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POJ 1269 Intersecting Lines --计算几何

时间:2014-11-19 07:19:24      阅读:209      评论:0      收藏:0      [点我收藏+]

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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:

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#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\n",GetLineIntersection(L1,L2).x,GetLineIntersection(L1,L2).y);
        }
        puts("END OF OUTPUT");
    }
    return 0;
}
View Code

 

POJ 1269 Intersecting Lines --计算几何

标签:style   blog   http   io   ar   color   os   sp   for   

原文地址:http://www.cnblogs.com/whatbeg/p/4107258.html

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