15.6 计算几何 ——【线段相交判断】

题目描述

平面直角坐标系中有一条线段AB和一条线段CD，求线段CD和线段AB的相交判断。

输入描述

输出描述

输入输出样例

输入：

3
6 2 3 4
4 5 2 2
7 7 10 1
7 7 3 9
1 6 3 5
2 5 1 4

输出：

2
1
0

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最终代码c/c++

#include
using namespace std;
const double pi = acos(-1.0);  //高精度圆周率
const double eps = 1e-8;        //偏差值

 //判断x是否等于0
int sgn(double x)
{ 
    if(fabs(x) < eps)  return 0;
    else return x<0?-1:1;
}

struct Point
{
    double x, y;
    void input(){ scanf("%lf%lf", &x, &y); }
    Point(){}
    Point(double x,double y):x(x),y(y){}
    Point operator + (Point B){return Point(x+B.x,y+B.y);}
    Point operator - (Point B){return Point(x-B.x,y-B.y);}
};

typedef Point Vector;                    //定义向量
double Cross(Point A,Point B){return A.x*B.y - A.y*B.x;} //叉积
double Dot(Vector A,Vector B){return A.x*B.x + A.y*B.y;} //点积
struct Line{
    Point p1,p2;       //线上的两个点
    Line(){}
    Line(Point p1,Point p2):p1(p1),p2(p2){}
};
//点和线段：0 点不在线段v上；1 点在线段v上
bool Point_on_seg(Point p, Line v){ 
           return sgn(Cross(p-v.p1, v.p2-v.p1)) == 0 && sgn(Dot(p - v.p1,p- v.p2)) <= 0;
}
//线段相交
bool Cross_segment(Point a,Point b,Point c,Point d){//Line1:ab,  Line2:cd
    double c1=Cross(b-a,c-a),c2=Cross(b-a,d-a);
    double d1=Cross(d-c,a-c),d2=Cross(d-c,b-c);
    return sgn(c1)*sgn(c2)<=0 && sgn(d1)*sgn(d2)<=0;  //1相交；0不相交
}
int main(){
    int t;     cin >> t;
    while(t--)
    {
        Point a, b, c,d;
        a.input(); b.input(); c.input();d.input();

        int pos=Cross_segment(a,b,c,d);
        if(pos==0) cout<<"0"<<endl; //不相交
        else{
            Line v1,v2;
            v1=Line(a,b),v2=Line(c,d);
            if(Point_on_seg(c,v1) || Point_on_seg(d,v1) || Point_on_seg(a,v2) || Point_on_seg(b,v2))   
                cout<<"1"<<endl;   //交点在端点上,不规范相交
            else
                cout<<"2"<<endl;   //交点不在端点上,规范相交

        }
    }
    return 0;
}

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过程理解