poj 1269 Intersecting Lines

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直线和直线的位置关系判断：

1）重叠：判断方法就是四点共线。

2）平行：判断的方法就是叉积为0 且四点不共线

3）相交：交点的坐标就是利用类似定比分点公式的方法；

Code：

/*判断直线和直线位置关系和求交点*/#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<vector>using namespace std;#define FOR(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)#define DOR(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)#define oo 1<<30#define eps 1e-8#define nMax 500#define pb push_back#define bug puts("OOOOh.....");#define zero(x) (((x)>0?(x):-(x))<eps)int dcmp(double x){    if(fabs(x)<eps) return 0;    return x>0?1:-1;}struct point {    double x,y;    point(double x=0,double y=0): x(x),y(y) {}    void read(){ scanf("%lf%lf",&x,&y); }    friend point operator -(point const& u,point const& v) {        return point(u.x-v.x,u.y-v.y);    }    friend double operator *(point const& u,point const& v) {        return u.x*v.y-u.y*v.x;    }    friend point operator *(double const& k,point const& v) {        return point(k*v.x,k*v.y);    }    friend point operator /(point const& u,double const& k){        return point(u.x/k,u.y/k);    }    friend int dots_online(point,point,point);};int dots_online(point a,point b,point c){ return dcmp((a-c)*(b-c))==0; }struct line{    point a,b;    line() {}    line(point a,point b): a(a),b(b) {}    void read() { a.read(),b.read(); }    friend int intersection(line,line);    friend int parallel(line,line);};int parallel(line l1,line l2){ return dcmp((l1.a-l1.b)*(l2.a-l2.b))==0; }int intersection(line l1,line l2,point &p){    if(dots_online(l1.a,l1.b,l2.a) && dots_online(l1.a,l1.b,l2.b)) return 1;    if(parallel(l1,l2)) return 0;    double s1 = (l1.a-l2.a)*(l2.b-l2.a);    double s2 = (l1.b-l2.a)*(l2.b-l2.a);    p = (s1*l1.b-s2*l1.a)/(s1-s2);    return 2;}int n;line l1,l2;point p;int main(){#ifndef ONLINE_JUDGE    freopen("input.txt","r",stdin);#endif    while(~scanf("%d",&n)){        puts("INTERSECTING LINES OUTPUT");        FOR(i,1,n) {            l1.read(),l2.read();            switch(intersection(l1,l2,p)){                case 0: puts("NONE");break;                case 1: puts("LINE");break;                case 2: printf("POINT %.2f %.2f\n",p.x,p.y);break;            }        }        puts("END OF OUTPUT");    }    return 0;}