如何在c ++中找到3D空间中的相邻点

Question

我在3D空间中有几个点，我想写一个条件来确定两个点是否相邻：它们是否只是整数点阵中的一个单位？

我有一个名为Point的结构，它包含x，y和z坐标。然后在main函数中，我设置点a，b，c，d，e的值并将它们推入向量。然后在for循环中我想检查两个点是否相邻。目前我只是检查它们是否在同一轴上，但我不知道如何继续。

struct Point{
   int x;
   int y;
   int z;
};

int main(){
    // just a couple of points
    struct Point a;
    struct Point b;
    struct Point c;
    struct Point d;
    struct Point e;

    vector<Point> pointVector;

    a.x = 0 ; a.y = 0; a.z = 0;
    b.x = 0; b.y = 0; b.z = -1;
    c.x = 1; c.y = 0; c.z = -1;
    d.x = 1; d.y = -1; d.z = -1;
    e.x = 2; e.y = -1; e.z = -1;

    pointVector.push_back(a);
    pointVector.push_back(b);
    pointVector.push_back(c);
    pointVector.push_back(d);
    pointVector.push_back(e);

    for(int i = 0; i < 5; i++){
       if(pointVector[i].x == pointVector[i+1].x ||  // how to set the condition to check if two points are adjacent?
          pointVector[i].y == pointVector[i+1].y || 
          pointVector[i].z == pointVector[i+1].z  
       ) cout << "adjacent" << endl;
       else
         cout << "not adjacent" << endl;


    }



    return 0;
}

通过邻近我的意思是这张照片中的东西：

Answer 1

非常简短：

for each pair of points:
    if two of the three coordinates are equal AND
              the other coordinate differs by 1:
        then mark the pair as adjacent.

通过点对迭代非常简单：第一个点a遍历索引0-（n-2）;第二个点b遍历从a的位置到结尾的索引，n-1。

给定整数坐标，检查邻接也很容易。

diff = abs(a.x - b.x) + 
       abs(a.y - b.y) + 
       abs(a.z - b.z)

diff = 1 iff 这些点相邻。