【转载】C#实现的最短路径分析

1 using System;  2 using System.Collections.Generic;  3 using System.Linq;  4 using System.Text;  5
6 namespace ConsoleApplication1  7 {  8     class Program  9  {  10         static int length = 6;  11         static string[] shortedPath = new string[length];  12         static int noPath = 2000;  13         static int MaxSize = 1000;  14         static int[,] G =
15  {  16             { noPath, noPath, 10, noPath, 30, 100 },  17             { noPath, noPath, 5, noPath, noPath, noPath },  18             { noPath, noPath, noPath, 50, noPath, noPath },  19             { noPath, noPath, noPath, noPath, noPath, 10 },  20             { noPath, noPath, noPath, 20, noPath, 60 },  21  { noPath, noPath, noPath, noPath, noPath, noPath }  22  };  23         static string[] PathResult = new string[length];  24
25         static int[] path1 = new int[length];  26         static int[,] path2 = new int[length, length];  27         static int[] distance2 = new int[length];  28
29         static void Main(string[] args)  30  {  31             int dist1 = getShortedPath(G, 0, 1, path1);  32             Console.WriteLine("点0到点5路径:");  33             for (int i = 0; i < path1.Length; i++)  34                 Console.Write(path1[i].ToString() + " ");  35             Console.WriteLine("长度:" + dist1);  36
37
38             Console.WriteLine("\r\n-----------------------------------------\r\n");  39
40             int[] pathdist = getShortedPath(G, 0, path2);  41             Console.WriteLine("点0到任意点的路径:");  42             for (int j = 0; j < pathdist.Length; j++)  43  {  44                 Console.WriteLine("点0到" + j + "的路径:");  45                 for (int i = 0; i < length; i++)  46                     Console.Write(path2[j, i].ToString() + " ");  47                 Console.WriteLine("长度:" + pathdist[j]);  48  }  49  Console.ReadKey();  50
51  }  52
53
54         //从某一源点出发，找到到某一结点的最短路径
55         static int getShortedPath(int[,]G, int start, int end,int [] path)  56  {  57             bool[] s = new bool[length]; //表示找到起始结点与当前结点间的最短路径
58             int min;  //最小距离临时变量
59             int curNode=0; //临时结点，记录当前正计算结点
60             int[] dist = new int[length];  61             int[] prev = new int[length];  62
63             //初始结点信息
64             for (int v = 0; v < length; v++)  65  {  66                 s[v] = false;  67                 dist[v] = G[start, v];  68                 if (dist[v] > MaxSize)  69                     prev[v] = 0;  70                 else
71                     prev[v] = start;  72  }  73             path[0] = end;  74             dist[start] = 0;  75             s[start] = true;  76             //主循环
77             for (int i = 1; i < length; i++)  78  {  79                 min = MaxSize;  80                 for (int w = 0; w < length; w++)  81  {  82                     if (!s[w] && dist[w] < min)  83  {  84                         curNode = w;  85                         min = dist[w];  86  }  87  }  88
89                 s[curNode] = true;  90                 for (int j = 0; j < length; j++)  91                     if (!s[j] && min + G[curNode, j] < dist[j])  92  {  93                         dist[j] = min + G[curNode, j];  94                         prev[j] = curNode;  95  }  96
97  }  98             //输出路径结点
99             int e = end, step = 0; 100             while (e != start) 101  { 102                 step++; 103                 path[step] = prev[e]; 104                 e = prev[e]; 105  } 106             for (int i = step; i > step/2; i--) 107  { 108                 int temp = path[step - i]; 109                 path[step - i] = path[i]; 110                 path[i] = temp; 111  } 112             return dist[end]; 113  } 114
115
116
117
118
119
120         //从某一源点出发，找到到所有结点的最短路径
121         static int[] getShortedPath(int[,] G, int start, int[,] path) 122  { 123             int[] PathID = new int[length];//路径（用编号表示）
124             bool[] s = new bool[length]; //表示找到起始结点与当前结点间的最短路径
125             int min;  //最小距离临时变量
126             int curNode = 0; //临时结点，记录当前正计算结点
127             int[] dist = new int[length]; 128             int[] prev = new int[length]; 129             //初始结点信息
130             for (int v = 0; v < length; v++) 131  { 132                 s[v] = false; 133                 dist[v] = G[start, v]; 134                 if (dist[v] > MaxSize) 135                     prev[v] = 0; 136                 else
137                     prev[v] = start; 138                 path[v,0] = v; 139  } 140
141             dist[start] = 0; 142             s[start] = true; 143             //主循环
144             for (int i = 1; i < length; i++) 145  { 146                 min = MaxSize; 147                 for (int w = 0; w < length; w++) 148  { 149                     if (!s[w] && dist[w] < min) 150  { 151                         curNode = w; 152                         min = dist[w]; 153  } 154  } 155
156                 s[curNode] = true; 157
158                 for (int j = 0; j < length; j++) 159                     if (!s[j] && min + G[curNode, j] < dist[j]) 160  { 161                         dist[j] = min + G[curNode, j]; 162                         prev[j] = curNode; 163  } 164
165
166  } 167             //输出路径结点
168             for (int k = 0; k < length; k++) 169  { 170                 int e = k, step = 0; 171                 while (e != start) 172  { 173                     step++; 174                     path[k, step] = prev[e]; 175                     e = prev[e]; 176  } 177                 for (int i = step; i > step / 2; i--) 178  { 179                     int temp = path[k, step - i]; 180                     path[k, step - i] = path[k, i]; 181                     path[k, i] = temp; 182  } 183  } 184             return dist; 185
186  } 187
188
189  } 190 }