广度优先：课程表

题源力扣，仅作记录
https://leetcode-cn.com/problems/course-schedule
题解参考自

选修 numCourse 门课程，记为 0 到 numCourse-1 。课程0的前修课程1，记为[0,1]。
给定课程总量以及它们的先决条件，判断是否可能完成所有课程的学习？
def canFinish(self, numCourses, prerequisites):
“”"
:type numCourses: int
:type prerequisites: List[List[int]]
:rtype: bool
“”"

解题思路

用数组 [ int ] 记录课程为 index 的前修课程数，初始化为0

before = [0 for _ in range(numCourses)]

用嵌套数组 [[ int ]] 记录课程为 index 的后续课程，初始化为空

after = [[] for _ in range(numCourses)]

根据 prerequisites 里的内容重置 before 和 after

for cla, bef in prerequisites:
	before[cla] += 1
	after[bef].append(cla)

queue 用于放置前修课程数为 0 的课程队列

queue = [] # 放前修课程数为0的课
for cla in range(numCourses):
	if before[cla] == 0: # 无前修课程
	queue.append(cla)

循环取出队列中的课程直至queue为空，每取出一门课，待修课程数numCourses - 1，其后续课程 after[[ ]] 相应的前修课程数 before[ ] - 1，如果得到的before[ ] = 0 ，该门课进入queue，循环继续

while queue:
	# 修完一门前修课程数为0的课
	cla = queue.pop(0) 
	numCourses -= 1 
	# 该门课的后续课程的前修课程数-1
	for after_cla in after[cla]:
		before[after_cla] -= 1
		if before[after_cla] == 0:
			queue.append(after_cla)

如果 numCourses = 0，则所有课程修完，返回真。

return numCourses == 0

完整代码

def canFinish(self, numCourses, prerequisites):
	"""
    :type numCourses: int
	:type prerequisites: List[List[int]]
	:rtype: bool
	"""
	# 第 index 门课的前修课程数，初始为0
	before = [0 for _ in range(numCourses)]
	# 修完第 index 门课后可修的课程集合，初始为空[]
	after = [[] for _ in range(numCourses)]

	for cla, bef in prerequisites:
		before[cla] += 1
		after[bef].append(cla)

    queue = [] # 放前修课程数为0的课

	for cla in range(numCourses):
		if before[cla] == 0: # 无前修课程
			queue.append(cla)

	while queue:
		# 修完一门前修课程数为0的课
		cla = queue.pop(0) 
		numCourses -= 1 
		# 该门课的后续课程的前修课程数-1
		for after_cla in after[cla]:
			before[after_cla] -= 1
			if before[after_cla] == 0:
				queue.append(after_cla)

	return numCourses == 0