 1;    }        // 有外公切线的情形        double ang=acos(rdiff/d);    a[cnt]=A.point(base+ang);    b[cnt]=B.point(base+ang);    cnt++;    a[cnt]=A.point(base-ang);    b[cnt]=B.point(base-ang);    cnt++;        if(dcmp(d-rsum)==0)     // 外切 有内公切线    {        a[cnt]=A.point(base);        b[cnt]=B.point(base+PI);        cnt++;    }        else  if(dcmp(d-rsum)>0)   // 外离   又有两条外公切线    {        double  ang_in=acos(rsum/d);        a[cnt]=A.point(base+ang_in);        b[cnt]=B.point(base+ang_in+PI);        cnt++;        a[cnt]=A.point(base-ang_in);        b[cnt]=B.point(base-ang_in+PI);        cnt++;    }        return cnt;}//  几何算法模板int  isPointInPolygon(Point p,Point * poly,int n){    int wn=0;    for(int i=0;i<n;i++)    {        if(OnSegment(p, poly[i], poly[(i+1)%n]))  return -1;        int k=dcmp(Cross(poly[(i+1)%n]-poly[i], p-poly[i]));        int d1=dcmp(poly[i].y-p.y);        int d2=dcmp(poly[(i+1)%n].y-p.y);                if(k>0&&d1<=0&&d2>0) wn++;        if(k<0&&d2<=0&&d1>0) wn--;            }        if(wn!=0)  return 1;    else   return 0;    }//  Andrew 算法求凸包int ConvexHull(Point *p,int n,Point *ch){    int m=0;    sort(p,p+n);        n=unique(p, p+n)-p;        for(int i=0;i<n;i++)    {        while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)  m--;        ch[m++]=p[i];    }        int k=m;        for(int i=n-2;i>=0;i--)    {       while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)   m--;        ch[m++]=p[i];    }        if(n>1) m--;        return m;         }Point p[2500];Point ch[2500];int main(){            int T;    cin>>T;    Point cur;    double w,h,j;    while(T--)    {   double area1=0;                int n;        int cnt=0;        cin>>n;        Circle C;        for(int i=0;i<n;i++)        {            C.c=read_point();            cin>>w>>h>>j;            area1+=w*h;                                               C.r=sqrt(w*w/4+h*h/4);                        double angle=atan(h/w);                        j=j/180*PI;            p[cnt++]=C.point(angle-j);            p[cnt++]=C.point(angle-j+PI);            p[cnt++]=C.point(-angle-j);            p[cnt++]=C.point(-angle-j+PI);            //            j=-j;//            p[cnt++]=C.c+Rotate(Vector(w/2,h/2),j);//            p[cnt++]=C.c+Rotate(Vector(-w/2,h/2),j);//            p[cnt++]=C.c+Rotate(Vector(w/2,-h/2),j);//            p[cnt++]=C.c+Rotate(Vector(-w/2,-h/2),j);                    }                        int m=ConvexHull(p, cnt, ch);                double area2=fabs(PolygonArea(ch, m));                double rate=area1/area2*100;                printf("%.1lf %%\n",rate);            }}
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