对dijkstra算法的自我理解，c#例子

 1   static void Main(string[] args)
 2         {
 3             double[,] path = getpath(10);//随机生成10个点之间的距离矩阵
 4             Dictionary<int, double> result = getr(path);//result<点，距离>
 5            
 6         }
 7         public static Dictionary<int, double> getr(double[,] da)
 8         {
 9             Dictionary<int, double> result = new Dictionary<int, double>();
10             result.Add(0, 0); //默认添加首点
11             int n =Convert.ToInt32(Math.Sqrt(da.Length));
12             double[] minlen =new double[n];//存储每次找到的最近点距离其它点的距离
13             int[] visit =new int[n]; //一共0到9点 计算过后的点为1
14             for(int i=0;i<n;i++){
15                 minlen[i] = da[0, i];//初始化首点与其它点距离
16             }
17             visit[0] = 1; //默认第一点为原点,已经添加到visit集合
18             int minj = 0; //遍历循环默认第一点最小
19             for (int i = 0; i < n; i++) {
20                 double min = double.PositiveInfinity;
21                 for (int j =0; j < n ; j++) {
22                     if (visit[j] == 0 && minlen[j] < min) {//遍历minlen 找到最小
23                         min = minlen[j];//j循环最小值
24                         minj = j; //第minj最小
25                     }
26                 }
27                 if (min == double.PositiveInfinity) return result;//测试数据有可能出现x点到其它点没有距离
28                 visit[minj] = 1; //第minj点已添加
29                 result.Add(minj,min);
30                 for (int j = 0; j < n ; j++)//重新计算minlen
31                 {
32                     if (visit[j] == 0 && minlen[minj] !=double.PositiveInfinity
33                             && da[minj,j] != double.PositiveInfinity
34                             && minlen[j] > (minlen[minj] + da[minj,j]))
35                     {
36                         minlen[j] = minlen[minj] + da[minj,j];
37                     }
38                 }
39             }
40             return result;
41         }
42         public static double[,] getpath(int num) {
43              Random rd=new Random();
44             double[,] path = new double[num, num];
45             for (int i = 0; i < num; i++)
46             {
47                 for (int j = 0; j < num; j++)
48                 {
49                     path[i, j] = rd.Next(1,100);//生成测试数据
50                     if (path[i, j] > 60 || i == j) path[i, j] = double.PositiveInfinity;
51                 }
52             }
53             return path;
54         }