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  • 【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 }
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  • 原文地址:https://www.cnblogs.com/cuish/p/2969442.html
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