简单三维几何，判断俩个三角形是否相交-白红宇的个人博客

发布日期：2021-10-08 15:48:56 浏览次数：23 分类：技术文章

本文共 3066 字，大约阅读时间需要 10 分钟。

简单三维几何，多贴些三维几何的基本的东西，一些基本函数，和定义，

判断俩个三角形是否相交，如果相交，那么必然有一个三角形的一条边经过另一个三角形的内部、边上或者顶点，

代码：

#include
   
    #include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          #include
           
            #include
            
             #include
             
              #include
              
               using namespace std;const double eps=1e-6;int dcmp(double x){ return fabs(x)
               
                0?1:-1);}struct Point3{ double x,y,z; Point3(double x,double y,double z):x(x),y(y),z(z){} Point3() {} void in() { cin>>x>>y>>z; }};typedef Point3 Vector3;Vector3 operator +(Vector3 A,Vector3 B){ return Vector3(A.x+B.x,A.y+B.y,A.z+B.z);}Vector3 operator -(Vector3 A,Vector3 B){ return Vector3(A.x-B.x,A.y-B.y,A.z-B.z);}Vector3 operator * (Vector3 A,double p){ return Vector3(A.x*p,A.y*p,A.z*p);}Vector3 operator / (Vector3 A,double p){ return Vector3(A.x/p,A.y/p,A.z/p);}bool operator == (Vector3 A,Vector3 B){ return !dcmp(A.x-B.x)&&!dcmp(A.y-B.y)&&!dcmp(A.z-B.z);}double Dot(Vector3 A,Vector3 B){ return A.x*B.x+A.y*B.y+A.z*B.z;}double Length(Vector3 A){ return sqrt(Dot(A,A));}double Angle(Vector3 A,Vector3 B){ return acos(Dot(A,B)/Length(A)/Length(B));}double DistanceToPlane(Point3 p,Point3 p0,Vector3 n){ return fabs(Dot(p-p0,n)/Length(n));}Point3 GetPlaneProjection(Point3 p,Point3 p0,Vector3 n){ double d=Dot(p-p0,n)/Length(n); return p+n*d;}Point3 LinePlaneIntersection(Point3 p1,Point3 p2,Point3 p3,Point3 p0,Vector3 n){ Vector3 v=p2-p1; double t=(Dot(n,p0-p1)/Dot(n,p2-p1)); return p1+v*t;}Vector3 Cross(Vector3 A,Vector3 B){ return Vector3(A.y*B.z-A.z*B.y,A.z*B.x-A.x*B.z,A.x*B.y-A.y*B.x);}double Area2(Point3 A,Point3 B,Point3 C){ return Length(Cross(B-A,C-A));}bool PointInTri(Point3 P,Point3 P0,Point3 P1,Point3 P2){ double area1=Area2(P,P0,P1); double area2=Area2(P,P1,P2); double area3=Area2(P,P2,P0); return dcmp(area1+area2+area3-Area2(P0,P1,P2))==0;}bool TriSegIntersection(Point3 P0,Point3 P1,Point3 P2,Point3 A,Point3 B,Point3 &P){ Vector3 n=Cross(P1-P0,P2-P0); if(dcmp(Dot(n,B-A)==0)) return false; else { double t=Dot(n,P0-A)/Dot(n,B-A); if(dcmp(t)<0||dcmp(t-1)>0) return false; P=A+(B-A)*t; return PointInTri(P,P0,P1,P2); }}double DistaceToLine(Point3 P,Point3 A,Point3 B){ Vector3 v1=B-A,v2=P-A; return Length(Cross(v1,v2)/Length(v1));}double DistaceToSegment(Point3 P,Point3 A,Point3 B){ if(A==B) return Length(P-A); Vector3 v1=B-A,v2=P-A,v3=P-B; if(dcmp(Dot(v1,v2))<0) return Length(v2); else if(dcmp(Dot(v1,v3))>0) return Length(v3); else return Length(Cross(v1,v2))/Length(v1);}double Volume6(Point3 A,Point3 B,Point3 C,Point3 D){ return Dot(D-A,Cross(B-A,C-A));}struct Face{ int v[3]; Vector3 normal(Point3 *P)const { return Cross(P[v[1]]-P[v[0]],P[v[2]]-P[v[0]]); } int cansee(Point3 *P,int i) const { return Dot(P[i]-P[v[0]],normal(P))>0?1:0; }};vector
                
                  CH3D(Point3 *P,int n){ int vis[100][100]; memset(vis,0,sizeof(vis)); vector
                 
                   cur; cur.push_back((Face) { {0,1,2}}); cur.push_back((Face) { {2,1,0}}); for(int i=3;i
                  
                    next; for(int j=0;j
                   
                    >t; while(t--) { Point3 T1[3],T2[3]; for(int i=0;i<3;i++) T1[i].in(); for(int i=0;i<3;i++) T2[i].in(); printf("%d\n",TriTriIntersection(T1,T2)?1:0); } return 0;}

转载地址：https://blog.csdn.net/ONE_PIECE_HMH/article/details/45771877 如侵犯您的版权，请留言回复原文章的地址，我们会给您删除此文章，给您带来不便请您谅解！

上一篇：三维几何之求空间俩线段的公垂线，以及分数类

下一篇：PSLG,直线切割凸多边形，和判断圆与多边形相交

发表评论

关于作者

喝酒易醉，品茶养心，人生如梦，品茶悟道，何以解忧？唯有杜康！

-- 愿君每日到此一游！

发表评论

最新留言

关于作者

推荐文章