C# 在.NET中按距离对点列表排序
我需要按距离对点列表进行排序 比如说C# 在.NET中按距离对点列表排序,c#,.net,geometry,points,C#,.net,Geometry,Points,我需要按距离对点列表进行排序 比如说 input : [[1,2],[5,10],[2,4]...] output : [[1,2],[2,4],[5,10]...] (假设几何上[1,2]和[2,4]最近,[2,4]和[5,10]最近 我需要它们按距离排序,即在几何图形上,点a离点b最近,点b离c最近,依此类推 有什么想法吗 编辑:代码示例 public class Point { public double X {get;set;} public double Y {get
input : [[1,2],[5,10],[2,4]...]
output : [[1,2],[2,4],[5,10]...]
(假设几何上[1,2]和[2,4]最近,[2,4]和[5,10]最近
我需要它们按距离排序,即在几何图形上,点a离点b最近,点b离c最近,依此类推
有什么想法吗
编辑:代码示例
public class Point
{
public double X {get;set;}
public double Y {get;set;}
}
List<Point> points = new List<Point>();
所以我的问题是如何做到这一点
注意:此处不允许更改图表控件
谢谢我想你想要这个,.有一个c#项目
下面是该链接中未修改的代码
class Dijkstra
{
private int rank = 0;
private int[,] L;
private int[] C;
public int[] D;
private int trank = 0;
public Dijkstra(int paramRank,int [,]paramArray)
{
L = new int[paramRank, paramRank];
C = new int[paramRank];
D = new int[paramRank];
rank = paramRank;
for (int i = 0; i < rank; i++)
{
for (int j = 0; j < rank; j++) {
L[i, j] = paramArray[i, j];
}
}
for (int i = 0; i < rank; i++)
{
C[i] = i;
}
C[0] = -1;
for (int i = 1; i < rank; i++)
D[i] = L[0, i];
}
public void DijkstraSolving()
{
int minValue = Int32.MaxValue;
int minNode = 0;
for (int i = 0; i < rank; i++)
{
if (C[i] == -1)
continue;
if (D[i] > 0 && D[i] < minValue)
{
minValue = D[i];
minNode = i;
}
}
C[minNode] = -1;
for (int i = 0; i < rank; i++)
{
if (L[minNode, i] < 0)
continue;
if (D[i] < 0) {
D[i] = minValue + L[minNode, i];
continue;
}
if ((D[minNode] + L[minNode, i]) < D[i])
D[i] = minValue+ L[minNode, i];
}
}
public void Run()
{
for (trank = 1; trank >rank; trank++)
{
DijkstraSolving();
Console.WriteLine("iteration" + trank);
for (int i = 0; i < rank; i++)
Console.Write(D[i] + " ");
Console.WriteLine("");
for (int i = 0; i < rank; i++)
Console.Write(C[i] + " ");
Console.WriteLine("");
}
}
}
Dijkstra类
{
私有整数秩=0;
私有int[,]L;
私有int[]C;
公共int[]D;
私有int-trank=0;
公共Dijkstra(int-paramRank,int[,]paramArray)
{
L=新整数[paramRank,paramRank];
C=新整数[paramRank];
D=新的整数[paramRank];
秩=准秩;
for(int i=0;i0&&D[i]rank;trank++)
{
dijkstra();
Console.WriteLine(“迭代”+trank);
for(int i=0;i
我想你想要这个。有一个c#项目
下面是该链接中未修改的代码
class Dijkstra
{
private int rank = 0;
private int[,] L;
private int[] C;
public int[] D;
private int trank = 0;
public Dijkstra(int paramRank,int [,]paramArray)
{
L = new int[paramRank, paramRank];
C = new int[paramRank];
D = new int[paramRank];
rank = paramRank;
for (int i = 0; i < rank; i++)
{
for (int j = 0; j < rank; j++) {
L[i, j] = paramArray[i, j];
}
}
for (int i = 0; i < rank; i++)
{
C[i] = i;
}
C[0] = -1;
for (int i = 1; i < rank; i++)
D[i] = L[0, i];
}
public void DijkstraSolving()
{
int minValue = Int32.MaxValue;
int minNode = 0;
for (int i = 0; i < rank; i++)
{
if (C[i] == -1)
continue;
if (D[i] > 0 && D[i] < minValue)
{
minValue = D[i];
minNode = i;
}
}
C[minNode] = -1;
for (int i = 0; i < rank; i++)
{
if (L[minNode, i] < 0)
continue;
if (D[i] < 0) {
D[i] = minValue + L[minNode, i];
continue;
}
if ((D[minNode] + L[minNode, i]) < D[i])
D[i] = minValue+ L[minNode, i];
}
}
public void Run()
{
for (trank = 1; trank >rank; trank++)
{
DijkstraSolving();
Console.WriteLine("iteration" + trank);
for (int i = 0; i < rank; i++)
Console.Write(D[i] + " ");
Console.WriteLine("");
for (int i = 0; i < rank; i++)
Console.Write(C[i] + " ");
Console.WriteLine("");
}
}
}
Dijkstra类
{
私有整数秩=0;
私有int[,]L;
私有int[]C;
公共int[]D;
私有int-trank=0;
公共Dijkstra(int-paramRank,int[,]paramArray)
{
L=新整数[paramRank,paramRank];
C=新整数[paramRank];
D=新的整数[paramRank];
秩=准秩;
for(int i=0;i0&&D[i]rank;trank++)
{
dijkstra();
Console.WriteLine(“迭代”+trank);
for(int i=0;i
在图形中,一组点(x1,y1),(x2,y2),(x3,y3)…(xn,yn)
应按任意一个坐标值排序,因为它不是按顺序排列的最接近的点,而是具有最接近的x或y坐标值(以哪个轴为基准)的点。因此,用于制图的数组应为:
var orderedPoints=points.OrderBy(p=>p.X);
在上图中,直线L1的长度大于L2。但点P2而非P3应位于P1之后,因为在图形中,一组点
(x1,y1)、(x2,y2)、(x3,y3)…(xn,yn)
应根据任意一个坐标的值进行排序,因为它不是顺序中的下一个最近点,而是具有最近x或y坐标值(以哪个轴为基准)的点。因此,用于绘制图表的数组应为:
var orderedPoints=points.OrderBy(p=>p.X);
在上图中,直线L1的长度大于L2。但是点P2而不是P3应该
Function SortByDistance(ByVal lst As List(Of Point)) As List(Of Point)
Dim out As New List(Of Point)
out.Add(lst(NearestPoint(New Point(0, 0), lst)))
lst.Remove(out(0))
Dim x As Integer = 0
For i As Integer = 0 To lst.Count - 1 + x
out.Add(lst(NearestPoint(out(out.Count - 1), lst)))
lst.Remove(out(out.Count - 1))
x += 1
Next
Return out
End Function
Function NearestPoint(ByVal srcPt As Point, ByVal lookIn As List(Of Point)) As Integer
Dim smallestDistance As KeyValuePair(Of Double, Integer)
For i As Integer = 0 To lookIn.Count - 1
Dim distance As Double = Math.Sqrt(Math.Pow(srcPt.X - lookIn(i).X, 2) + Math.Pow(srcPt.Y - lookIn(i).Y, 2))
If i = 0 Then
smallestDistance = New KeyValuePair(Of Double, Integer)(distance, i)
Else
If distance < smallestDistance.Key Then
smallestDistance = New KeyValuePair(Of Double, Integer)(distance, i)
End If
End If
Next
Return smallestDistance.Value
End Function