There are a total of n courses you have to take, labeled from 0 to n-1.
Some courses may have direct prerequisites, for example, to take course 0 you have first to take course 1, which is expressed as a pair: [1,0]
Given the total number of courses n, a list of direct prerequisite pairs and a list of queries pairs.
You should answer for each queries[i] whether the course queries[i][0] is a prerequisite of the course queries[i][1] or not.
Return a list of boolean, the answers to the given queries.
Please note that 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.
Example 1:
Input: n = 2, prerequisites = [[1,0]], queries = [[0,1],[1,0]] Output: [false,true] Explanation: course 0 is not a prerequisite of course 1 but the opposite is true.Example 2:
Input: n = 2, prerequisites = [], queries = [[1,0],[0,1]] Output: [false,false] Explanation: There are no prerequisites and each course is independent.Example 3:
Input: n = 3, prerequisites = [[1,2],[1,0],[2,0]], queries = [[1,0],[1,2]] Output: [true,true]Example 4:
Input: n = 3, prerequisites = [[1,0],[2,0]], queries = [[0,1],[2,0]] Output: [false,true]Example 5:
Input: n = 5, prerequisites = [[0,1],[1,2],[2,3],[3,4]], queries = [[0,4],[4,0],[1,3],[3,0]] Output: [true,false,true,false]
Constraints:
2 <= n <= 1000 <= prerequisite.length <= (n * (n - 1) / 2)0 <= prerequisite[i][0], prerequisite[i][1] < nprerequisite[i][0] != prerequisite[i][1]- The prerequisites graph has no cycles.
- The prerequisites graph has no repeated edges.
1 <= queries.length <= 10^4queries[i][0] != queries[i][1]
课程安排IV。题意是一共有 N 门课程需要修，他们的编号是从0到 N - 1。其中有一些课程是有直接的先修课程prerequisites的，会以数对形式表示。现在给你课程总数 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。
这道题我给出一个比较朴素的解法，还是涉及到拓扑排序。对于这道题，你需要一个入度表 indegree 记录课程的入度和两个hashmap，一个 adj 哈希表用来记录课程之间的邻接关系；一个 preMap 哈希表用来记录课程之间的先修关系。首先对于每一个课程 i ，我们需要为这个课程在邻接表和先修表里创建一个hashset以记录课程之间的关系。
接着遍历直接先修课程数对列表 prerequisite，记录课程的入度，同时也记录各个课程的邻接关系。接着还是跟之前的课程安排的题一样，把入度为0的课程加入queue，意思是从不需要先修课程的课开始找起，去创建课程之间的先修关系。从queue弹出一个课程cur的时候，从邻接哈希表 adj 拿到cur的所有邻接课程，同时因为cur的入度为0所以对于每一个 next 课程而言，cur课程一定是next的先修课程，所以我们在 preMap 里找到 next ，并且把 cur 加入 next 的先修课程里。同时如果 cur 自己也有先修课程，也需要把他们统统加入 next 的先修课程的hashset里。
最后对于需要判断的 queries，因为queries[0]是先修课程，queries[1]是后继课程，所以判断的方式是看看 preMap 里key为queries[0]的hashset是否包含queries[1]。
时间O(n^2)
空间O(n)
Java实现
1 class Solution { 2 public List<Boolean> checkIfPrerequisite(int n, int[][] prerequisites, int[][] queries) { 3 int[] indegree = new int[n]; 4 HashMap<Integer, Set<Integer>> adj = new HashMap<>(); 5 HashMap<Integer, Set<Integer>> preMap = new HashMap<>(); 6 for (int i = 0; i < n; i++) { 7 preMap.put(i, new HashSet<>()); 8 adj.put(i, new HashSet<>()); 9 } 10 for (int[] pre : prerequisites) { 11 indegree[pre[1]]++; 12 adj.get(pre[0]).add(pre[1]); 13 } 14 Queue<Integer> queue = new LinkedList<>(); 15 for (int i = 0; i < n; i++) { 16 if (indegree[i] == 0) { 17 queue.offer(i); 18 } 19 } 20 while (!queue.isEmpty()) { 21 int cur = queue.poll(); 22 Set<Integer> set = adj.get(cur); 23 for (int next : set) { 24 preMap.get(next).add(cur); 25 preMap.get(next).addAll(preMap.get(cur)); 26 indegree[next]--; 27 if (indegree[next] == 0) { 28 queue.offer(next); 29 } 30 } 31 } 32 List<Boolean> res = new ArrayList<>(); 33 for (int[] pair : queries) { 34 Set<Integer> set = preMap.get(pair[1]); 35 if (set.contains(pair[0])) { 36 res.add(true); 37 } else { 38 res.add(false); 39 } 40 } 41 return res; 42 } 43 }
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