我正在从一本书中学习寻路，但我发现了一些奇怪的陈述。
为了完整性，我将包含大部分代码，但可以跳到第二部分（搜索（））
template <class graph_type >
class Graph_SearchDijkstra
{
private:
//create typedefs for the node and edge types used by the graph
typedef typename graph_type::EdgeType Edge;
typedef typename graph_type::NodeType Node;
private:
const graph_type & m_Graph;
//this vector contains the edges that comprise the shortest path tree -
//a directed sub-tree of the graph that encapsulates the best paths from
//every node on the SPT to the source node.
std::vector<const Edge*> m_ShortestPathTree;
//this is indexed into by node index and holds the total cost of the best
//path found so far to the given node. For example, m_CostToThisNode[5]
//will hold the total cost of all the edges that comprise the best path
//to node 5 found so far in the search (if node 5 is present and has
//been visited of course).
std::vector<double> m_CostToThisNode;
//this is an indexed (by node) vector of "parent" edges leading to nodes
//connected to the SPT but that have not been added to the SPT yet.
std::vector<const Edge*> m_SearchFrontier;
int m_iSource;
int m_iTarget;
void Search();
public:
Graph_SearchDijkstra(const graph_type& graph,
int source,
int target = -1):m_Graph(graph),
m_ShortestPathTree(graph.NumNodes()),
m_SearchFrontier(graph.NumNodes()),
m_CostToThisNode(graph.NumNodes()),
m_iSource(source),
m_iTarget(target)
{
Search();
}
//returns the vector of edges defining the SPT. If a target is given
//in the constructor, then this will be the SPT comprising all the nodes
//examined before the target is found, else it will contain all the nodes
//in the graph.
std::vector<const Edge*> GetAllPaths()const;
//returns a vector of node indexes comprising the shortest path
//from the source to the target. It calculates the path by working
//backward through the SPT from the target node.
std::list<int> GetPathToTarget()const;
//returns the total cost to the target
double GetCostToTarget()const;
};
搜索（）：
template <class graph_type>
void Graph_SearchDijkstra<graph_type>::Search()
{
//create an indexed priority queue that sorts smallest to largest
//(front to back). Note that the maximum number of elements the iPQ
//may contain is NumNodes(). This is because no node can be represented
// on the queue more than once.
IndexedPriorityQLow<double> pq(m_CostToThisNode, m_Graph.NumNodes());
//put the source node on the queue
pq.insert(m_iSource);
//while the queue is not empty
while(!pq.empty())
{
//get the lowest cost node from the queue. Don't forget, the return value
//is a *node index*, not the node itself. This node is the node not already
//on the SPT that is the closest to the source node
int NextClosestNode = pq.Pop();
//move this edge from the search frontier to the shortest path tree
m_ShortestPathTree[NextClosestNode] = m_SearchFrontier[NextClosestNode];
//if the target has been found exit
if (NextClosestNode == m_iTarget) return;
//now to relax the edges. For each edge connected to the next closest node
graph_type::ConstEdgeIterator ConstEdgeItr(m_Graph, NextClosestNode);
for (const Edge* pE=ConstEdgeItr.begin();
!ConstEdgeItr.end();
pE=ConstEdgeItr.next())
{
//the total cost to the node this edge points to is the cost to the
//current node plus the cost of the edge connecting them.
double NewCost = m_CostToThisNode[NextClosestNode] + pE->Cost();
//if this edge has never been on the frontier make a note of the cost
//to reach the node it points to, then add the edge to the frontier
//and the destination node to the PQ.
if (m_SearchFrontier[pE->To()] == 0)
{
m_CostToThisNode[pE->To()] = NewCost;
pq.insert(pE->To());
m_SearchFrontier[pE->To()] = pE;
}
//else test to see if the cost to reach the destination node via the
//current node is cheaper than the cheapest cost found so far. If
//this path is cheaper we assign the new cost to the destination
//node, update its entry in the PQ to reflect the change, and add the
//edge to the frontier
else if ( (NewCost < m_CostToThisNode[pE->To()]) &&
(m_ShortestPathTree[pE->To()] == 0) )
{
m_CostToThisNode[pE->To()] = NewCost;
//because the cost is less than it was previously, the PQ must be
//resorted to account for this.
pq.ChangePriority(pE->To());
m_SearchFrontier[pE->To()] = pE;
}
}
}
}
我没有得到的是这部分：
//else test to see if the cost to reach the destination node via the
//current node is cheaper than the cheapest cost found so far. If
//this path is cheaper we assign the new cost to the destination
//node, update its entry in the PQ to reflect the change, and add the
//edge to the frontier
else if ( (NewCost < m_CostToThisNode[pE->To()]) &&
(m_ShortestPathTree[pE->To()] == 0) )
如果新成本低于已经找到的成本，那么我们为什么还要测试该节点是否尚未添加到SPT中？这似乎超过了检查的目的？
仅供参考，在m_ShortestPathTree[pE->To()] == 0中，容器是一个向量，每个索引都有一个指向边（或NULL）的指针（索引代表一个节点）
答案 0 :(得分：3)
想象一下下面的图表：
S --5-- A --2-- F
\ /
-3 -4
\ /
B
您希望从S转到F。首先，让我告诉你Dijkstra的算法假设图中没有带有负权重的循环。在我的示例中，此循环为S -> B -> A -> S或更简单，S -> B -> S
如果你有这样一个循环，你可以无限循环，并且F的成本会越来越低。这就是Dijkstra算法无法接受的原因。
现在，您如何检查？它就像你发布的代码一样。每次要更新节点的权重时，除了检查权重是否变小之外，还要检查它是否不在接受列表中。如果是，那么你必须有一个负重量循环。否则，你怎么能最终前进到一个已经被接受的权重较小的节点？
让我们按照示例图上的算法（[]中的节点被接受）：
没有 if if question：
Starting Weights: S(0), A(inf), B(inf), F(inf)
- Accept S
New weights: [S(0)], A(5), B(-3), F(inf)
- Accept B
New weights: [S(-3)], A(-7), [B(-3)], F(inf)
- Accept A
New weights: [S(-3)], [A(-7)], [B(-11)], F(-5)
- Accept B again
New weights: [S(-14)], [A(-18)], [B(-11)], F(-5)
- Accept A again
... infinite loop
if if question：
Starting Weights: S(0), A(inf), B(inf), F(inf)
- Accept S
New weights: [S(0)], A(5), B(-3), F(inf)
- Accept B (doesn't change S)
New weights: [S(0)], A(-7), [B(-3)], F(inf)
- Accept A (doesn't change S or B
New weights: [S(0)], [A(-7)], [B(-3)], F(-5)
- Accept F