static bool countNeighbor3D(Point3D_int p, int branch_max)
{
int i;
int count = 0;
Point3D_int p3;
for (i = 0; i < D3; i++)
{
p3.x = p.x + dx[i];
p3.y = p.y + dy[i];
p3.z = p.z + dz[i];
if (inBound3D(p3) && !isEmpty3D(p3))
{
count++;
}
}
return(count < branch_max);
}
//branch_max表示路径的交叉点处的最多连接多少个节点，至少为2，数值越大，分支越多
static void search3D(int branch_max)
{
int i, j;
int[] r = new int[D3];
Random rand = new Random();
Stack <Point3D_int> s = new Stack <Point3D_int>();
Point3D_int p, p3;
p.x = 0;
p.y = 0;
p.z = 0;
s.Push(p);
while (s.Count > 0)
{
p = s.Pop();
if (isEmpty3D(p) && countNeighbor3D(p, branch_max))
{
map3D[p.x, p.y, p.z] = 1;
for (j = 0; j < D3; j++)//重置随机方向数组，令到数组的值为从0至D3-1
{
r[j] = j;
}
for (i = D3; i > 0; i--) //随机方向数组的大小从D3开始递减至0
{
j = rand.Next() % i; //每次从随机方向数组取一个方向索引j (j>=0&&j<=i-1)
p3 = new Point3D_int();
p3.x = p.x + dx[r[j]];
p3.y = p.y + dy[r[j]];
p3.z = p.z + dz[r[j]];
if (j != i - 1)//如果j不是随机方向数组的最后一个元素，则将最后一个元素i-1保存到位置j
{
r[j] = r[i - 1];
}
if (inBound3D(p3) && isEmpty3D(p3) && countNeighbor3D(p3, branch_max))
{
s.Push(p3);
}
}
}
}
}
static bool isEmpty3D(Point3D_int p)
{
return(map3D[p.x, p.y, p.z] == 0);
}
static bool inBound3D(Point3D_int p)
{
return(p.x >= 0 && p.x < xmax && p.y >= 0 && p.y < ymax && p.z >= 0 && p.z < zmax);
}
