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？
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：
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’；
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]}；
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])；
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：
1 using System;
2 using System.Collections.Generic;
3 using System.Linq;
4
5 namespace GraphAlgorithmTesting
6 {
7 class Program
8 {
9 static void Main(string[] args)
10 {
11 // build a directed and negative weighted graph
12 Graph directedGraph1 = new Graph(5);
13 directedGraph1.AddEdge(0, 1, -1);
14 directedGraph1.AddEdge(0, 2, 4);
15 directedGraph1.AddEdge(1, 2, 3);
16 directedGraph1.AddEdge(1, 3, 2);
17 directedGraph1.AddEdge(1, 4, 2);
18 directedGraph1.AddEdge(3, 2, 5);
19 directedGraph1.AddEdge(3, 1, 1);
20 directedGraph1.AddEdge(4, 3, -3);
21
22 Console.WriteLine();
23 Console.WriteLine("Graph Vertex Count : {0}", directedGraph1.VertexCount);
24 Console.WriteLine("Graph Edge Count : {0}", directedGraph1.EdgeCount);
25 Console.WriteLine();
26
27 int[,] distSet1 = directedGraph1.Johnsons();
28 PrintSolution(directedGraph1, distSet1);
29
30 // build a directed and positive weighted graph
31 Graph directedGraph2 = new Graph(4);
32 directedGraph2.AddEdge(0, 1, 5);
33 directedGraph2.AddEdge(0, 3, 10);
34 directedGraph2.AddEdge(1, 2, 3);
35 directedGraph2.AddEdge(2, 3, 1);
36
37 Console.WriteLine();
38 Console.WriteLine("Graph Vertex Count : {0}", directedGraph2.VertexCount);
39 Console.WriteLine("Graph Edge Count : {0}", directedGraph2.EdgeCount);
40 Console.WriteLine();
41
42 int[,] distSet2 = directedGraph2.Johnsons();
43 PrintSolution(directedGraph2, distSet2);
44
45 Console.ReadKey();
46 }
47
48 private static void PrintSolution(Graph g, int[,] distSet)
49 {
50 Console.Write("\t");
51 for (int i = 0; i < g.VertexCount; i++)
52 {
53 Console.Write(i + "\t");
54 }
55 Console.WriteLine();
56 Console.Write("\t");
57 for (int i = 0; i < g.VertexCount; i++)
58 {
59 Console.Write("-" + "\t");
60 }
61 Console.WriteLine();
62 for (int i = 0; i < g.VertexCount; i++)
63 {
64 Console.Write(i + "|\t");
65 for (int j = 0; j < g.VertexCount; j++)
66 {
67 if (distSet[i, j] == int.MaxValue)
68 {
69 Console.Write("INF" + "\t");
70 }
71 else
72 {
73 Console.Write(distSet[i, j] + "\t");
74 }
75 }
76 Console.WriteLine();
77 }
78 }
79
80 class Edge
81 {
82 public Edge(int begin, int end, int weight)
83 {
84 this.Begin = begin;
85 this.End = end;
86 this.Weight = weight;
87 }
88
89 public int Begin { get; private set; }
90 public int End { get; private set; }
91 public int Weight { get; private set; }
92
93 public void Reweight(int newWeight)
94 {
95 this.Weight = newWeight;
96 }
97
98 public override string ToString()
99 {
100 return string.Format(
101 "Begin[{0}], End[{1}], Weight[{2}]",
102 Begin, End, Weight);
103 }
104 }
105
106 class Graph
107 {
108 private Dictionary<int, List<Edge>> _adjacentEdges
109 = new Dictionary<int, List<Edge>>();
110
111 public Graph(int vertexCount)
112 {
113 this.VertexCount = vertexCount;
114 }
115
116 public int VertexCount { get; private set; }
117
118 public int EdgeCount
