UVa11978福岛核爆问题：圆与多边形的交集面积计算及二分法应用

题目地址：pdf版本

和上一个题目几乎是一模一样的，就不说思路了。

但是呢，这次是用for 50次写的二分，然后，容易wa的一点是，要把p[n] 也进行平移，漏掉了就wa了啊

代码：

#include
#include
#include
#include
#include
const  long double eps=1e-10;
const long double PI=acos(-1.0);

using namespace std;


struct Point{
    long double x;
    long double y;
    Point(long double x=0,long double y=0):x(x),y(y){}
    void operator<<(Point &A) {cout<eps)-(x<-eps); }
int sgn(long 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,long double p) { return Vector(A.x*p,A.y*p);  }

Vector  operator /(Vector A,long double p) {return Vector(A.x/p,A.y/p);}



ostream &operator<<(ostream & out,Point & P) { out<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;
    long double t=Dot(v,v1)/Dot(v,v);
    
    return  A+v*t;
}

bool  SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2)
{
    long double c1=Cross(b1-a1, a2-a1);
    long double c2=Cross(b2-a1, a2-a1);
    long double c3=Cross(a1-b1, b2-b1);
    long 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;
}

long double PolygonArea(Point *p,int n)
{
    long double area=0;
    
    for(int i=1;i &sol)
{
    long double a=L.v.x;
    long double b=L.p.x-C.c.x;
    long double c=L.v.y;
    long double d=L.p.y-C.c.y;
    
    long double e=a*a+c*c;
    long double f=2*(a*b+c*d);
    long double g=b*b+d*d-C.r*C.r;
    
    long 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

long double angle(Vector v)  {return atan2(v.y,v.x);}

int getCircleCircleIntersection(Circle C1,Circle C2,vector &sol)
{
    long 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
    
    long double a=angle(C2.c-C1.c);
    long 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;
    
    long 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
    {
        
        long 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.r0)   // 外离   又有两条外公切线
    {
        long 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;
}



Point Zero=Point(0,0);

long double  common_area(Circle C,Point A,Point B)
{
   // if(A==B)  return 0;
    if(A==C.c||B==C.c)  return 0;

    long double  OA=Length(A-C.c),OB=Length(B-C.c);
    long double  d=DistanceToLine(Zero, A, B);

    int sg=sgn(Cross(A,B));
    if(sg==0)  return 0;

    long double angle=Angle(A,B);

    if(dcmp(OA-C.r)<=0&&dcmp(OB-C.r)<=0)
    {
        return Cross(A,B)/2;

    }

    else if(dcmp(OA-C.r)>=0&&dcmp(OB-C.r)>=0&&dcmp(d-C.r)>=0)
    {
        return  sg*C.r*C.r*angle/2;

    }
    else if (dcmp(OA-C.r)>=0&&dcmp(OB-C.r)>=0&&dcmp(d-C.r)<0)
    {


        Point prj=GetLineProjection(Zero, A, B);

        if(OnSegment(prj, A, B))
        {
        vector p;
        Line L=Line(A,B-A);
        long double t1,t2;
        getLineCircleIntersection(L,C, t1, t2, p);

        long double s1=0;
        s1=C.r*C.r*angle/2;

        long double s2=0;
        s2=C.r*C.r*Angle(p[0],p[1])/2;
        s2-=fabs(Cross(p[0],p[1])/2);
        s1=s1-s2;

        return  sg*s1;
        }

        else
        {
            return sg*C.r*C.r*angle/2;
        }

    }
    else
    {
       
            if(dcmp(OB-C.r)<0)
            {
               
                Point temp=A;
                A=B;
                B=temp;
            }

         Point inter_point;
        
            long double t1,t2;
            Line L=Line(A,B-A);
            vector inter;
            getLineCircleIntersection(L, C, t1, t2,inter);
        
                    if(OnSegment(inter[0], A, B))
                inter_point=inter[0];
            else
            {
                inter_point=inter[1];

            }

        
//        两种方法求交点都可以
//        Point  prj=GetLineProjection(Zero, A, B);
//        
//        Vector v=B-A;
//        v=v/Length(v);
//        long long double mov=sqrt(C.r*C.r-d*d);
//        
//        if(OnSegment(prj+v*mov, A, B))
//        {
//            inter_point=prj+v*mov;
//        }
//        else
//        {
//            inter_point=prj+Vector(-v.x,-v.y)*mov;
//        }
        
        
            long double s=fabs(Cross(inter_point, A)/2);
            s+=C.r*C.r*Angle(inter_point,B)/2;

            return s*sg;


    }
}

Point p[5500];

int main()
{
    int n;
    Circle C;
    int index=0;
    int T;
    cin>>T;
    long  double  P;
    while(cin>>n)
    {
        for(int i=0;i>P;
        
        
        for(int i=0;i<=n;i++)   // wa!
            p[i]=p[i]-C.c;
        
    
        
        
        C.c=Zero;
        
       
    
        long double l=0,r=1e4;
        long double mid=l;
        
        for(int tim=0;tim<50;tim++)
        {
            mid=l+(r-l)/2;
            
            C.r=mid;
            
            long double ans=0;
        
           for(int i=0;i=area*P/100.0)  r=mid;
            else l=mid;
        }
        
       // if(mid-floor(mid)>0.5) mid+=1;
        printf("Case %d: %.0Lf\n",++index,mid);
        
        
    }

}