HDU 5130 求圆和简单多边形公共部分面积_圆和多边形面积求和-优快云博客

本文介绍了一种方法，用于计算圆和简单多边形（可凸可凹）之间的公共面积。该方法涉及将多边形点与圆和多边形的交点整合到一个有序点集中，并通过计算三角形的面积来估算公共区域。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

/*
模板：
求圆和简单多边形（可凸可凹，简单多边形是不自交的多边形）的公共面积
原理：
有序点集=多边形的点+圆和多边形的交点
然后以圆心为一点，每次按顺序（逆时针或者顺时针）取点集中两点，构成三角形，
比较三角形和对应弧形的绝对值面积大小，取绝对值较小的那块面积，按照逆时针为正面积，
顺时针为负面积求得所有三角形的有向面积和即可，最后取绝对值
*/
struct GetCirclePolyIntersectionArea{
	Circle cir;
	double Scir;

	Point p[MAXN];
	int tail;
	GetCirclePolyIntersectionArea(){tail=0;}
	GetCirclePolyIntersectionArea(Circle cir):cir(cir){
		Scir = PI*cir.r*cir.r;
		tail=0;
	}
	//tp[]是多边形的点集，n是点的个数。tp[]必须满足点是按顺时针或者逆时针排序的
	void solve(Point tp[],int n){
		for(int i=0;i<n;i++){
            p[tail++]=tp[i];//p[]是囊括了圆和多边形交点的点集，也是按顺时针或者逆时针排序的
            Line line = Line(tp[i],tp[(i+1)%n] - tp[i]);
            double t1,t2;
			vector<Point > sol;
            sol.clear();
            getLineCircleIntersection(line , cir , t1,t2,sol);

            for(int j=0;j<sol.size();j++){
                p[tail++]=sol[j];
            }
        }
        double res=0;
        for(int i=0;i<tail;i++){
            Point O = cir.c;
            double ang = Angle(p[(i+1)%tail]-O , p[i]-O);
            if( dcmp( Cross( p[i]-O , p[(i+1)%tail]-O)) > 0 ) ang*=1;
            else ang*=-1;
            double Sshan = ang/(2*PI)*Scir;
            double Strian = Area(O , p[i] ,p[(i+1)%tail] );
            if(dcmp( abs(Sshan) - abs(Strian))<=0  ){
                res += Sshan;
            }else res += Strian;
        }
        printf("Case %d: %.10f\n",Cas++,abs(res));//.10f
	}
};


/*完整代码*/
#include <iostream>
#include <cstdio>
#include <stack>
#include <cstring>
#include <queue>
#include <algorithm>
#include <cmath>
#include <map>
using namespace std;
#define N 300
typedef pair<int,int> PII;
int Cas=1;
const int INF=0x3f3f3f3f;
void Open()
{
    #ifndef ONLINE_JUDGE
        freopen("C:/OJ/in.txt","r",stdin);
        //freopen("D:/my.txt","w",stdout);
    #endif // ONLINE_JUDGE
}
const double eps = 1e-10;
const int MAXN=1111;
struct Point{
    double x,y;
    Point(double x=0 ,double y=0):x(x),y(y){}
    void read(){
        scanf("%lf%lf",&x,&y);
    }
    friend ostream &operator<< (ostream& output , Point& p){
        output<<"("<<p.x<<","<<p.y<<")";
        return output;
    }

}tp[MAXN];
typedef Point Vector;

Vector operator - (Point A, Point B){
    return Vector(A.x - B.x , A.y - B.y);
}
Vector operator + (Point A, Point B){
    return Vector(A.x + B.x , A.y + B.y);
}
Vector operator * (Point A, double p){
    return Vector(A.x * p , A.y * p);
}
Vector operator / (Point A, double p){
    return Vector(A.x / p , A.y / p);
}

