C#实现的最短路径分析

using system;
 using system.collections.generic;
 using system.linq;
 using system.text;

 namespace consoleapplication1
 {
     class program
     {
         static int length = 6;
         static string[] shortedpath = new string[length];
         static int nopath = 2000;
         static int maxsize = 1000;
         static int[,] g =
         {
             { nopath, nopath, 10, nopath, 30, 100 },
             { nopath, nopath, 5, nopath, nopath, nopath },
             { nopath, nopath, nopath, 50, nopath, nopath },
             { nopath, nopath, nopath, nopath, nopath, 10 },
             { nopath, nopath, nopath, 20, nopath, 60 },
             { nopath, nopath, nopath, nopath, nopath, nopath }
         };
         static string[] pathresult = new string[length];

         static int[] path1 = new int[length];
         static int[,] path2 = new int[length, length];
         static int[] distance2 = new int[length];

         static void main(string[] args)
         {
             int dist1 = getshortedpath(g, 0, 1, path1);
             console.writeline("点0到点5路径:");
             for (int i = 0; i < path1.length; i++)
                 console.write(path1[i].tostring() + " "); 
             console.writeline("长度:" + dist1);

 
             console.writeline("\r\n-----------------------------------------\r\n");

             int[] pathdist = getshortedpath(g, 0, path2);
             console.writeline("点0到任意点的路径:");
             for (int j = 0; j < pathdist.length; j++)
             {
                 console.writeline("点0到" + j + "的路径:");
                 for (int i = 0; i < length; i++)
                     console.write(path2[j, i].tostring() + " ");
                 console.writeline("长度:" + pathdist[j]);
             }
             console.readkey();

         }

 
         //从某一源点出发，找到到某一结点的最短路径
         static int getshortedpath(int[,]g, int start, int end,int [] path)
         {
             bool[] s = new bool[length]; //表示找到起始结点与当前结点间的最短路径
             int min;  //最小距离临时变量
             int curnode=0; //临时结点，记录当前正计算结点
             int[] dist = new int[length];
             int[] prev = new int[length];

             //初始结点信息
             for (int v = 0; v < length; v++)
             {
                 s[v] = false;
                 dist[v] = g[start, v];
                 if (dist[v] > maxsize)
                     prev[v] = 0;
                 else
                     prev[v] = start;
             }
             path[0] = end;
             dist[start] = 0;
             s[start] = true;
             //主循环
             for (int i = 1; i < length; i++)
             {
                 min = maxsize;
                 for (int w = 0; w < length; w++)
                 {
                     if (!s[w] && dist[w] < min)
                     {
                         curnode = w;
                         min = dist[w];
                     }
                 }

                 s[curnode] = true;
                 for (int j = 0; j < length; j++)
                     if (!s[j] && min + g[curnode, j] < dist[j])
                     {
                         dist[j] = min + g[curnode, j];
                         prev[j] = curnode;
                     }

             }
             //输出路径结点
             int e = end, step = 0;
             while (e != start)
             {
                 step++;
                 path[step] = prev[e];
                 e = prev[e];
             }
             for (int i = step; i > step/2; i--)
             {
                 int temp = path[step - i];
                 path[step - i] = path[i];
                 path[i] = temp;
             }
             return dist[end];
         }

 

 

 
         //从某一源点出发，找到到所有结点的最短路径
         static int[] getshortedpath(int[,] g, int start, int[,] path)
         {
             int[] pathid = new int[length];//路径（用编号表示）
             bool[] s = new bool[length]; //表示找到起始结点与当前结点间的最短路径
             int min;  //最小距离临时变量
             int curnode = 0; //临时结点，记录当前正计算结点
             int[] dist = new int[length];
             int[] prev = new int[length];
             //初始结点信息
             for (int v = 0; v < length; v++)
             {
                 s[v] = false;
                 dist[v] = g[start, v];
                 if (dist[v] > maxsize)
                     prev[v] = 0;
                 else
                     prev[v] = start;
                 path[v,0] = v;
             }

             dist[start] = 0;
             s[start] = true;
             //主循环
             for (int i = 1; i < length; i++)
             {
                 min = maxsize;
                 for (int w = 0; w < length; w++)
                 {
                     if (!s[w] && dist[w] < min)
                     {
                         curnode = w;
                         min = dist[w];
                     }
                 }

                 s[curnode] = true;

                 for (int j = 0; j < length; j++)
                     if (!s[j] && min + g[curnode, j] < dist[j])
                     {
                         dist[j] = min + g[curnode, j];
                         prev[j] = curnode;
                     }

 
             }
             //输出路径结点
             for (int k = 0; k < length; k++)
             {
                 int e = k, step = 0;
                 while (e != start)
                 {
                     step++;
                     path[k, step] = prev[e];
                     e = prev[e];
                 }
                 for (int i = step; i > step / 2; i--)
                 {
                     int temp = path[k, step - i];
                     path[k, step - i] = path[k, i];
                     path[k, i] = temp;
                 }
             }
             return dist;

         }

 
     }
 }