python - 拓扑排序中的 indegrees 用 Kahn 算法解决 CouseSchedule

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 -----> v , 节点 u 有一个出度而节点 v 有一个入度。

所以我认为，in_degrees[u] += 1 应该改为 in_degrees[v] += 1因为如果存在 (u,v)，那么 v 至少有一个入射事件和一个入度

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

但是，原始解决方案有效。

我的理解有什么问题吗？

Answer 1

看上面那行； 图[v].append(u)。边缘实际上与您的假设和输入格式相反。这是因为对于拓扑排序，我们希望没有依赖项/传入边的事物在结果顺序的前面结束，因此我们根据解释来引导边，“是要求”而不是“需要”。例如。输入对 (0,1) 意味着 0 需要 1，因此在图中我们绘制一条有向边 (1,0)，以便 1 在我们的排序中可以排在 0 之前。因此 0 通过考虑这对获得度数。