经过其他首都的首都之间的最短路径

Question

我正在尝试为大学作业开发一些代码，我有一个算法可以为我提供图中两个节点之间的最短路径。请注意，节点是指有首都的国家。

有人能向我解释一下我如何通过首都(国家(列表开发出从A国到B国的最短路径吗？

我已经实现了一种方法，它也可以给我两个地理点之间的距离。

我最初的想法是根据首都到国家A的距离来排列首都列表，然后求出国家A和列表第一个、列表第一个和列表第三个之间最短路径的所有距离。显然这是不正确的。

public double shortestPathCapitals2(List<String> capitais, Pais pOrig, Pais pDest) {
double dist = 0;
LinkedList<Pais> shortPath = new LinkedList<Pais>();
LinkedList<String> temp = new LinkedList<>(capitais);
temp.addFirst(pOrig.getCapital());
temp.addLast(pDest.getCapital());
Collections.sort(temp, (c1, c2) -> (int) (distance(pOrig, shortestPathCapitals2(c2)) - distance(pOrig, obterPaisPorCapital(c1))));
for (int i = 0; i < temp.size() - 1; i++) {
Pais p1 = obterPaisPorCapital(temp.get(i));
Pais p2 = obterPaisPorCapital(temp.get(i + 1));
dist += shortestPath(p1, p2, shortPath);
shortPath.clear();
}
return dist;
}

谢谢。

Answer 1

问题描述：

给定一个具有顶点V和边E的图。我们想要找到Va和Vb之间的路径P，使得：

• 路径包含{V0，V1，..}(V的某个子集(
• P中边上的权重之和是最小的

伪代码：

function findPath(startVertex, endVertex, verticesToBeVisited, currentPath)
// check if we have reached the destination
if startVertex == endVertex:
/*
* there are multiple ways of reaching the destination
* calculate the length of the past (also called the cost)
* if the cost is lower than the current minimum, store the path
*/
cost = calculateCost(currentPath)
if cost  < currentMinCost:
currentMinCost = cost
currentMinPath = currentPath            
else:
/*
* if we have not reached the destination
* we need to try all possible next hops
* this algorithm uses recursion to do so
*/
for every vertex Vn that is a neighbour of startVertex:
/*
* this check prevents us from going
* Paris --> Rome --> Paris --> Rome (endlessly)
*/
if currentPath contains Vn:
continue
// add the next hop to our path
currentPath += Vn
// if this vertex needed to be visit, cross it out in the list
if verticesToBeVisited contains Vn:
verticesToBeVisited -= Vn
// recursion
findPath(Vn, endVertex, verticesToBeVisited, currentPath)
// clean up
if verticesToBeVisited contained Vn:
verticesToBeVisited += Vn
currentPath -= Vn