119 {
120 get
121 {
122 return _adjacentEdges.Values.SelectMany(e => e).Count();
123 }
124 }
125
126 public void AddEdge(int begin, int end, int weight)
127 {
128 if (!_adjacentEdges.ContainsKey(begin))
129 {
130 var edges = new List<Edge>();
131 _adjacentEdges.Add(begin, edges);
132 }
133
134 _adjacentEdges[begin].Add(new Edge(begin, end, weight));
135 }
136
137 public void AddEdge(Edge edge)
138 {
139 AddEdge(edge.Begin, edge.End, edge.Weight);
140 }
141
142 public void AddEdges(IEnumerable<Edge> edges)
143 {
144 foreach (var edge in edges)
145 {
146 AddEdge(edge);
147 }
148 }
149
150 public IEnumerable<Edge> GetAllEdges()
151 {
152 return _adjacentEdges.Values.SelectMany(e => e);
153 }
154
155 public int[,] Johnsons()
156 {
157 // distSet[,] will be the output matrix that will finally have the shortest
158 // distances between every pair of vertices
159 int[,] distSet = new int[VertexCount, VertexCount];
160
161 for (int i = 0; i < VertexCount; i++)
162 {
163 for (int j = 0; j < VertexCount; j++)
164 {
165 distSet[i, j] = int.MaxValue;
166 }
167 }
168 for (int i = 0; i < VertexCount; i++)
169 {
170 distSet[i, i] = 0;
171 }
172
173 // step 1: add new vertex s and connect to all vertices
174 Graph g = new Graph(this.VertexCount + 1);
175 g.AddEdges(this.GetAllEdges());
176
177 int s = this.VertexCount;
178 for (int i = 0; i < this.VertexCount; i++)
179 {
180 g.AddEdge(s, i, 0);
181 }
182
183 // step 2: use Bellman-Ford to evaluate shortest paths from s
184 int[] h = g.BellmanFord(s);
185
186 // step 3: re-weighting edges of the original graph
187 // w(u, v) = w(u, v) + (h[u] - h[v])
188 foreach (var edge in this.GetAllEdges())
189 {
190 edge.Reweight(edge.Weight + (h[edge.Begin] - h[edge.End]));
191 }
192
193 // step 4: use Dijkstra for each edges
194 for (int begin = 0; begin < this.VertexCount; begin++)
195 {
196 int[] dist = this.Dijkstra(begin);
197 for (int end = 0; end < dist.Length; end++)
198 {
199 if (dist[end] != int.MaxValue)
200 {
201 distSet[begin, end] = dist[end] - (h[begin] - h[end]);
202 }
203 }
204 }
205
206 return distSet;
207 }
208
209 public int[,] FloydWarshell()
210 {
211 /* distSet[,] will be the output matrix that will finally have the shortest
212 distances between every pair of vertices */
213 int[,] distSet = new int[VertexCount, VertexCount];
214
215 for (int i = 0; i < VertexCount; i++)
216 {
217 for (int j = 0; j < VertexCount; j++)
218 {
219 distSet[i, j] = int.MaxValue;
220 }
221 }
222 for (int i = 0; i < VertexCount; i++)
223 {
224 distSet[i, i] = 0;
225 }
226
227 /* Initialize the solution matrix same as input graph matrix. Or
228 we can say the initial values of shortest distances are based
229 on shortest paths considering no intermediate vertex. */
230 foreach (var edge in _adjacentEdges.Values.SelectMany(e => e))
231 {
232 distSet[edge.Begin, edge.End] = edge.Weight;
233 }
234
235 /* Add all vertices one by one to the set of intermediate vertices.
236 ---> Before start of a iteration, we have shortest distances between all
237 pairs of vertices such that the shortest distances consider only the
238 vertices in set {0, 1, 2, .. k-1} as intermediate vertices.
