课程表系列

相关题目：
207. 课程表
210. 课程表 II
630. 课程表 III
1462. 课程表 IV

class CourseSchedule:
    """
    207.课程表
    https://leetcode.cn/problems/course-schedule/
    """
    def __init__(self):
        # 记录⼀次递归堆栈中的节点
        self.onPath = []
        # 记录遍历过的节点，防⽌⾛回头路
        self.visited = []
        # 记录图中是否有环
        self.hasCycle = False

    def buildGraph(self, numCourses: int, prerequisites: List[List[int]]) -> List[List[int]]:

        # 注意这两种新建对象的区别，前者是传的引用，后者是拷贝一个新的变量
        # graph = [[]] * numCourses
        graph = [[] for _ in range(numCourses)]

        for edge in prerequisites:
            src = edge[1]
            dst = edge[0]
            graph[src].append(dst)

        return graph

    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        """
        dfs
        :param numCourses:
        :param prerequisites:
        :return:
        """
        graph = self.buildGraph(numCourses, prerequisites)
        self.visited = [False] * numCourses
        self.onPath = [False] * numCourses

        for i in range(numCourses):
            self.traverse(graph, i)

        return not self.hasCycle

    def traverse(self, graph, i):
        if self.onPath[i]:
            self.hasCycle = True
            return

        if self.visited[i]:
            return

        self.visited[i] = True
        self.onPath[i] = True

        for t in graph[i]:
            self.traverse(graph, t)

        self.onPath[i] = False

    def canFinish2(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        """
        bfs
        :param numCourses:
        :param prerequisites:
        :return:
        """
        graph = [[] for _ in range(numCourses)]
        indegree = [0] * numCourses

        for edge in prerequisites:
            src = edge[1]
            dst = edge[0]
            graph[src].append(dst)
            indegree[dst] += 1

        queue = []
        for i in range(numCourses):
            if indegree[i] == 0:
                queue.append(i)

        visited = 0
        while queue:
            cur = queue.pop(0)
            visited += 1
            for v in graph[cur]:
                indegree[v] -= 1
                if indegree[v] == 0:
                    queue.append(v)

        # 最后只需要判断已访问的课程数是否等于课程总数即可
        return visited == numCourses
        

# 210. 课程表 II
class Solution:
    def __init__(self):
        self.hascycle = False
        self.postorder = []

    def buildGraph(self, numCourses: int, prerequisites: List[List[int]]) -> List[List[int]]:

        # 注意这两种新建对象的区别，前者是传的引用，后者是拷贝一个新的变量
        # graph = [[]] * numCourses
        graph = [[] for _ in range(numCourses)]

        for edge in prerequisites:
            src = edge[1]
            dst = edge[0]
            graph[src].append(dst)

        return graph

    def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
        graph = self.buildGraph(numCourses, prerequisites)
        self.indegree = [0] * numCourses
        # 计算入度，和环检测算法相同
        for edge in prerequisites:
            dst = edge[0]
            self.indegree[dst] += 1

        # 根据入度初始化队列中的节点，和环检测算法相同
        queue = []
        for i in range(numCourses):
            if self.indegree[i] == 0:
                queue.append(i)

        res = [0] * numCourses
        # 记录遍历节点的顺序
        count = 0
        while queue:
            cur = queue.pop(0)
            # 弹出节点的顺序即为拓扑排序结果
            res[count] = cur
            count += 1
            for neighbor in graph[cur]:
                self.indegree[neighbor] -= 1
                if self.indegree[neighbor] == 0:
                    queue.append(neighbor)
        
        # 存在环
        if count != numCourses:
            return []

        return res

# 630. 课程表 III
# 这道题本质上用得是贪心算法或动态规划，和其他几道课程表题的解题思路不太一样。
import heapq
class Solution:
    def scheduleCourse(self, courses: List[List[int]]) -> int:
        courses.sort(key=lambda c: c[1])

        q = list()
        # 优先队列中所有课程的总时间
        total = 0

        # 从大根堆中依次取出 t值最大的那门课程 
        for ti, di in courses:
            # 如果当前队列中所有课程的总时间与ti之和小于等于di，就把ti加入优先级队列
            if total + ti <= di:
                total += ti
                # Python 默认是小根堆
                heapq.heappush(q, -ti)
            # 如果 当前队列中所有课程的总时间与ti之和大于di，就找出堆顶的元素，和当前元素t1进行比较，
            # 如果大于ti，就替换掉堆顶元素
            elif q and -q[0] > ti:
                total -= -q[0] - ti
                heapq.heappop(q)
                heapq.heappush(q, -ti)

        return len(q)

# 1462. 课程表 IV
class Solution:
    def checkIfPrerequisite(self, numCourses: int, prerequisites: List[List[int]], queries: List[List[int]]) -> List[bool]:
        g = [[] for _ in range(numCourses)]
        indgree = [0] * numCourses
        isPre = [[False] * numCourses for _ in range(numCourses)]
        for p in prerequisites:
            indgree[p[1]] += 1
            g[p[0]].append(p[1])

        q = []
        # 将入度为0的节点加入队列
        for i in range(numCourses):
            if indgree[i] == 0:
                q.append(i)

        while q:
            cur = q.pop(0)
            for ne in g[cur]:
                isPre[cur][ne] = True
                for i in range(numCourses):
                    isPre[i][ne] = isPre[i][ne] or isPre[i][cur]
                    
                indgree[ne] -= 1
                if indgree[ne] == 0:
                    q.append(ne)
        res = []
        for query in queries:
            res.append(isPre[query[0]][query[1]])
        return res