Johnson 全源最短路径算法

解决单源最短路径问题（Single Source Shortest Paths Problem）的算法包括：

• Dijkstra 单源最短路径算法：时间复杂度为 O(E + VlogV)，要求权值非负；
• Bellman-Ford 单源最短路径算法：时间复杂度为 O(VE)，适用于带负权值情况；

对于全源最短路径问题（All-Pairs Shortest Paths Problem），可以认为是单源最短路径问题的推广，即分别以每个顶点作为源顶点并求其至其它顶点的最短距离。例如，对每个顶点应用 Bellman-Ford 算法，则可得到所有顶点间的最短路径的运行时间为 O(V²E)，由于图中顶点都是连通的，而边的数量可能会比顶点更多，这个时间没有比 Floyd-Warshall 全源最短路径算法 O(V³) 更优。那么，再试下对每个顶点应用 Dijkstra 算法，则可得到所有顶点间的最短路径的运行时间为 O(VE + V²logV)，看起来优于 Floyd-Warshall 算法的 O(V³)，所以看起来使用基于 Dijkstra 算法的改进方案好像更好，但问题是 Dijkstra 算法要求图中所有边的权值非负，不适合通用的情况。

在 1977 年，Donald B. Johnson 提出了对所有边的权值进行 "re-weight" 的算法，使得边的权值非负，进而可以使用 Dijkstra 算法进行最短路径的计算。

我们先自己思考下如何进行 "re-weight" 操作，比如，简单地对每条边的权值加上一个较大的正数，使其非负，是否可行？

   1     1     1
s-----a-----b-----c
               /
             /
     \______/
         4

比如上面的图中，共 4 条边，权值分别为 1,1,1,4。当前 s --> c 的最短路径是 {s-a, a-b, b-c} 即 1+1+1=3。而如果将所有边的权值加 1，则最短路径就会变成 {s-c} 的 5，因为 2+2+2=6，实际上导致了最短路径的变化，显然是错误的。

那么，Johnson 算法是如何对边的权值进行 "re-weight" 的呢？以下面的图 G 为例，有 4 个顶点和 5 条边。

首先，新增一个源顶点 4，并使其与所有顶点连通，新边赋权值为 0，如下图所示。

使用 Bellman-Ford 算法 计算新的顶点到所有其它顶点的最短路径，则从 4 至 {0, 1, 2, 3} 的最短路径分别是 {0, -5, -1, 0}。即有 h[] = {0, -5, -1, 0}。当得到这个 h[] 信息后，将新增的顶点 4 移除，然后使用如下公式对所有边的权值进行 "re-weight"：

w(u, v) = w(u, v) + (h[u] - h[v]).

则可得到下图中的结果：

此时，所有边的权值已经被 "re-weight" 为非负。此时，就可以利用 Dijkstra 算法对每个顶点分别进行最短路径的计算了。

Johnson 算法描述如下：

1. 给定图 G = (V, E)，增加一个新的顶点 s，使 s 指向图 G 中的所有顶点都建立连接，设新的图为 G’；
2. 对图 G’ 中顶点 s 使用 Bellman-Ford 算法计算单源最短路径，得到结果 h[] = {h[0], h[1], .. h[V-1]}；
3. 对原图 G 中的所有边进行 "re-weight"，即对于每个边 (u, v)，其新的权值为 w(u, v) + (h[u] - h[v])；
4. 移除新增的顶点 s，对每个顶点运行 Dijkstra 算法求得最短路径；

Johnson 算法的运行时间为 O(V²logV + VE)。

Johnson 算法伪码实现如下：

Johnson 算法 C# 代码实现如下：

using System;
using System.Collections.Generic;
using System.Linq;

namespace GraphAlgorithmTesting
{
  class Program
  {
    static void Main(string[] args)
    {
      // build a directed and negative weighted graph
      Graph directedGraph1 = new Graph(5);
      directedGraph1.AddEdge(0, 1, -1);
      directedGraph1.AddEdge(0, 2, 4);
      directedGraph1.AddEdge(1, 2, 3);
      directedGraph1.AddEdge(1, 3, 2);
      directedGraph1.AddEdge(1, 4, 2);
      directedGraph1.AddEdge(3, 2, 5);
      directedGraph1.AddEdge(3, 1, 1);
      directedGraph1.AddEdge(4, 3, -3);