239 ---> After the end of a iteration, vertex no. k is added to the set of
240 intermediate vertices and the set becomes {0, 1, 2, .. k} */
241 for (int k = 0; k < VertexCount; k++)
242 {
243 // Pick all vertices as source one by one
244 for (int i = 0; i < VertexCount; i++)
245 {
246 // Pick all vertices as destination for the above picked source
247 for (int j = 0; j < VertexCount; j++)
248 {
249 // If vertex k is on the shortest path from
250 // i to j, then update the value of distSet[i,j]
251 if (distSet[i, k] != int.MaxValue
252 && distSet[k, j] != int.MaxValue
253 && distSet[i, k] + distSet[k, j] < distSet[i, j])
254 {
255 distSet[i, j] = distSet[i, k] + distSet[k, j];
256 }
257 }
258 }
259 }
260
261 return distSet;
262 }
263
264 public int[] BellmanFord(int source)
265 {
266 // distSet[i] will hold the shortest distance from source to i
267 int[] distSet = new int[VertexCount];
268
269 // Step 1: Initialize distances from source to all other vertices as INFINITE
270 for (int i = 0; i < VertexCount; i++)
271 {
272 distSet[i] = int.MaxValue;
273 }
274 distSet[source] = 0;
275
276 // Step 2: Relax all edges |V| - 1 times. A simple shortest path from source
277 // to any other vertex can have at-most |V| - 1 edges
278 for (int i = 1; i <= VertexCount - 1; i++)
279 {
280 foreach (var edge in _adjacentEdges.Values.SelectMany(e => e))
281 {
282 int u = edge.Begin;
283 int v = edge.End;
284 int weight = edge.Weight;
285
286 if (distSet[u] != int.MaxValue
287 && distSet[u] + weight < distSet[v])
288 {
289 distSet[v] = distSet[u] + weight;
290 }
291 }
292 }
293
294 // Step 3: check for negative-weight cycles. The above step guarantees
295 // shortest distances if graph doesn't contain negative weight cycle.
296 // If we get a shorter path, then there is a cycle.
297 foreach (var edge in _adjacentEdges.Values.SelectMany(e => e))
298 {
299 int u = edge.Begin;
300 int v = edge.End;
301 int weight = edge.Weight;
302
303 if (distSet[u] != int.MaxValue
304 && distSet[u] + weight < distSet[v])
305 {
306 Console.WriteLine("Graph contains negative weight cycle.");
307 }
308 }
309
310 return distSet;
311 }
312
313 public int[] Dijkstra(int source)
314 {
315 // dist[i] will hold the shortest distance from source to i
316 int[] distSet = new int[VertexCount];
317
318 // sptSet[i] will true if vertex i is included in shortest
319 // path tree or shortest distance from source to i is finalized
320 bool[] sptSet = new bool[VertexCount];
321
322 // initialize all distances as INFINITE and stpSet[] as false
323 for (int i = 0; i < VertexCount; i++)
324 {
325 distSet[i] = int.MaxValue;
326 sptSet[i] = false;
327 }
328
329 // distance of source vertex from itself is always 0
330 distSet[source] = 0;
331
332 // find shortest path for all vertices
333 for (int i = 0; i < VertexCount - 1; i++)
334 {
335 // pick the minimum distance vertex from the set of vertices not
336 // yet processed. u is always equal to source in first iteration.
337 int u = CalculateMinDistance(distSet, sptSet);
338
339 // mark the picked vertex as processed
340 sptSet[u] = true;
341
342 // update dist value of the adjacent vertices of the picked vertex.
343 for (int v = 0; v < VertexCount; v++)
344 {
345 // update dist[v] only if is not in sptSet, there is an edge from
346 // u to v, and total weight of path from source to v through u is
347 // smaller than current value of dist[v]
348 if (!sptSet[v]
349 && distSet[u] != int.MaxValue
350 && _adjacentEdges.ContainsKey(u)
351 && _adjacentEdges[u].Exists(e => e.End == v))
352 {
353 int d = _adjacentEdges[u].Single(e => e.End == v).Weight;
354 if (distSet[u] + d < distSet[v])
355 {
356 distSet[v] = distSet[u] + d;
357 }
358 }
359 }
360 }
361
362 return distSet;
363 }
364
365 private int CalculateMinDistance(int[] distSet, bool[] sptSet)
366 {
367 int minDistance = int.MaxValue;
368 int minDistanceIndex = -1;
369
370 for (int v = 0; v < VertexCount; v++)
371 {
372 if (!sptSet[v] && distSet[v] <= minDistance)
373 {
374 minDistance = distSet[v];
375 minDistanceIndex = v;
376 }
377 }
378
379 return minDistanceIndex;
380 }
381 }
382 }
383 }运行结果如下：