在图算法中找到最短路径
Find shortest path in graph algorithm
我刚刚看过这个视频:https://youtu.be/2E7MmKv0Y24?t=1335
它讨论了一种在 DAG 中找到源和给定顶点之间的最短距离的算法:
1.Sort the graph in topological order and express the graph in its linear form
2.Initialize all of the vertex to infinity except the source, which is initialized to 0
3.Iterate from the source to the rightmost vertex. For every vertex u, update the distance of all of its neighbor v to min((distance between source and v), (distance between source and u) + (distance between u and v))
大约在22:00,教授说这个算法适用于负边但图不能包含循环,但我认为该算法适用于包含非负循环的图。我说的对吗?
..., but i think the algorithm works for graphs that contain non-negative cycle. Am I right?
是的,你是对的。有关详细信息,请参阅 this post。
Another question is why do I need to topologically sort the array first? Why can't I just loop through every neighbour and calculate the distance to them?
如果我正确理解了这个问题,你不能只去任何下一个节点,因为可能有一个更短的方法到这个节点首先使用另一个节点(例如到达一个节点的成本是 5,还有另一种方法到使用成本为 1 + 1 = 2 的两个节点的节点;如果您不先排序,在这种情况下您将使用错误的路径)
我意识到我显然是不正确的。如果图形有循环,则无法进行拓扑排序。
我刚刚看过这个视频:https://youtu.be/2E7MmKv0Y24?t=1335
它讨论了一种在 DAG 中找到源和给定顶点之间的最短距离的算法:
1.Sort the graph in topological order and express the graph in its linear form
2.Initialize all of the vertex to infinity except the source, which is initialized to 0
3.Iterate from the source to the rightmost vertex. For every vertex u, update the distance of all of its neighbor v to min((distance between source and v), (distance between source and u) + (distance between u and v))
大约在22:00,教授说这个算法适用于负边但图不能包含循环,但我认为该算法适用于包含非负循环的图。我说的对吗?
..., but i think the algorithm works for graphs that contain non-negative cycle. Am I right?
是的,你是对的。有关详细信息,请参阅 this post。
Another question is why do I need to topologically sort the array first? Why can't I just loop through every neighbour and calculate the distance to them?
如果我正确理解了这个问题,你不能只去任何下一个节点,因为可能有一个更短的方法到这个节点首先使用另一个节点(例如到达一个节点的成本是 5,还有另一种方法到使用成本为 1 + 1 = 2 的两个节点的节点;如果您不先排序,在这种情况下您将使用错误的路径)
我意识到我显然是不正确的。如果图形有循环,则无法进行拓扑排序。