      Console.WriteLine();
      Console.WriteLine("Graph Vertex Count : {0}", directedGraph1.VertexCount);
      Console.WriteLine("Graph Edge Count : {0}", directedGraph1.EdgeCount);
      Console.WriteLine();

      int[,] distSet1 = directedGraph1.Johnsons();
      PrintSolution(directedGraph1, distSet1);

      // build a directed and positive weighted graph
      Graph directedGraph2 = new Graph(4);
      directedGraph2.AddEdge(0, 1, 5);
      directedGraph2.AddEdge(0, 3, 10);
      directedGraph2.AddEdge(1, 2, 3);
      directedGraph2.AddEdge(2, 3, 1);

      Console.WriteLine();
      Console.WriteLine("Graph Vertex Count : {0}", directedGraph2.VertexCount);
      Console.WriteLine("Graph Edge Count : {0}", directedGraph2.EdgeCount);
      Console.WriteLine();

      int[,] distSet2 = directedGraph2.Johnsons();
      PrintSolution(directedGraph2, distSet2);

      Console.ReadKey();
    }

    private static void PrintSolution(Graph g, int[,] distSet)
    {
      Console.Write("	");
      for (int i = 0; i < g.VertexCount; i++)
      {
        Console.Write(i + "	");
      }
      Console.WriteLine();
      Console.Write("	");
      for (int i = 0; i < g.VertexCount; i++)
      {
        Console.Write("-" + "	");
      }
      Console.WriteLine();
      for (int i = 0; i < g.VertexCount; i++)
      {
        Console.Write(i + "|	");
        for (int j = 0; j < g.VertexCount; j++)
        {
          if (distSet[i, j] == int.MaxValue)
          {
            Console.Write("INF" + "	");
          }
          else
          {
            Console.Write(distSet[i, j] + "	");
          }
        }
        Console.WriteLine();
      }
    }

    class Edge
    {
      public Edge(int begin, int end, int weight)
      {
        this.Begin = begin;
        this.End = end;
        this.Weight = weight;
      }

      public int Begin { get; private set; }
      public int End { get; private set; }
      public int Weight { get; private set; }

      public void Reweight(int newWeight)
      {
        this.Weight = newWeight;
      }

      public override string ToString()
      {
        return string.Format(
          "Begin[{0}], End[{1}], Weight[{2}]",
          Begin, End, Weight);
      }
    }

    class Graph
    {
      private Dictionary<int, List<Edge>> _adjacentEdges
        = new Dictionary<int, List<Edge>>();

      public Graph(int vertexCount)
      {
        this.VertexCount = vertexCount;
      }

      public int VertexCount { get; private set; }

      public int EdgeCount
      {
        get
        {
          return _adjacentEdges.Values.SelectMany(e => e).Count();
        }
      }

      public void AddEdge(int begin, int end, int weight)
      {
        if (!_adjacentEdges.ContainsKey(begin))
        {
          var edges = new List<Edge>();
          _adjacentEdges.Add(begin, edges);
        }

        _adjacentEdges[begin].Add(new Edge(begin, end, weight));
      }

      public void AddEdge(Edge edge)
      {
        AddEdge(edge.Begin, edge.End, edge.Weight);
      }

      public void AddEdges(IEnumerable<Edge> edges)
      {
        foreach (var edge in edges)
        {
          AddEdge(edge);
        }
      }

      public IEnumerable<Edge> GetAllEdges()
      {
        return _adjacentEdges.Values.SelectMany(e => e);
      }

      public int[,] Johnsons()
      {
        // distSet[,] will be the output matrix that will finally have the shortest 
        // distances between every pair of vertices
        int[,] distSet = new int[VertexCount, VertexCount];

        for (int i = 0; i < VertexCount; i++)
        {
          for (int j = 0; j < VertexCount; j++)
          {
            distSet[i, j] = int.MaxValue;
          }
        }
        for (int i = 0; i < VertexCount; i++)
        {
          distSet[i, i] = 0;
        }