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 - A.y*B.x;
}
double Length(Vector A){
    return sqrt(Dot(A,A));
}
double Area(Point A,Point B,Point C){
    return 0.5*Cross(B-A,C-A);
}
double Angle(Vector A, Vector B){
    return acos(Dot(A,B) / Length(A) / Length(B) );
}

int dcmp(double x){
    if(x>eps) return 1;
    else if(x<-eps) return -1;
    else return 0;
}

struct Line{
    Point P;
    Vector v;
    double ang;
    Line(){}
    Line(Point P, Vector v):P(P) ,v(v){ang = atan2(v.y , v.x);}
    bool operator < (const Line& L) const{
        return ang<L.ang;
    }
    Point point(double t){return P+v*t;}
};
struct Circle{
    Point c;
    double r;
    Circle(){}
    Circle(Point c, double r):c(c) , r(r){}
    Point point(double a){
        return Point(c.x + cos(a)*r ,c.y + sin(a)*r);
    }
};

int getLineCircleIntersection(Line L ,Circle C, double& t1, double& t2 , vector<Point>& sol){
    double a = L.v.x , b = L.P.x - C.c.x , c = L.v.y , d=L.P.y - C.c.y;
    double e = a*a + c* c , f= 2*(a*b + c*d) , 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);
        if(dcmp(t1)>0 && dcmp(t1-1)<0)
            sol.push_back(L.point(t1));
        return 1;
    }
    t1 = (-f - sqrt(delta)) / (2*e);
    t2 = (-f + sqrt(delta)) / (2*e);

    if(t1>t2) swap(t1,t2);

    if(dcmp(t1)>0 && dcmp(t1-1)<0)
        sol.push_back(L.point(t1));
    if(dcmp(t2)>0 && dcmp(t2-1)<0)
        sol.push_back(L.point(t2));
//    if(sol.size()==2){
//        cerr<<"2 sol:";
//        cerr<<t1<<" "<<t2<<endl;
//    }
    return (int)sol.size();
}

Point p[MAXN+MAXN];
int tail=0;
double DOU(double x){
    return x*x;
}
const double PI = acos(-1.0);

struct GetCirclePolyIntersectionArea{
	Circle cir;
	double Scir;

	Point p[MAXN];
	int tail;
	GetCirclePolyIntersectionArea(){tail=0;}
	GetCirclePolyIntersectionArea(Circle cir):cir(cir){
		Scir = PI*cir.r*cir.r;
		tail=0;
	}
	//tp[]是多边形的点集，n是点的个数。tp[]必须满足点是按顺时针或者逆时针排序的
	void solve(Point tp[],int n){
		for(int i=0;i<n;i++){
            p[tail++]=tp[i];//p[]是囊括了圆和多边形交点的点集，也是按顺时针或者逆时针排序的
            Line line = Line(tp[i],tp[(i+1)%n] - tp[i]);
            double t1,t2;
			vector<Point > sol;
            sol.clear();
            getLineCircleIntersection(line , cir , t1,t2,sol);

            for(int j=0;j<sol.size();j++){
                p[tail++]=sol[j];
            }
        }
        double res=0;
        for(int i=0;i<tail;i++){
            Point O = cir.c;
            double ang = Angle(p[(i+1)%tail]-O , p[i]-O);
            if( dcmp( Cross( p[i]-O , p[(i+1)%tail]-O)) > 0 ) ang*=1;
            else ang*=-1;
            double Sshan = ang/(2*PI)*Scir;
            double Strian = Area(O , p[i] ,p[(i+1)%tail] );
            if(dcmp( abs(Sshan) - abs(Strian))<=0  ){
                res += Sshan;
            }else res += Strian;
        }
        printf("Case %d: %.10f\n",Cas++,abs(res));//.10f
	}
};

int main()
{
    //Open();
    int n;
    double K;

    while(scanf("%d%lf",&n,&K)==2){
        tail=0;
        for(int i=0;i<n;i++){
            tp[i].read();
        }
        double ax,ay,bx,by;
        scanf("%lf%lf%lf%lf",&ax,&ay,&bx,&by);

        double mu = K*K-1;
        Point O=Point(-(bx - K * K * ax) / mu , -(by - K * K * ay) / mu);
        double R = sqrt( DOU((bx - K * K * ax)/(mu)) + DOU((by - K * K * ay) / (mu)) - (K * K * (ax*ax + ay*ay)) / mu + (bx*bx + by*by) / mu );
        Circle cir = Circle(O,R);

        GetCirclePolyIntersectionArea gcp=GetCirclePolyIntersectionArea(cir);
        gcp.solve(tp,n);
    }
    return 0;
}