【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
",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
",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");
}