计算几何-hdoj-1221-Rectangle and Circle

Rectangle and Circle

 

Problem Description

Given a rectangle and a circle in the coordinate system(two edges of the rectangle are parallel with the X-axis, and the other two are parallel with the Y-axis), you have to tell if their borders intersect.

Note: we call them intersect even if they are just tangent. The circle is located by its centre and radius, and the rectangle is located by one of its diagonal.

 

Input

The first line of input is a positive integer P which indicates the number of test cases. Then P test cases follow. Each test cases consists of seven real numbers, they are X,Y,R,X1,Y1,X2,Y2. That means the centre of a circle is (X,Y) and the radius of the circle is R, and one of the rectangle's diagonal is (X1,Y1)-(X2,Y2).

Output

For each test case, if the rectangle and the circle intersects, just output "YES" in a single line, or you should output "NO" in a single line.

Sample Input

2 1 1 1 1 2 4 3 1 1 1 1 3 4 4.5

Sample Output

YES NO

 

Source

杭州电子科技大学第三届程序设计大赛

 

分析：圆与矩形有交点的充分必要条件是：点到四个线段的最短距离dmin<=radius && 点到四个线段的最长距离dmax>=radius。

//1221 ac 
#include 
#include 
#include 
#include  
using namespace std;
struct Point{
	double x,y;
	Point(double x=0,double y=0);
};
typedef Point Vector;
Point::Point(double x,double y){
	this->x=x,this->y=y;
}
//向量与向量的+ - 点积 叉积   
Vector operator +(const Vector& A,const Vector& B){
	return Vector(A.x+B.x,A.y+B.y);
}
Vector operator -(const Vector& A,const Vector& B){
	return Vector(A.x-B.x,A.y-B.y);
}
double Dot(const Vector& A,const Vector& B){
	return (A.x*B.x+A.y*B.y);
}
double Cross(const Vector& A,const Vector& B){//叉积结果仍为向量，这里是模 
	return (A.x*B.y-B.x*A.y);
}
//向量间关系运算符
bool operator < (const Vector& A,const Vector& B){
	if(A.x0) return Length(v3);
	else return fabs(Cross(v1,v2))/Length(v1);
}
double DistanceMaxToSegment(Point P,Point A,Point B){//点到线段最远距离 
	Vector v2=P-A,v3=P-B;
	return min(Length(v2),Length(v3));
}


//线段是否“正规相交”，即交点不在四个端点上 
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){
	double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1),
	       c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
} 
bool OnSegment(Point p,Point a1,Point a2){ //点是否在线段上 
	return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}

int ncase;
Point circle,v1,v2,v3,v4;
double radius;
void f_init(){
	cin>>circle.x>>circle.y>>radius;
   	cin>>v1.x>>v1.y>>v3.x>>v3.y;
   	v2.x=v3.x,v2.y=v1.y;
   	v4.x=v1.x,v4.y=v3.y;
}
bool f_calc(){
	double dis[4],dis_max[4];
	dis[0]=DistanceToSegment(circle,v1,v2);
	dis[1]=DistanceToSegment(circle,v2,v3);
	dis[2]=DistanceToSegment(circle,v3,v4);
	dis[3]=DistanceToSegment(circle,v4,v1);
	sort(dis,dis+4);
	dis_max[0]=DistanceMaxToSegment(circle,v1,v2);
	dis_max[1]=DistanceMaxToSegment(circle,v2,v3);
	dis_max[2]=DistanceMaxToSegment(circle,v3,v4);
	dis_max[3]=DistanceMaxToSegment(circle,v4,v1);
	sort(dis_max,dis_max+4);
	if(dis[0]radius||dcmp(dis_max[3]-radius)==0)
	    return true;
	
	return false;
}
int main(int argc, char *argv[])
{
    //ifstream in("in.txt"); cin.rdbuf(in.rdbuf());
    cin>>ncase;
    while(ncase--){
    	f_init();
    	if(f_calc()) cout<<"YES\n";
    	else cout<<"NO\n";
    }
    return 0;
}