原文链接: https://leetcode-cn.com/problems/course-schedule-iv

英文原文

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course ai first if you want to take course bi.

• For example, the pair [0, 1] indicates that you have to take course 0 before you can take course 1.

Prerequisites can also be indirect. If course a is a prerequisite of course b, and course b is a prerequisite of course c, then course a is a prerequisite of course c.

You are also given an array queries where queries[j] = [uj, vj]. For the jth query, you should answer whether course uj is a prerequisite of course vj or not.

Return a boolean array answer, where answer[j] is the answer to the jth query.

 

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]], queries = [[0,1],[1,0]]
Output: [false,true]
Explanation: The pair [1, 0] indicates that you have to take course 1 before you can take course 0.
Course 0 is not a prerequisite of course 1, but the opposite is true.

Example 2:

Input: numCourses = 2, prerequisites = [], queries = [[1,0],[0,1]]
Output: [false,false]
Explanation: There are no prerequisites, and each course is independent.

Example 3:

Input: numCourses = 3, prerequisites = [[1,2],[1,0],[2,0]], queries = [[1,0],[1,2]]
Output: [true,true]

 

Constraints:

• 2 <= numCourses <= 100
• 0 <= prerequisites.length <= (numCourses * (numCourses - 1) / 2)
• prerequisites[i].length == 2
• 0 <= ai, bi <= n - 1
• ai != bi
• All the pairs [ai, bi] are unique.
• The prerequisites graph has no cycles.
• 1 <= queries.length <= 104
• 0 <= ui, vi <= n - 1
• ui != vi

中文题目

你总共需要上 n 门课，课程编号依次为 0 到 n-1 。

有的课会有直接的先修课程，比如如果想上课程 0 ，你必须先上课程 1 ，那么会以 [1,0] 数对的形式给出先修课程数对。

给你课程总数 n 和一个直接先修课程数对列表 prerequisite 和一个查询对列表 queries 。

对于每个查询对 queries[i] ，请判断 queries[i][0] 是否是 queries[i][1] 的先修课程。

请返回一个布尔值列表，列表中每个元素依次分别对应 queries 每个查询对的判断结果。

注意：如果课程 a 是课程 b 的先修课程且课程 b 是课程 c 的先修课程，那么课程 a 也是课程 c 的先修课程。

 

示例 1：

输入：n = 2, prerequisites = [[1,0]], queries = [[0,1],[1,0]]
输出：[false,true]
解释：课程 0 不是课程 1 的先修课程，但课程 1 是课程 0 的先修课程。

示例 2：

输入：n = 2, prerequisites = [], queries = [[1,0],[0,1]]
输出：[false,false]
解释：没有先修课程对，所以每门课程之间是独立的。

示例 3：

输入：n = 3, prerequisites = [[1,2],[1,0],[2,0]], queries = [[1,0],[1,2]]
输出：[true,true]

示例 4：

输入：n = 3, prerequisites = [[1,0],[2,0]], queries = [[0,1],[2,0]]
输出：[false,true]

示例 5：

输入：n = 5, prerequisites = [[0,1],[1,2],[2,3],[3,4]], queries = [[0,4],[4,0],[1,3],[3,0]]
输出：[true,false,true,false]

 

提示：

• 2 <= n <= 100
• 0 <= prerequisite.length <= (n * (n - 1) / 2)
• 0 <= prerequisite[i][0], prerequisite[i][1] < n
• prerequisite[i][0] != prerequisite[i][1]
• 先修课程图中没有环。
• 先修课程图中没有重复的边。
• 1 <= queries.length <= 10^4
• queries[i][0] != queries[i][1]

通过代码

高赞题解

解题思路

执行用时 :
952 ms, 在所有 Python3 提交中击败了100.00%的用户
内存消耗 :
15.6 MB, 在所有 Python3 提交中击败了100.00%的用户

借鉴了大神们的代码，其实就是打表，可惜一开始就想复杂了，还需努力呀！
只需要判断 二维表格中，i行j列是不是先导关系就好！

代码

class Solution:
    def checkIfPrerequisite(self, n: int, prerequisites: List[List[int]], queries: List[List[int]]) -> List[bool]:
        dp = [[False] * n for _ in range(n)]       
        for p, c in prerequisites:
            dp[p][c] = True

        for k in range(n):
            for i in range(n):
                for j in range(n):
                    if dp[i][k] and dp[k][j]:
                        dp[i][j] = True
        ans = []
        for i, j in queries:
            ans.append(dp[i][j])
        return ans

统计信息

通过次数	提交次数	AC比率
6803	15735	43.2%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言