解决方案是暴力破解。您可以进行一些优化以加快速度,有些很简单,有些非常复杂。我怀疑您无法在台式计算机上使其足够快地工作,即使您能够这样做,我也不知道如何做到。然而,以下是暴力破解的原理。
注意:如果您想要一个准确的答案,Dijkstra和任何其他最短路径算法都不适用于此问题。
Start at a root node *root*
Let D[i] = longest path from node *root* to node i. D[*root*] = 0, and the others are also 0.
void getLongestPath(node, currSum)
{
if node is visited
return;
mark node as visited;
if D[node] < currSum
D[node] = currSum;
for each child i of node do
getLongestPath(i, currSum + EdgeWeight(i, node));
mark node as not visited;
}
让我们手动在这个图上运行它:1 - 2 (4), 1 - 3 (100), 2 - 3 (5), 3 - 5 (200), 3 - 4 (7), 4 - 5 (1000)
Let the root be 1. We call getLongestPath(1, 0);
2 is marked as visited and getLongestPath(2, 4); is called
D[2] = 0 < currSum = 4 so D[2] = 4.
3 is marked as visited and getLongestPath(3, 4 + 5); is called
D[3] = 0 < currSum = 9 so D[3] = 9.
4 is marked as visited and getLongestPath(4, 9 + 7); is called
D[4] = 0 < currSum = 16 so D[4] = 16.
5 is marked as visited and getLongestPath(5, 16 + 1000); is called
D[5] = 0 < currSum = 1016 so D[5] = 1016.
getLongestPath(3, 1016 + 200); is called, but node 3 is marked as visited, so nothing happens.
Node 5 has no more child nodes, so the function marks 5 as not visited and backtracks to 4. The backtracking will happen until node 1 is hit, which will end up setting D[3] = 100 and updating more nodes.
这是迭代方式的大致实现(未经测试,仅为基本思路):
Let st be a stack, the rest remains unchanged;
void getLongestPath(root)
{
st.push(pair(root, 0));
while st is not empty
{
topStack = st.top();
if topStack.node is visited
goto end;
mark topStack.node as visited;
if D[topStack.node] < topStack.sum
D[topStack.node = topStack.sum;
if topStack.node has a remaining child (*)
st.push(pair(nextchild of topStack.node, topStack.sum + edge cost of topStack.node - nextchild))
end:
mark topStack.node as not visited
st.pop();
}
}
(*) - 这是一个小问题 - 您需要为每个节点保留下一个子节点的指针,因为它可能在 while 循环的不同迭代中甚至重置自身(当您将 topStack.node 节点从堆栈中弹出时,指针会重置自身,因此请确保重置它)。这在链表上最容易实现,但是您应该使用 int[] 列表或 vector 列表,以便能够存储指针并具有随机访问,因为您将需要它。例如,您可以保持 next[i] = node i 在其邻接列表中的下一个子节点,并相应地更新它。您可能会遇到一些边缘情况,并且可能需要不同的 end: 情况:一种正常情况,一种是当您访问已经访问过的节点时发生的情况,在这种情况下,指针不需要被重置。也许将 visited 条件移动到您决定将某些内容推送到堆栈之前,以避免这种情况。