poj 1269

Posted 2021-02-10 mxang

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。

 1 #include <cstdio>
 2 #include <cmath>
 3 #define db double
 4 using namespace std;
 5 const db eps=1e-6;
 6 const db pi = acos(-1);
 7 int sign(db k){
 8     if (k>eps) return 1; else if (k<-eps) return -1; return 0;
 9 }
10 int cmp(db k1,db k2){return sign(k1-k2);}
11 struct point{
12     db x,y;
13     point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
14     point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
15     point operator * (db k1) const{return (point){x*k1,y*k1};}
16     point operator / (db k1) const{return (point){x/k1,y/k1};}
17 };
18 db dot(point k1,point k2){
19     return k1.x*k2.x+k1.y*k2.y;
20 }
21 db cross(point k1,point k2){
22     return k1.x*k2.y-k1.y*k2.x;
23 }
24 int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}
25 int inmid(point k1,point k2,point k3){//k3在[k1,k2]
26     return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);
27 }
28 point getLL (point k1,point k2,point k3,point k4){//两直线交点
29     db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3);
30     return (k1*w2+k2*w1)/(w1+w2);
31 }
32 bool onS(point k1,point k2,point q){//q在[k1,k2]
33     return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==0;
34 }
35 int checkLL(point k1,point k2,point k3,point k4){//求两条直线是否 (平行||重合)
36     return cmp(cross(k3-k1,k4-k1),cross(k3-k2,k4-k2))==0;
37 }
38 struct Line{
39     point p[2];
40 };
41 
42 db xl,yl,x2,y2;
43 int n;
44 point p[5];
45 int main(){
46     scanf("%d",&n);
47     printf("INTERSECTING LINES OUTPUT
");
48     while (n--){
49         for(int i=1;i<=4;i++){
50             scanf("%lf%lf",&p[i].x,&p[i].y);
51         }
52         if(checkLL(p[1],p[2],p[3],p[4])){
53             if(sign(cross(p[1]-p[3],p[2]-p[3]))==0){
54                 printf("LINE
");
55             } else{
56                 printf("NONE
");
57             }
58         } else{
59             point tmp = getLL(p[1],p[2],p[3],p[4]);
60             printf("POINT %.2f %.2f
",tmp.x,tmp.y);
61         }
62     }
63     printf("END OF OUTPUT
");
64 }

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