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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-8using namespace std;#define N 100017struct 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 endsint 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\n",GetLineIntersection(L1,L2).x,GetLineIntersection(L1,L2).y);        }        puts("END OF OUTPUT");    }    return 0;}
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POJ 1269 Intersecting Lines --计算几何