数组按条件分组

[liangming1128]

在C#中存在这样的一些点1、2、3、4、5、6、7、8、9、10
1、先将这些相邻点之间的距离算出来，如果小于50，就将他们分到一个数组中

应怎样分组？？？？怎样算？？



求答案？？？

[636f6c696e]

题目有歧义。
不管怎么样，一个循环搞定

[rekym]

int[] num={5,8,12,52,330,340,780,785}

list<int> newnum=new list<int>();

for(int i=0;i<num.length-1++)
{
if(num[i+1]-num[i]>50)
newnum.add(num[i+1]);
}

[eclipse_xu]

可不可以重载下+运算符 然后判断result的大小？

[liangming1128]

public Bitmap DrawMap(string jingDu1, string weiDu1)
{
int Lx = -1, Ly = -1;
double jingDu2 = Convert.ToDouble(jingDu1);
double weiDu2 = Convert.ToDouble(weiDu1);
Lx = Convert.ToInt32(jingDu2) / 200;
Ly = Convert.ToInt32(weiDu2) / 120;
string sqlstr4 = "select LONGITUDE,LATITUDE from t_mrdata ";
DataTable dt1 = dbexec.GetDataTable(sqlstr4, null);
Bitmap M = new Bitmap(320, 240);
Graphics ghs = Graphics.FromImage(M);
Pen myPen = new Pen(Color.Black, 1);
System.Drawing.Drawing2D.AdjustableArrowCap aac;
aac = new System.Drawing.Drawing2D.AdjustableArrowCap(3, 3);
myPen.CustomEndCap = aac;
int n = dt1.Rows.Count;
float sum1 = 0, sum2 = 0;
int k=1;int count=0,m=0;
Point[] MapPoints = new Point[n];
Point[] MapPoints4 = new Point[n];
Point[] MapPoints2 = new Point[n];
Point[] MapPoints3 = new Point[n];
List <float> add=new List<float>();
Point p2;

for (int i = 0; i <= dt1.Rows.Count - 1; i++)
{
if (dt1.Rows[i][0] != DBNull.Value || dt1.Rows[i][1] != DBNull.Value)
{
string a = dt1.Rows[i][0].ToString();
string b = dt1.Rows[i][1].ToString();

if (a != String.Empty && b != String.Empty)
{
double jingDu = Convert.ToDouble(a);
double weiDu = Convert.ToDouble(b);
p2 = new Point(Convert.ToInt32(jingDu), Convert.ToInt32(weiDu));
MapPoints2[i] = p2;
int a1 = Convert.ToInt32(jingDu) / 200;
int a2 = Convert.ToInt32(weiDu) / 120;
Point p1 = new Point(a1, a2);
MapPoints[i] = p1;
}
else
{

p2 = new Point(-100000, -100000);

MapPoints2[i] = p2;
}
}
}

//判断相邻点之间的距离小于50
for (int i = 0; i < MapPoints2.Length ; i++)
{


double distance = (MapPoints2[i + 1].X - MapPoints2[i].X) * (MapPoints2[i + 1].X - MapPoints2[i].X) + (MapPoints2[i + 1].Y - MapPoints2[i].Y) * (MapPoints2[i + 1].Y - MapPoints2[i].Y);

if (Math.Sqrt(distance) <= 50)
{

Point p3 = new Point(MapPoints2[i].X, MapPoints2[i].Y);

MapPoints3[k] = p3;

k++;

}
else
{

for (int t = 0; t < MapPoints3.Length; t++)
{
sum1 = +MapPoints3[i].X;
sum2 = +MapPoints3[i].Y;
count++;
}
Point p1 = new Point(Convert.ToInt32(sum1 / count / 200), Convert.ToInt32(sum2 / count / 120));


MapPoints[m] = p1;
m++;

k = 0; count = 0;

sum1 = 0; sum2 = 0;

}


}



//画点

for (int j = 0; j < MapPoints.Length; j++)
{



ghs.FillEllipse(new SolidBrush(Color.Red), MapPoints[j].X, MapPoints[j].Y, 8, 8);



}
//画线
for (int j = 0; j < MapPoints.Length - 1; j++)
{


ghs.DrawLine(myPen, MapPoints[j], MapPoints[j + 1]);


}


return M;
}
测试数据：经度、纬度（200,120）、（200,120）、（300,120）、（300,120）…………这样的点很多

[liangming1128]

GPS图形中由于坐标点太多，因此将位置相近的点设为一个抄表点。对抄表点的分组算法暂定如下：在某一抄表段内，按照抄表顺序由小到大，如果某两点间距离大于一定值，则作为分割点。例如，某抄表段有10个电表，抄表顺序为1,2,3…9,10，如果第5点和第6点间的距离大于50m，则1-5为一个抄表点，6-10为一个抄表点。图形中只显示抄表点，不显示每个电表的位置。

[liangming1128]

[Quote=引用 2 楼 rekym 的回复:]
int[] num={5,8,12,52,330,340,780,785}

list<int> newnum=new list<int>();

for(int i=0;i<num.length-1++)
{
if(num[i+1]-num[i]>50)
newnum.add(num[i+1]);
}
[/Quote]
这个算法有错误

[shine_fly]

......