蛮力法03
——可视化,是如此的充满魅力。
最近距离问题:平面中n(n>1)个点,求距离最近的两个点。
分析:设定集合 pts[n] , minLen=INT_MAX;
for i=0, to n
{
for j=i+1, to n //计算两点间的距离不需要重复计算,所以全部计算过的顶点需要排除
{
if minLen > Dis(pts[i],pts[j])
{
//求出了当前最近的连个点距离
}
}
}
/// <summary>
/// 最近点对距离计算
/// </summary>
class PointPairCompute
{
private int imgWidth;
private int imgHeight;
private int createCount;
private List<Point> ptLst;
private readonly int radius;
public PointPairCompute(int width,int height,int count)
{
this.imgWidth = width;
this.imgHeight = height;
this.createCount = count;
this.ptLst = new List<Point>();
this.radius = 2;
}
/// <summary>
/// 获取随机生成点的图片
/// </summary>
/// <returns>图片</returns>
public Image GetRandomImag()
{
//随机产生点
RandomCratePtLst();
//绘制图片
Bitmap map = new Bitmap(this.imgWidth, this.imgHeight);
Graphics gfx = Graphics.FromImage(map);
gfx.Clear(Color.Gray);
foreach (Point item in this.ptLst)
{
//map.SetPixel(item.X, item.Y, Color.Red);
gfx.FillEllipse(Brushes.Red, item.X - radius, item.Y - radius, radius * 2, radius * 2);
}
gfx.Dispose();
return map;
}
/// <summary>
/// 获取点对最近距离结果图片
/// </summary>
/// <returns>结果图片</returns>
public Image GetPointPariResult()
{
if (this.ptLst.Count <2)
{
return null;
}
int leftIndex = -1;
int rightIndex = -1;
ComputerPointPari(ref leftIndex, ref rightIndex);
Bitmap map = new Bitmap(this.imgWidth, this.imgHeight);
Graphics gfx = Graphics.FromImage(map);
gfx.Clear(Color.White);
Pen pen = new Pen(Color.Green, 4);
gfx.DrawLine(pen, ptLst[leftIndex], ptLst[rightIndex]);
foreach (Point item in this.ptLst)
{
//map.SetPixel(item.X, item.Y, Color.Red);
gfx.FillEllipse(Brushes.Red, item.X - radius, item.Y - radius, radius * 2, radius * 2);
}
gfx.Dispose();
return map;
}
//随机生成点
private void RandomCratePtLst()
{
this.ptLst.Clear();
Random rd = new Random();
for (int i = 0; i < this.createCount; i++)
{
this.ptLst.Add(new Point(rd.Next(0, this.imgWidth), rd.Next(0, this.imgHeight)));
}
}
//蛮力计算最短距离
private void ComputerPointPari(ref int leftIndex, ref int rightIndex)
{
int minLen = Int32.MaxValue;
for (int i = 0; i < this.ptLst.Count; ++i)
{
for (int j = i+1; j < this.ptLst.Count; j++)
{
int tempLen = (ptLst[i].X - ptLst[j].X) * (ptLst[i].X - ptLst[j].X) +
(ptLst[i].Y - ptLst[j].Y) * (ptLst[i].Y - ptLst[j].Y);
if (minLen > tempLen)
{
minLen = tempLen;
leftIndex = i;
rightIndex = j;
}
}
}
}
}