uva 11275 3D Triangles

题意：三维空间中，给出两个三角形的左边，问是否相交。

面积法判断点在三角形内：

 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<memory.h>
 #include<cstdlib>
 #include<vector>
 #define clc(a,b) memset(a,b,sizeof(a))
 #define LL long long int
 #define up(i,x,y) for(i=x;i<=y;i++)
 #define w(a) while(a)
 const double inf=0x3f3f3f3f;
 const double PI = acos(-1.0);
 using namespace std;

 struct Point3
 {
     double x, y, z;
     Point3(double x=, double y=, double z=):x(x),y(y),z(z) { }
 };

 typedef Point3 Vector3;

 Vector3 operator + (const Vector3& A, const Vector3& B)
 {
     return Vector3(A.x+B.x, A.y+B.y, A.z+B.z);
 }
 Vector3 operator - (const Point3& A, const Point3& B)
 {
     return Vector3(A.x-B.x, A.y-B.y, A.z-B.z);
 }
 Vector3 operator * (const Vector3& A, double p)
 {
     return Vector3(A.x*p, A.y*p, A.z*p);
 }
 Vector3 operator / (const Vector3& A, double p)
 {
     return Vector3(A.x/p, A.y/p, A.z/p);
 }

 const double eps = 1e-;
 int dcmp(double x)
 {
     if(fabs(x) < eps) return ;
     else return x <  ? - : ;
 }

 double Dot(const Vector3& A, const Vector3& B)
 {
     return A.x*B.x + A.y*B.y + A.z*B.z;
 }
 double Length(const Vector3& A)
 {
     return sqrt(Dot(A, A));
 }
 double Angle(const Vector3& A, const Vector3& B)
 {
     return acos(Dot(A, B) / Length(A) / Length(B));
 }
 Vector3 Cross(const Vector3& A, const 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(const Point3& A, const Point3& B, const Point3& C)
 {
     return Length(Cross(B-A, C-A));
 }

 Point3 read_point3()
 {
     Point3 p;
     scanf("%lf%lf%lf", &p.x, &p.y, &p.z);
     return p;
 }

 // 点在三角形P0, P1, P2中
 bool PointInTri(const Point3& P, const Point3& P0, const Point3& P1, const 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)) == ;
 }

 // 三角形P0P1P2是否和线段AB相交
 bool TriSegIntersection(const Point3& P0, const Point3& P1, const Point3& P2, const Point3& A, const Point3& B, Point3& P)
 {
     Vector3 n = Cross(P1-P0, P2-P0);
     if(dcmp(Dot(n, B-A)) == ) return false; // 线段A-B和平面P0P1P2平行或共面
     else   // 平面A和直线P1-P2有惟一交点
     {
         double t = Dot(n, P0-A) / Dot(n, B-A);
         if(dcmp(t) <  || dcmp(t-) > ) return false; // 不在线段AB上
         P = A + (B-A)*t; // 交点
         return PointInTri(P, P0, P1, P2);
     }
 }

 bool TriTriIntersection(Point3* T1, Point3* T2)
 {
     Point3 P;
     for(int i = ; i < ; i++)
     {
         if(TriSegIntersection(T1[], T1[], T1[], T2[i], T2[(i+)%], P)) return true;
         if(TriSegIntersection(T2[], T2[], T2[], T1[i], T1[(i+)%], P)) return true;
     }
     return false;
 }

 int main()
 {
     int T;
     scanf("%d", &T);
     while(T--)
     {
         Point3 T1[], T2[];
         for(int i = ; i < ; i++) T1[i] = read_point3();
         for(int i = ; i < ; i++) T2[i] = read_point3();
         printf("%d\n", TriTriIntersection(T1, T2) ?  : );
     }
     return ;
 }

用Barycentric坐标法判断点在三角形内（重心法）

 #include<cstdio>
 #include<cmath>
 using namespace std;

 struct Point3
 {
     double x, y, z;
     Point3(double x=, double y=, double z=):x(x),y(y),z(z) { }
 };

 typedef Point3 Vector3;

 Vector3 operator + (const Vector3& A, const Vector3& B)
 {
     return Vector3(A.x+B.x, A.y+B.y, A.z+B.z);
 }
 Vector3 operator - (const Point3& A, const Point3& B)
 {
     return Vector3(A.x-B.x, A.y-B.y, A.z-B.z);
 }
 Vector3 operator * (const Vector3& A, double p)
 {
     return Vector3(A.x*p, A.y*p, A.z*p);
 }
 Vector3 operator / (const Vector3& A, double p)
 {
     return Vector3(A.x/p, A.y/p, A.z/p);
 }

 const double eps = 1e-;
 int dcmp(double x)
 {
     if(fabs(x) < eps) return ;
     else return x <  ? - : ;
 }

 double Dot(const Vector3& A, const Vector3& B)
 {
     return A.x*B.x + A.y*B.y + A.z*B.z;
 }
 double Length(const Vector3& A)
 {
     return sqrt(Dot(A, A));
 }
 double Angle(const Vector3& A, const Vector3& B)
 {
     return acos(Dot(A, B) / Length(A) / Length(B));
 }
 Vector3 Cross(const Vector3& A, const 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(const Point3& A, const Point3& B, const Point3& C)
 {
     return Length(Cross(B-A, C-A));
 }

