问题描述：对于Point[] points坐标数组（Point是坐标点，包含X与Y），请计算两两坐标点之间的距离。
思路：既然是算距离，没有别的要求，只需要最简单的曼哈顿距离就行。
测试数据：
Random random = new Random();
int n = 3000;
Point[] points = new Point[n];
for (int i = 0; i < n; i++)
{points[i] = new Point(random.Next(10), random.Next(10));
}方法一：采用普通遍历法
Stopwatch sw = Stopwatch.StartNew();
int m = (n * n - n) / 2;
List<int> list2 = new List<int>(m);
for (int i = 0; i < n; i++)
{Point p= points[i];for (int j = i + 1; j < n; j++){Point p2 = points[j];int dif = Math.Abs(p.X - p2.X) + Math.Abs(p.Y - p2.Y);list2.Add(dif);}
}
sw.Stop();
Console.WriteLine($"普通耗时：{sw.ElapsedMilliseconds}ms");方法二：采用SIMD加速
public static int[] CalculateDistance(Point[] points){int n = points.Length;int m = (n * n - n) >> 1;int[] ints = new int[m];int k = 0;int vsize = Vector128<int>.Count;if (n <= vsize)// 如果数据量小于等于vsize采用指针处理{fixed (Point* ptr = points){for (int i = 0; i < n; i++){Point* p1 = ptr + i;for (int j = i + 1; j < n; j++){Point* p2 = ptr + j;int dif = Math.Abs(p1->X - p2->X) + Math.Abs(p1->Y - p2->Y);ints[k++] = dif;}}return ints;}}int t = 0;int[] xs = new int[n];int[] ys = new int[n];fixed (Point* ptr = points){fixed (int* xptr = xs, yptr = ys, rptr = ints){// 分离x与yfor (int i = 0; i < n; i++){Point* p = ptr + i;*(xptr + i) = p->X;*(yptr + i) += p->Y;}// 开始向量计算for (int i = 0; i < n; i += vsize){Vector128<int> vx1 = *(Vector128<int>*)(xptr + i);Vector128<int> vy1 = *(Vector128<int>*)(yptr + i);for (int j = i + 1; j < n; j++){Vector128<int> vx2 = *(Vector128<int>*)(xptr + j);Vector128<int> vy2 = *(Vector128<int>*)(yptr + j);// 向量运算var vDifference = Sse2.Add(Vector128.Abs(Sse2.Subtract(vx1, vx2)), Vector128.Abs(Sse2.Subtract(vy1, vy2)));int* maskPtr = (int*)&vDifference;int h = n - j;for (int a = 0; a < vsize; a++){if (a >= h) break;*(rptr + t++) = *(maskPtr + a);}}}}}return ints;}注：根据CPU选择合适的Vector，比如Vector256、Vector512等等。本文只用Vector128.
测试结果（Release环境）：
| 数据量 | 3000 | 5000 | 7000 | 10000 | 30000 |
| 普通耗时（ms） | 43 | 127 | 211 | 538 | 3872 |
| 向量耗时（ms） | 19 | 56 | 68 | 190 | 1092 |
结论：明显向量运算性能优于普通计算。