【转载】C#实现的最短路径分析

C#实现的最短路径分析,转载的

  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 }

 

posted on 2013-03-19 16:22  崔好好  阅读(630)  评论(0编辑  收藏  举报

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