【POJ 1269】判断两直线相交

题

利用叉积解方程

#include <cstdio>
#define MAX 1<<31
#define dd double
int xmult(dd x1,dd y1,dd x2,dd y2,dd x,dd y){
	return (x1-x)*(y2-y)-(x2-x)*(y1-y);
}
int main(){
	int n;
	dd x1,y1,x2,y2,x3,y3,x4,y4;
	scanf("%d",&n);
	puts("INTERSECTING LINES OUTPUT");
	while(n--){
		scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
		dd a1=y1-y2;
		dd b1=x2-x1;
		dd c1=xmult(x1,y1,x2,y2,0,0);
		dd a2=y3-y4;
		dd b2=x4-x3;
		dd c2=xmult(x3,y3,x4,y4,0,0);
		if(a1*b2==a2*b1){
			if(xmult(x1,y1,x2,y2,x3,y3)==0)
				puts("LINE");
			else puts("NONE");
		}
		else {
			dd cx,cy;
			cx=(b1*c2-c1*b2)/(a1*b2-a2*b1);
			cy=(a1*c2-c1*a2)/(b1*a2-b2*a1);
			printf("POINT %.2f %.2f\n",cx,cy);
		}
	}
	puts("END OF OUTPUT");
}

利用点斜式解方程

#include <cstdio>
#define MAX 1<<31
#define dd double
struct P{
	dd x,y;
	void input(){
		scanf("%lf%lf",&x,&y);
	}
	void output(){
		printf("POINT %.2f %.2f\n",x,y);
	}
};
struct L{
	P s,e;
	void input(){
		s.input(),e.input();
	}
	dd k(){
		if(s.x==e.x)return MAX;
		return (s.y-e.y)/(s.x-e.x);
	}
}l1,l2;
int xmult(P a,P b,P o){
	return (a.x-o.x)*(b.y-o.y)-(b.x-o.x)*(a.y-o.y);
}
void getCross(L a,L b){
	P c;
	dd ka=a.k(),kb=b.k();
	if(ka==MAX){//a是竖直的
		c.x=a.s.x;
		c.y=(c.x-b.s.x)*kb+b.s.y;
	}
	else{
		if(kb==MAX)
			c.x=b.s.x;
		else
			c.x=(a.s.y-b.s.y-ka*a.s.x+kb*b.s.x)/(kb-ka);
		c.y=(c.x-a.s.x)*ka+a.s.y;
	}
	c.output();
}
int main(){
	int n;
	scanf("%d",&n);
	puts("INTERSECTING LINES OUTPUT");
	while(n--){
		l1.input(),l2.input();
		dd ka=l1.k(),kb=l2.k();
		if(ka==kb){
			if(xmult(l1.s,l1.e,l2.s)==0)
				puts("LINE");
			else puts("NONE");
		}
		else
			getCross(l1,l2);
	}
	puts("END OF OUTPUT");
}