جدول المقررات I وII

نظرة عامة على المسألة

نمذجة الرسم البياني

from collections import defaultdict

def build_graph(n, prerequisites):
    graph = defaultdict(list)
    in_degree = [0] * n
    for a, b in prerequisites:  # b must come before a
        graph[b].append(a)
        in_degree[a] += 1
    return graph, in_degree

graph, ind = build_graph(4, [[1,0],[2,0],[3,1],[3,2]])
print('In-degrees:', ind)   # [0, 1, 1, 2]
print('Graph edges:', dict(graph))

Course Schedule I: حل Kahn

from collections import deque, defaultdict

def canFinish(numCourses, prerequisites):
    graph = defaultdict(list)
    in_degree = [0] * numCourses
    for a, b in prerequisites:
        graph[b].append(a)
        in_degree[a] += 1
    
    queue = deque(i for i in range(numCourses) if in_degree[i] == 0)
    count = 0
    
    while queue:
        course = queue.popleft()
        count += 1
        for nxt in graph[course]:
            in_degree[nxt] -= 1
            if in_degree[nxt] == 0:
                queue.append(nxt)
    
    return count == numCourses

print(canFinish(2, [[1,0]]))        # True
print(canFinish(2, [[1,0],[0,1]])) # False

Course Schedule II: إعادة الترتيب

from collections import deque, defaultdict

def findOrder(numCourses, prerequisites):
    graph = defaultdict(list)
    in_degree = [0] * numCourses
    for a, b in prerequisites:
        graph[b].append(a)
        in_degree[a] += 1
    
    queue = deque(i for i in range(numCourses) if in_degree[i] == 0)
    order = []
    
    while queue:
        course = queue.popleft()
        order.append(course)
        for nxt in graph[course]:
            in_degree[nxt] -= 1
            if in_degree[nxt] == 0:
                queue.append(nxt)
    
    return order if len(order) == numCourses else []

print(findOrder(4, [[1,0],[2,0],[3,1],[3,2]]))

Course Schedule باستخدام DFS

from collections import defaultdict

def canFinish_dfs(numCourses, prerequisites):
    graph = defaultdict(list)
    for a, b in prerequisites:
        graph[b].append(a)
    
    # 0=unvisited, 1=in-progress, 2=done
    state = [0] * numCourses
    
    def has_cycle(course):
        if state[course] == 1: return True  # back edge
        if state[course] == 2: return False # already cleared
        state[course] = 1
        for nxt in graph[course]:
            if has_cycle(nxt):
                return True
        state[course] = 2
        return False
    
    return not any(has_cycle(i) for i in range(numCourses))

print(canFinish_dfs(2, [[1,0]]))        # True
print(canFinish_dfs(2, [[1,0],[0,1]])) # False

أهمية اتجاه الحواف

Course Schedule III: الصيغة الجشعة

التعامل مع العقد المعزولة

# Example: 4 courses, but only courses 0 and 1 have a prerequisite relationship
# Courses 2 and 3 are isolated - they should appear in the output
from collections import deque, defaultdict

def findOrder_isolated(numCourses, prerequisites):
    graph = defaultdict(list)
    in_degree = [0] * numCourses  # initialise ALL nodes
    for a, b in prerequisites:
        graph[b].append(a)
        in_degree[a] += 1
    queue = deque(i for i in range(numCourses) if in_degree[i] == 0)
    order = []
    while queue:
        c = queue.popleft(); order.append(c)
        for nxt in graph[c]:
            in_degree[nxt] -= 1
            if in_degree[nxt] == 0: queue.append(nxt)
    return order if len(order) == numCourses else []

print(findOrder_isolated(4, [[1,0]]))  # [0,1,2,3] or [2,3,0,1] etc.

زمن إكمال المقررات بالتوازي

استراتيجية التواصل في المقابلة

اختبار شامل

from collections import deque, defaultdict

def findOrder(numCourses, prerequisites):
    graph = defaultdict(list)
    in_degree = [0] * numCourses
    for a, b in prerequisites:
        graph[b].append(a)
        in_degree[a] += 1
    queue = deque(i for i in range(numCourses) if in_degree[i] == 0)
    order = []
    while queue:
        c = queue.popleft(); order.append(c)
        for nxt in graph[c]:
            in_degree[nxt] -= 1
            if in_degree[nxt] == 0: queue.append(nxt)
    return order if len(order) == numCourses else []

print(findOrder(1, []))                    # [0]
print(findOrder(2, [[0,1]]))              # [1, 0]
print(findOrder(3, [[1,0],[2,1]]))        # [0, 1, 2]
print(findOrder(3, [[1,0],[0,1]]))        # [] cycle

اختبار سريع

مراجعة الدرس