Kahn算法求解CouseSchedule的拓扑排序中的索引

Question

我正在学习解决leetcode中的拓扑排序问题

There are a total of n courses you have to take, labeled from 0 to n-1.

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?

Example 1:

Input: 2, [[1,0]] 
Output: true
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0. So it is possible.

Example 2:

Input: 2, [[1,0],[0,1]]
Output: false
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0, and to take course 0 you should
             also have finished course 1. So it is impossible.

Note:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

我在讨论区阅读了以下拓扑排序解决方案

class Solution5:
    def canFinish(self,numCourses, prerequirements):
    """
    :type numCourse: int
    :type prerequirements: List[List[int]]
    :rtype:bool
    """
    if not prerequirements: return True 
    count = []

    in_degrees = defaultdict(int)
    graph = defaultdict(list)

    for u, v in prerequirements:
        graph[v].append(u)
        in_degrees[u] += 1 #Confused here

    queue = [u for u in graph if in_degrees[u]==0]

    while queue:
        s = queue.pop()
        count.append(s)
        for v in graph(s):
            in_degrees[v] -= 1
            if in_degrees[v] ==0:
                queue.append(v)
    #check there exist a circle
    for u in in_degrees:
        if in_degrees[u]:
            return False 
    return True 

我对in_degrees[u] += 1感到困惑

for u, v in prerequirements:
    graph[v].append(u)
    in_degrees[u] += 1 #Confused here

对于有向边（u，v），u----->；节点u有一个outdegree，而节点v有一个indegree

所以我认为in_degrees[u] += 1应该改成in_degrees[v] += 1 因为如果存在（u，v），那么v至少有一个传入事件和一个indegree

In Degree: This is applicable only for directed graph. This represents the number of edges incoming to a vertex.

然而，最初的解决方案是有效的

我的理解有什么问题

Answer 1

看看上面的那条线graph[v].append(u)。这些边实际上与您的假设和输入格式的方向相反。这是因为对于拓扑排序，我们希望没有依赖项/传入边的东西最终位于结果顺序的前面，因此我们根据解释“is requirement for”而不是“requires”来引导边。输入对（0,1）意味着0需要1，所以在图中我们画了一条有向边（1,0），这样在我们的排序中1可以先于0。因此，0从考虑这一对中获得了独立性