图解-迪杰斯特拉算法（找最短路径）Dijkstra's Algorithm (finding shortestpaths)

转自：http://www.mathcs.emory.edu/~cheung/Courses/171/Syllabus/11-Graph/dijkstra2.html

 

一. 图解迪杰斯特拉

• Before showing you the pseudo code, I will first execute the Dijkstra's Algorithm on the following graph:

  Find the shortest paths from the node 1 to all other nodes in the graph:

  ---

• The execution of the Dijkstra's Algorithm:

  Initilaization: Label the source node (node 1) as reached Label all the other nodes as unreached Use each edge from the source node (node 1) as the shortest path to nodes that you can reach immediately Result: (The reached node(s) have a cyan circle as marking. The unreached nodes are unmarked) Iteration (1): Find the unreached node m that has the shortest path from the source node: m = 2 (with path length = 3) Added the edge you used to reach m to the shortest path Label the node m as reached Recompute the shortest paths of nodes that can be reached via m (if possible) We can reach the node 3 via node 2 through a shorter path !!! Result at the end of the iteration: Iteration (2): Find the unreached node m that has the shortest path from the source node: m = 3 (with path length = 5) (Node 0 will work also, but I picked node 3) Added the edge you used to reach m to the shortest path Label the node m as reached Recompute the shortest paths of nodes that can be reached via m (if possible) We can reach the node 4 via node 3 through a path of length 8 !!! Result at the end of the iteration: Iteration (3): Find the unreached node m that has the shortest path from the source node: m = 0 (with path length = 5) Added the edge you used to reach m to the shortest path Label the node m as reached Recompute the shortest paths of nodes that can be reached via m (if possible) There are no improvements.... Result at the end of the iteration: Iteration (4): Find the unreached node m that has the shortest path from the source node: m = 4 (with path length = 8) Added the edge you used to reach m to the shortest path Label the node m as reached Recompute the shortest paths of nodes that can be reached via m (if possible) Again, no improvements.... (because there are no more unreached nodes !!!) Result at the end of the iteration: Done !!!

二. 算法伪代码

• Psuedo code of the Dijkstra's Algorithm:

  /* ------------------------ Initialization ------------------------ */ ReachSet := {S} (Shortest path from S to all other nodes) for ( all other nodes n ≠ S ) do { D[n] = linkcost(S,n); // D[n] = length of shortest path S -> n if ( LinkCost[s][i] < infinite ) predNode[i] = S; // Record the "predecessor edge" to node n // This info. says: edge S -> n is used to go to n } /* ----------------------- Find min. cost path ----------------------- */ while ( ReachSet ≠ all nodes ) do { Find the node m ∈ (N - ReachSet) that has smallest value for D[m] /* ================================================ Label node m as "reached" ================================================ */ ReachSet := ReachSet ∪ {m} // shortest path to m has been found // We don't have to do anything to add the edge used to // get to m to the shortest path // The variable "predNode[m] kept track of this info ! /* --------------------------------------------------------------- Recompute the shortest paths of nodes that can be reached via m --------------------------------------------------------------- */ for ( each node n ∈ (N - ReachSet) ) do { /* ================================================================== How to discover possible better path to n via m: current shortest length = D[n] S <-------> n \ / \ / D[m] \ / linkcost(m,n) \ / \ / m We have a better shortest path is: D[m] + linkcost(m,n) < D[n] ================================================================== */ if ( D[m] + linkcost(m,n) < D[n] ) { D[n] = D[m] + linkcost(m,n); // Update the length predNode[n] = m; // Use m -> n to reach n ! } } }

三.Code in java 

Dijkstra's Algorithm in Java:

public void Dijkstra(int s) // s = the source node { int i, j, m, n, k; int[] D = new int[NNodes]; // Current distance int[] predNode = new int[NNodes]; // shortest path edges /* ============================================== Initialize: ReachSet = {s} UnReachSet = all other nodes ============================================== */ for ( i = 0; i < NNodes; i++ ) Reached[i] = false; Reached[s] = true; // ReachSet = {s} /* ======================================================== Use edge from the source node as the shortest path to nodes that you can reach immediately ======================================================== */ predNode[s] = s; for ( i = 0; i < NNodes; i++ ) { D[i] = LinkCost[s][i]; // Min. distance if ( LinkCost[s][i] < infinite ) predNode[i] = s; // edge of shortest path } /* ======================================================== The Dijkstra's Algorithm (need to loop NNodes-1 times) ======================================================== */ for ( k = 0; k < NNodes-1; k++ ) { /* ============================================================= Find the first unreached node m ============================================================= */ for ( m = 0; m < NNodes; m++ ) if ( ! Reached[m] ) break; /* ============================================================= Check if there is a unreached node with shorter path length ============================================================= */ for ( n = m+1; n < NNodes; n++ ) { if ( Reached[n] == false && D[n] < D[m] ) m = n; // A better node is found, update m } // Now: m = unreach node with smallest value for D[m] /* ========================== Add m to ReachSet ========================== */ Reached[m] = true; /* ------------------------------------------------------ Find possible "short cut" through m (better shortest paths through m) ------------------------------------------------------ */ for ( n = 0; n < NNodes; n++ ) { if ( Reached[n] == false ) { if ( D[m] + LinkCost[m][n] < D[n] ) { D[n] = D[m] + LinkCost[m][n]; // Shorter path found ! predNode[n] = m; // Record the shortest path link } } } } }

完整运行Code 参见：

http://www.mathcs.emory.edu/~cheung/Courses/171/Syllabus/11-Graph/Progs/Dijkstra

Output:

Shortest path from node B A B C D E -------------------- A * 5 10 * * B 5 * 3 11 * C 10 3 * 2 * D * 11 2 * 3 E * * * 3 * ReachSet = B --- initial state D[A] = 5, D[C] = 3, D[D] = 11, D[E] = * --- add m = C ReachSet = B C D[A] = 5, D[D] = 5, D[E] = * --- add m = A ReachSet = A B C D[D] = 5, D[E] = * --- add m = D ReachSet = A B C D D[E] = 8 --- add m = E ReachSet = A B C D E Shortest paths: B --> A B --> B B --> C C --> D D --> E

四.分析

1.  时间复杂度O(n²)

2.  适用于权值为非负的图单源（从一个节点到其它节点）最短路径查找

3. 如果求所有N个节点之间的最短路径，N个节点都要执行该算法。时间复杂度为O(n³)