Dijkstra算法求最短路径(C#版)

行如下图的路径,(V0是中心):



经过该算法后转化为下图



using System;
using System.Collections;
using System.Text;

namespace Greedy
{  
    class Marx
    {
        private int[] distance;       
        private int row;
        private ArrayList ways = new ArrayList();

        public Marx(int n,params int[] d)
        {
            this.row = n;
            distance = new int[row * row];
            for (int i = 0; i < row * row; i++)
            {
                this.distance[i] = d[i];             
            }
            for (int i = 0; i < this.row; i++)  //有row个点，则从中心到各点的路有row-1条
            {
                ArrayList w = new ArrayList();
                int j = 0;
                w.Add(j);
                ways.Add(w);
            }
        }
        //------------------------------
        public void Find_way()
        {
            ArrayList S = new ArrayList(1);
            ArrayList Sr = new ArrayList(1);
            int []Indexof_distance=new int[this.row];
           
            for(int i=0; i < row; i++)
            {
                Indexof_distance[i]=i;
            }

            S.Add( Indexof_distance[0] );       

            for (int i = 0; i < this.row; i++)
            {
                Sr.Add( Indexof_distance[i] );
            }
            Sr.RemoveAt(0);
            int[] D = new int[this.row];    //存放中心点到每个点的距离
          
         //---------------以上已经初始化了，S和Sr(里边放的都是点的编号)------------------
            int Count = this.row - 1;
            while (Count>0)
            {
                //假定中心点的编号是0的贪吃法求路径
                for (int i = 0; i < row; i++)
                    D[i] = this.distance[i];

                int min_num = (int)Sr[0];  //距中心点的最小距离点编号

                foreach (int s in Sr)
                {
                    if (D[s] < D[min_num]) min_num = s;
                }
       
                //以上可以排序优化
                S.Add(min_num);
                Sr.Remove(min_num);
                //-----------把最新包含进来的点也加到路径中-------------
                ((ArrayList)ways[min_num]).Add(min_num);
                //-----------------------------------------------
                foreach (int element in Sr)
                {
                    int position = element * (this.row) + min_num;
                    bool exchange = false;      //有交换标志

                    if (D[element] < D[min_num] + this.distance[position])
                        D[element] = D[element];
                    else
                    {
                        D[element] = this.distance[position] + D[min_num];
                        exchange = true;
                    }
                    //修改距离矩阵                  
                    this.distance[element] = D[element];
                    position = element * this.row;     
                    this.distance[position] = D[element];

                    //修改路径---------------
                    if (exchange == true)
                    {                      
                        ((ArrayList)ways[element]).Clear();
                        foreach (int point in (ArrayList)ways[min_num])
                            ((ArrayList)ways[element]).Add(point);
                    }
                }
                --Count;
           }
        }
        //----------------------------------------------------
        public void Display()
        {        
            //------中心到各点的最短路径----------
            Console.WriteLine("中心到各点的最短路径如下: \n\n");
            int sum_d_index = 0;
            foreach(ArrayList mother in ways)
            {
                foreach (int child in mother)
                    Console.Write("V{0} -- ", child+1);
                Console.WriteLine("    路径长 {0}",distance[sum_d_index++]);
            }
        }
    }

    class MainEnterPoint
    {  
        static void Main(string[] args)
        {
            int r;     //列数
            Console.Write("请输入点个数（含配送中心点）: ");
            Int32.TryParse(Console.ReadLine(), out r);
            Console.WriteLine("各点分别为: \n");
            for (int i = 0; i < r; i++)
                Console.Write("V{0} ", i);
            Console.Write("  假定第一个点是配送中心");
            Console.WriteLine("\n\n输入各点之间的距离(无通径的用个大整数表示)\n");

            int[] a = new int[r * r];           
            int da;

            for (int i = 0; i < r; i++)
            {
                for (int j = i + 1; j < r; j++)
                {
                    Console.Write("V{0} 到 V{1}的距离是:  ",i,j);
                    Int32.TryParse(Console.ReadLine(), out da);
                    a[i * r + j] = da;
                    Console.WriteLine();
                }
            }
            //----完善距离矩阵(距离矩阵其实可以是个上三角矩阵，
            //----但为了处理方便，还是将其完整成一个对称阵)-----------
            for (int i = 0; i < r; i++)
            {
                for (int j = 0; j < r; j++)
                {
                    if (i == j)
                    {
                        a[i * r + j] = 0;
                    }
                    a[j * r + i] = a[i * r + j];
                }
            }     
            Marx m=new Marx(r,a);
            Console.WriteLine();
            m.Find_way();
            m.Display();
        }
    }
}

//该程序不但能够算出从中心到各点的最短路径距离,而且把路径也保存了下来.