任务 · 算法困难 Python ·

Course order

There are n courses numbered 0..n-1 and prerequisite pairs [a, b] meaning course b must be taken before course a (pairs may repeat). Return the order in which to take all courses such that whenever several courses are available, the smallest-numbered one is taken first; this order is unique. Return [] if it's impossible to take all courses. Up to 100,000 courses and 300,000 pairs.

函数签名

def course_order(n: int, prereqs: list[list[int]]) -> list[int]:

示例

course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])  # -> [0, 1, 2, 3]
course_order(2, [[0, 1], [1, 0]])  # -> []
course_order(3, [])  # -> [0, 1, 2]

另有 7 个隐藏测试和 1 个限时海量数据测试；其输入数据不公开。任务以英文提示词提供给模型。

第 1 名Z.ai GLM 5.3已解决 · 通过 11/11 个测试生成耗时 1.3 秒 · 执行耗时 555 毫秒 · $0.0003

✓	course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])预期 [0, 1, 2, 3]	4 毫秒
✓	course_order(2, [[0, 1], [1, 0]])预期 []	<1 毫秒
✓	course_order(3, [])预期 [0, 1, 2]	<1 毫秒

✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 546 毫秒

代码

import heapq

def course_order(n: int, prereqs: list[list[int]]) -> list[int]:
    adj = [[] for _ in range(n)]
    indeg = [0] * n
    for a, b in prereqs:
        adj[b].append(a)
        indeg[a] += 1
    heap = [i for i in range(n) if indeg[i] == 0]
    heapq.heapify(heap)
    res = []
    while heap:
        c = heapq.heappop(heap)
        res.append(c)
        for nxt in adj[c]:
            indeg[nxt] -= 1
            if indeg[nxt] == 0:
                heapq.heappush(heap, nxt)
    return res if len(res) == n else []

第 1 名OpenAI GPT-6 Luna已解决 · 通过 11/11 个测试生成耗时 2.1 秒 · 执行耗时 580 毫秒 · $0.0001

✓	course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])预期 [0, 1, 2, 3]	<1 毫秒
✓	course_order(2, [[0, 1], [1, 0]])预期 []	<1 毫秒
✓	course_order(3, [])预期 [0, 1, 2]	<1 毫秒

✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 572 毫秒

代码

import heapq

def course_order(n: int, prereqs: list[list[int]]) -> list[int]:
    graph = [[] for _ in range(n)]
    indegree = [0] * n
    for a, b in prereqs:
        graph[b].append(a)
        indegree[a] += 1

    available = [i for i, degree in enumerate(indegree) if degree == 0]
    heapq.heapify(available)
    order = []

    while available:
        course = heapq.heappop(available)
        order.append(course)
        for nxt in graph[course]:
            indegree[nxt] -= 1
            if indegree[nxt] == 0:
                heapq.heappush(available, nxt)

    return order if len(order) == n else []

第 3 名SpaceXAI Grok Build 0.1已解决 · 通过 11/11 个测试生成耗时 16 秒 · 执行耗时 831 毫秒 · $0.0052

✓	course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])预期 [0, 1, 2, 3]	<1 毫秒
✓	course_order(2, [[0, 1], [1, 0]])预期 []	1 毫秒
✓	course_order(3, [])预期 [0, 1, 2]	<1 毫秒

✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 823 毫秒

代码

import heapq
from collections import defaultdict

def course_order(n: int, prereqs: list[list[int]]) -> list[int]:
    if n == 0:
        return []
    graph = defaultdict(set)
    indegree = [0] * n
    for a, b in prereqs:
        if a not in graph[b]:
            graph[b].add(a)
            indegree[a] += 1
    heap = [i for i in range(n) if indegree[i] == 0]
    heapq.heapify(heap)
    order = []
    while heap:
        node = heapq.heappop(heap)
        order.append(node)
        for nei in graph[node]:
            indegree[nei] -= 1
            if indegree[nei] == 0:
                heapq.heappush(heap, nei)
    if len(order) == n:
        return order
    else:
        return []

第 3 名MoonshotAI Kimi K2.7 Code已解决 · 通过 11/11 个测试生成耗时 43 秒 · 执行耗时 917 毫秒 · $0.01

✓	course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])预期 [0, 1, 2, 3]	1 毫秒
✓	course_order(2, [[0, 1], [1, 0]])预期 []	<1 毫秒
✓	course_order(3, [])预期 [0, 1, 2]	<1 毫秒

✓ 2 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 908 毫秒

代码

import heapq

def course_order(n: int, prereqs: list[list[int]]) -> list[int]:
    adj = [[] for _ in range(n)]
    indeg = [0] * n
    seen = set()

    for a, b in prereqs:
        if not (0 <= a < n and 0 <= b < n):
            continue
        key = (b, a)
        if key in seen:
            continue
        seen.add(key)
        adj[b].append(a)
        indeg[a] += 1

    heap = [i for i, d in enumerate(indeg) if d == 0]
    heapq.heapify(heap)

    order = []
    while heap:
        u = heapq.heappop(heap)
        order.append(u)
        for v in adj[u]:
            indeg[v] -= 1
            if indeg[v] == 0:
                heapq.heappush(heap, v)

    return order if len(order) == n else []

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