poj 1269

水题。判断两条直线位置关系。

考虑平行的情况，那么 四边形的面积会相等，重合的话，四边形的面积相等且为0.

除去这两种就一定有交点。

 #include <cstdio>
 #include <cmath>
 #define db double
 using namespace std;
 const db eps=1e-;
 const db pi = acos(-);
 int sign(db k){
     if (k>eps) return ; else if (k<-eps) return -; return ;
 }
 int cmp(db k1,db k2){return sign(k1-k2);}
 struct point{
     db x,y;
     point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
     point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
     point operator * (db k1) const{return (point){x*k1,y*k1};}
     point operator / (db k1) const{return (point){x/k1,y/k1};}
 };
 db dot(point k1,point k2){
     return k1.x*k2.x+k1.y*k2.y;
 }
 db cross(point k1,point k2){
     return k1.x*k2.y-k1.y*k2.x;
 }
 int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=;}
 int inmid(point k1,point k2,point k3){//k3在[k1,k2]
     return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);
 }
 point getLL (point k1,point k2,point k3,point k4){//两直线交点
     db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3);
     return (k1*w2+k2*w1)/(w1+w2);
 }
 bool onS(point k1,point k2,point q){//q在[k1,k2]
     return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==;
 }
 int checkLL(point k1,point k2,point k3,point k4){//求两条直线是否 (平行||重合)
     return cmp(cross(k3-k1,k4-k1),cross(k3-k2,k4-k2))==;
 }
 struct Line{
     point p[];
 };

 db xl,yl,x2,y2;
 int n;
 point p[];
 int main(){
     scanf("%d",&n);
     printf("INTERSECTING LINES OUTPUT\n");
     while (n--){
         for(int i=;i<=;i++){
             scanf("%lf%lf",&p[i].x,&p[i].y);
         }
         if(checkLL(p[],p[],p[],p[])){
             if(sign(cross(p[]-p[],p[]-p[]))==){
                 printf("LINE\n");
             } else{
                 printf("NONE\n");
             }
         } else{
             point tmp = getLL(p[],p[],p[],p[]);
             printf("POINT %.2f %.2f\n",tmp.x,tmp.y);
         }
     }
     printf("END OF OUTPUT\n");
 }