LeetCode解题报告（484）-- 1462. Course Schedule IV

Problem

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:

1
2
3

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

Example 3:

1 2	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

Analysis

这是课程表系列题的第4题，背景还是一样的，不同的是prerequisite的关系是有传递性的，同时题目还给出了queries，每个query[i, j]查询i是否j的prerequisite。对于这种query类型的题目，我们的处理思路一般是求解出所有的答案，然后直接返回，一般是求解出一个矩阵便于快速访问。这里就需要求解出isReach[i][j]代表i是否j的prerequisite。我们以每个节点作为BFS的起点都走一遍，就能构造出isReach了。

Solution

无。

Code

class Solution {
public:
    vector<bool> checkIfPrerequisite(int numCourses, vector<vector<int>>& prerequisites, vector<vector<int>>& queries) {
        // build graph
        vector<vector<int>> graph(numCourses);
        
        for (vector<int> &pre: prerequisites) {
            graph[pre[0]].push_back(pre[1]);
        }
        
        vector<vector<bool>> isReach(numCourses, vector<bool>(numCourses, false));
        
        for (int i = 0; i < numCourses; ++i) {
            queue<int> q;
            q.push(i);
            
            while (!q.empty()) {
                int size = q.size();
                for (int k = 0; k < size; ++k) {
                    int cur = q.front();
                    q.pop();
                    
                    for (int next: graph[cur]) {
                        if (isReach[i][next]) {
                            continue;
                        }
                        isReach[i][next] = true;
                        q.push(next);
                    }
                }
            }
        }
        
        vector<bool> result;
        for (vector<int> &query: queries) {
            result.push_back(isReach[query[0]][query[1]]);
        }
        return result;
    }
};

Summary

这道题目就是多次BFS，没有特别的难点，主要是根据queries这种题目特点确定出要计算出一个isReach二维数组。这道题目的分享到这里，感谢你的支持！