        // step 1: add new vertex s and connect to all vertices
        Graph g = new Graph(this.VertexCount + 1);
        g.AddEdges(this.GetAllEdges());

        int s = this.VertexCount;
        for (int i = 0; i < this.VertexCount; i++)
        {
          g.AddEdge(s, i, 0);
        }

        // step 2: use Bellman-Ford to evaluate shortest paths from s
        int[] h = g.BellmanFord(s);

        // step 3: re-weighting edges of the original graph
        //         w(u, v) = w(u, v) + (h[u] - h[v])
        foreach (var edge in this.GetAllEdges())
        {
          edge.Reweight(edge.Weight + (h[edge.Begin] - h[edge.End]));
        }

        // step 4: use Dijkstra for each edges
        for (int begin = 0; begin < this.VertexCount; begin++)
        {
          int[] dist = this.Dijkstra(begin);
          for (int end = 0; end < dist.Length; end++)
          {
            if (dist[end] != int.MaxValue)
            {
              distSet[begin, end] = dist[end] - (h[begin] - h[end]);
            }
          }
        }

        return distSet;
      }

      public int[,] FloydWarshell()
      {
        /* distSet[,] will be the output matrix that will finally have the shortest 
           distances between every pair of vertices */
        int[,] distSet = new int[VertexCount, VertexCount];

        for (int i = 0; i < VertexCount; i++)
        {
          for (int j = 0; j < VertexCount; j++)
          {
            distSet[i, j] = int.MaxValue;
          }
        }
        for (int i = 0; i < VertexCount; i++)
        {
          distSet[i, i] = 0;
        }

        /* Initialize the solution matrix same as input graph matrix. Or 
           we can say the initial values of shortest distances are based
           on shortest paths considering no intermediate vertex. */
        foreach (var edge in _adjacentEdges.Values.SelectMany(e => e))
        {
          distSet[edge.Begin, edge.End] = edge.Weight;
        }

        /* Add all vertices one by one to the set of intermediate vertices.
          ---> Before start of a iteration, we have shortest distances between all
          pairs of vertices such that the shortest distances consider only the
          vertices in set {0, 1, 2, .. k-1} as intermediate vertices.
          ---> After the end of a iteration, vertex no. k is added to the set of
          intermediate vertices and the set becomes {0, 1, 2, .. k} */
        for (int k = 0; k < VertexCount; k++)
        {
          // Pick all vertices as source one by one
          for (int i = 0; i < VertexCount; i++)
          {
            // Pick all vertices as destination for the above picked source
            for (int j = 0; j < VertexCount; j++)
            {
              // If vertex k is on the shortest path from
              // i to j, then update the value of distSet[i,j]
              if (distSet[i, k] != int.MaxValue
                && distSet[k, j] != int.MaxValue
                && distSet[i, k] + distSet[k, j] < distSet[i, j])
              {
                distSet[i, j] = distSet[i, k] + distSet[k, j];
              }
            }
          }
        }

        return distSet;
      }

      public int[] BellmanFord(int source)
      {
        // distSet[i] will hold the shortest distance from source to i
        int[] distSet = new int[VertexCount];

        // Step 1: Initialize distances from source to all other vertices as INFINITE
        for (int i = 0; i < VertexCount; i++)
        {
          distSet[i] = int.MaxValue;
        }
        distSet[source] = 0;

        // Step 2: Relax all edges |V| - 1 times. A simple shortest path from source
        // to any other vertex can have at-most |V| - 1 edges
        for (int i = 1; i <= VertexCount - 1; i++)
        {
          foreach (var edge in _adjacentEdges.Values.SelectMany(e => e))
          {
            int u = edge.Begin;
            int v = edge.End;
            int weight = edge.Weight;

            if (distSet[u] != int.MaxValue
              && distSet[u] + weight < distSet[v])
            {
              distSet[v] = distSet[u] + weight;
            }
          }
        }

