最短路径(Dijkstra)算法 之 .Net 4.0 Parallel 实现

Program.cs

using System;using System.Collections.Generic;using System.Linq;using System.Text;using System.Threading.Tasks;namespace ParallelDijkstra{    class Program    {        static void Main(string[] args)        {            int[,] c=new int[5,5];            Parallel.For(0, 5, (i) =>                {                    Parallel.For(0, 5, (j) =>                        {                            c[i,j] = Dijkstra.MaxInt;                        }                     );                }            );            c[0, 1] = c[1, 0] = 10;            c[0, 3] = c[3, 0] = 30;            c[0, 4] = c[4, 0] = 100;            c[1, 2] = c[2, 1] = 50;            c[2, 4] = c[4, 2] = 10;            c[3, 2] = c[2, 3] = 20;            c[3, 4] = c[4, 3] = 60;            int[] dist;            int[] prev;            Dijkstra.Execute(5, 0, c, out dist, out prev);            string path = Dijkstra.ParsePath(prev,0,4);            Console.WriteLine(path);            Console.ReadKey();        }    }}

Dijkstra.cs

using System;using System.Collections.Generic;using System.Linq;using System.Text;using System.Threading.Tasks;namespace ParallelDijkstra{    public class Dijkstra    {        public static int MaxInt = 999999;        /// <summary>        /// 执行最短路径分析        /// </summary>        /// <param name="n">结点个数</param>        /// <param name="v">源点</param>        /// <param name="c">路径权值矩阵</param>        /// <param name="dist">从源点出发至各节点的最短路径值列表</param>        /// <param name="prev">记录每一个结点的前一个最短路径结点</param>        public static void Execute(int n, int v, int[,] c,out int[] dist, out int[] prev)        {            int[] tmpDist=new int[n];            int[] tmpPrev=new int[n];            bool[] s = new bool[100];       // 判断是否已存入该点到S集合中            Parallel.For(0, n, (i) =>                {                    tmpDist[i] = c[v,i];                    s[i] = false;                    if (tmpDist[i] == MaxInt)                        tmpPrev[i] = -1;                    else                        tmpPrev[i] = v;                }            );            tmpDist[v] = 0;            s[v] = true;            //依次将未放入S集合的结点中,取dist[]最小值的结点，放入集合S之中            //一旦S包含了所有V中顶点，dist就记录了从源点到所有其他顶点之间的最短路径长度            Parallel.For(1, n, (i) =>                {                    int tmp = MaxInt;                    int u = v;                    // 找出当前未使用的点j的dist[j]最小值                    Parallel.For(0, n, (j) =>                        {                            if (!s[j] && tmpDist[j] < tmp)                            {                                u = j;          //u保存当前邻接点中距离最小的点的号码                                tmp=tmpDist[j];                            }                        }                    );                    s[u] = true;                //表示u点已存入S集合中                    //更新dist                    Parallel.For(0, n, (j) =>                        {                            if (!s[j] && c[u,j] < MaxInt)                            {                                int newDist = tmpDist[u] + c[u,j];                                if (newDist < tmpDist[j])                                {                                    tmpDist[j] = newDist;                                    tmpPrev[j] = u;                                }                            }                        }                    );                }            );            dist=tmpDist;            prev=tmpPrev;        }        /// <summary>        /// 获取从点v至点u的最短路径描述字符串        /// </summary>        /// <param name="prev">记录每一个结点的前一个最短路径结点</param>        /// <param name="v">起点</param>        /// <param name="u">终点</param>        /// <returns>从点v至点u的最短路径描述字符串</returns>        public static String ParsePath(int[] prev, int v, int u)        {            String path = "没有有效路径.";            int tot=0;            List<int> que = new List<int>();            que.Add(u);            tot++;            int tmp=prev[u];            while (tmp != v && tmp != -1)            {                que.Add(tmp);                tot++;                tmp=prev[tmp];            }            if (tmp != -1)            {                que.Add(v);                path=String.Empty;                for (int i = tot; i > 0; i--)                {                    path += que[i].ToString() + "-->";                }                path += que[0].ToString();            }            return path;        }    }}

输出结果：

0-->3-->2-->4