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  • POJ 1269 Intersecting Lines --计算几何

    题意: 二维平面,给两条线段,判断形成的直线是否重合,或是相交于一点,或是不相交。

    解法: 简单几何。 

    重合: 叉积为0,且一条线段的一个端点到另一条直线的距离为0

    不相交: 不满足重合的情况下叉积为0

    相交于一点: 直线相交的模板

    代码:

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <cmath>
    #include <algorithm>
    #define pi acos(-1.0)
    #define eps 1e-8
    using namespace std;
    #define N 100017
    
    struct Point{
        double x,y;
        Point(double x=0, double y=0):x(x),y(y) {}
        void input() { scanf("%lf%lf",&x,&y); }
    };
    typedef Point Vector;
    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); }
        void input() { scanf("%lf%lf%lf",&c.x,&c.y,&r); }
    };
    struct Line{
        Point p;
        Vector v;
        double ang;
        Line(){}
        Line(Point p, Vector v):p(p),v(v) { ang = atan2(v.y,v.x); }
        Point point(double t) { return Point(p.x + t*v.x, p.y + t*v.y); }
        bool operator < (const Line &L)const { return ang < L.ang; }
    };
    int dcmp(double x) {
        if(x < -eps) return -1;
        if(x > eps) return 1;
        return 0;
    }
    template <class T> T sqr(T x) { return x * x;}
    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); }
    bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
    bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
    bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
    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 Length(Vector A) { return sqrt(Dot(A, A)); }
    double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
    double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
    Vector VectorUnit(Vector x){ return x / Length(x);}
    Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);}
    double angle(Vector v) { return atan2(v.y, v.x); }
    
    bool OnSegment(Point P, Point A, Point B) {
        return dcmp(Cross(A-P,B-P)) == 0 && dcmp(Dot(A-P,B-P)) < 0;
    }
    double DistanceToSeg(Point P, Point A, Point B)
    {
        if(A == B) return Length(P-A);
        Vector v1 = B-A, v2 = P-A, v3 = P-B;
        if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
        if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
        return fabs(Cross(v1, v2)) / Length(v1);
    }
    double DistanceToLine(Point P, Point A, Point B){
        Vector v1 = B-A, v2 = P-A;
        return fabs(Cross(v1,v2)) / Length(v1);
    }
    Point GetLineIntersection(Line A, Line B){
        Vector u = A.p - B.p;
        double t = Cross(B.v, u) / Cross(A.v, B.v);
        return A.p + A.v*t;
    }
    
    //data segment
    //data ends
    
    int main()
    {
        Point A,B,C,D;
        int n,i,j;
        scanf("%d",&n);
        {
            puts("INTERSECTING LINES OUTPUT");
            for(i=1;i<=n;i++)
            {
                scanf("%lf%lf%lf%lf",&A.x,&A.y,&B.x,&B.y);
                scanf("%lf%lf%lf%lf",&C.x,&C.y,&D.x,&D.y);
                Line L1 = Line(A,B-A);
                Line L2 = Line(C,D-C);
                if(dcmp(Cross(L1.v,L2.v)) == 0 && dcmp(DistanceToLine(A,C,D)) == 0)
                    puts("LINE");
                else if(dcmp(Cross(L1.v,L2.v)) == 0)
                    puts("NONE");
                else
                    printf("POINT %.2f %.2f
    ",GetLineIntersection(L1,L2).x,GetLineIntersection(L1,L2).y);
            }
            puts("END OF OUTPUT");
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/whatbeg/p/4107258.html
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