 Point3 read_point3()
 {
     Point3 p;
     scanf("%lf%lf%lf", &p.x, &p.y, &p.z);
     return p;
 }

 // 点在三角形P0, P1, P2中
 // http://www.blackpawn.com/texts/pointinpoly/default.html
 bool PointInTri(const Point3& P, const Point3& P0, const Point3& P1, const Point3& P2)
 {
     Vector3 v0 = P2 - P0;
     Vector3 v1 = P1 - P0;
     Vector3 v2 = P - P0;

     // Compute dot products
     double dot00 = Dot(v0, v0);
     double dot01 = Dot(v0, v1);
     double dot02 = Dot(v0, v2);
     double dot11 = Dot(v1, v1);
     double dot12 = Dot(v1, v2);

     // Compute barycentric coordinates
     double invDenom =  / (dot00 * dot11 - dot01 * dot01);
     double u = (dot11 * dot02 - dot01 * dot12) * invDenom;
     double v = (dot00 * dot12 - dot01 * dot02) * invDenom;

     // Check if point is in triangle
     return (dcmp(u) >= ) && (dcmp(v) >= ) && (dcmp(u + v - ) <= );
 }

 // 三角形P0P1P2是否和线段AB相交
 bool TriSegIntersection(const Point3& P0, const Point3& P1, const Point3& P2, const Point3& A, const Point3& B, Point3& P)
 {
     Vector3 n = Cross(P1-P0, P2-P0);
     if(dcmp(Dot(n, B-A)) == ) return false; // 线段A-B和平面P0P1P2平行或共面
     else   // 平面A和直线P1-P2有惟一交点
     {
         double t = Dot(n, P0-A) / Dot(n, B-A);
         if(dcmp(t) <  || dcmp(t-) > ) return false; // 不在线段AB上
         P = A + (B-A)*t; // 交点
         return PointInTri(P, P0, P1, P2);
     }
 }

 bool TriTriIntersection(Point3* T1, Point3* T2)
 {
     Point3 P;
     for(int i = ; i < ; i++)
     {
         if(TriSegIntersection(T1[], T1[], T1[], T2[i], T2[(i+)%], P)) return true;
         if(TriSegIntersection(T2[], T2[], T2[], T1[i], T1[(i+)%], P)) return true;
     }
     return false;
 }

 int main()
 {
     int T;
     scanf("%d", &T);
     while(T--)
     {
         Point3 T1[], T2[];
         for(int i = ; i < ; i++) T1[i] = read_point3();
         for(int i = ; i < ; i++) T2[i] = read_point3();
         printf("%d\n", TriTriIntersection(T1, T2) ?  : );
     }
     return ;
 }

证明可以参考http://www.cnblogs.com/graphics/archive/2010/08/05/1793393.html

三角形的三个点在同一个平面上，如果选中其中一个点，其他两个点不过是相对该点的位移而已，比如选择点A作为起点，那么点B相当于在AB方向移动一段距离得到，而点C相当于在AC方向移动一段距离得到。

所以对于平面内任意一点，都可以由如下方程来表示

P = A +  u * (C – A) + v * (B - A) // 方程1

如果系数u或v为负值，那么相当于朝相反的方向移动，即BA或CA方向。那么如果想让P位于三角形ABC内部，u和v必须满足什么条件呢？有如下三个条件

u >= 0

v >= 0

u + v <= 1

几个边界情况，当u = 0且v = 0时，就是点A，当u = 0,v = 1时，就是点B，而当u = 1, v = 0时，就是点C

整理方程1得到P – A = u(C - A) + v(B - A)

令v0 = C – A, v1 = B – A, v2 = P – A，则v2 = u * v0 + v * v1，现在是一个方程，两个未知数，无法解出u和v，将等式两边分别点乘v0和v1的到两个等式

(v2) • v0 = (u * v0 + v * v1) • v0

(v2) • v1 = (u * v0 + v * v1) • v1

注意到这里u和v是数，而v0，v1和v2是向量，所以可以将点积展开得到下面的式子。

v2 • v0 = u * (v0 • v0) + v * (v1 • v0)  // 式1

v2 • v1 = u * (v0 • v1) + v * (v1• v1)   // 式2

解这个方程得到

u = ((v1•v1)(v2•v0)-(v1•v0)(v2•v1)) / ((v0•v0)(v1•v1) - (v0•v1)(v1•v0))

v = ((v0•v0)(v2•v1)-(v0•v1)(v2•v0)) / ((v0•v0)(v1•v1) - (v0•v1)(v1•v0))