        // Step 3: check for negative-weight cycles.  The above step guarantees
        // shortest distances if graph doesn't contain negative weight cycle.
        // If we get a shorter path, then there is a cycle.
        foreach (var edge in _adjacentEdges.Values.SelectMany(e => e))
        {
          int u = edge.Begin;
          int v = edge.End;
          int weight = edge.Weight;

          if (distSet[u] != int.MaxValue
            && distSet[u] + weight < distSet[v])
          {
            Console.WriteLine("Graph contains negative weight cycle.");
          }
        }

        return distSet;
      }

      public int[] Dijkstra(int source)
      {
        // dist[i] will hold the shortest distance from source to i
        int[] distSet = new int[VertexCount];

        // sptSet[i] will true if vertex i is included in shortest
        // path tree or shortest distance from source to i is finalized
        bool[] sptSet = new bool[VertexCount];

        // initialize all distances as INFINITE and stpSet[] as false
        for (int i = 0; i < VertexCount; i++)
        {
          distSet[i] = int.MaxValue;
          sptSet[i] = false;
        }

        // distance of source vertex from itself is always 0
        distSet[source] = 0;

        // find shortest path for all vertices
        for (int i = 0; i < VertexCount - 1; i++)
        {
          // pick the minimum distance vertex from the set of vertices not
          // yet processed. u is always equal to source in first iteration.
          int u = CalculateMinDistance(distSet, sptSet);

          // mark the picked vertex as processed
          sptSet[u] = true;

          // update dist value of the adjacent vertices of the picked vertex.
          for (int v = 0; v < VertexCount; v++)
          {
            // update dist[v] only if is not in sptSet, there is an edge from 
            // u to v, and total weight of path from source to  v through u is 
            // smaller than current value of dist[v]
            if (!sptSet[v]
              && distSet[u] != int.MaxValue
              && _adjacentEdges.ContainsKey(u)
              && _adjacentEdges[u].Exists(e => e.End == v))
            {
              int d = _adjacentEdges[u].Single(e => e.End == v).Weight;
              if (distSet[u] + d < distSet[v])
              {
                distSet[v] = distSet[u] + d;
              }
            }
          }
        }

        return distSet;
      }

      private int CalculateMinDistance(int[] distSet, bool[] sptSet)
      {
        int minDistance = int.MaxValue;
        int minDistanceIndex = -1;

        for (int v = 0; v < VertexCount; v++)
        {
          if (!sptSet[v] && distSet[v] <= minDistance)
          {
            minDistance = distSet[v];
            minDistanceIndex = v;
          }
        }

        return minDistanceIndex;
      }
    }
  }
}

运行结果如下：

参考资料

• 广度优先搜索
• 深度优先搜索
• Breadth First Traversal for a Graph
• Depth First Traversal for a Graph
• Dijkstra 单源最短路径算法
• Bellman-Ford 单源最短路径算法
• Bellman–Ford algorithm
• Introduction To Algorithm
• Floyd-Warshall's algorithm
• Bellman-Ford algorithm for single-source shortest paths
• Dynamic Programming | Set 23 (Bellman–Ford Algorithm)
• Dynamic Programming | Set 16 (Floyd Warshall Algorithm)
• Johnson’s algorithm for All-pairs shortest paths
• Floyd-Warshall's algorithm
• 最短路径算法--Dijkstra算法，Bellmanford算法，Floyd算法，Johnson算法
• QuickGraph, Graph Data Structures And Algorithms for .NET
• CHAPTER 26: ALL-PAIRS SHORTEST PATHS

本篇文章《Johnson 全源最短路径算法》由 Dennis Gao 发表自博客园，未经作者本人同意禁止任何形式的转载，任何自动或人为的爬虫转载行为均为耍流氓。