LeetCode210. Course Schedule II-优快云博客

本文深入解析了课程调度问题，通过利用队列进行BFS拓扑排序，解决课程依赖关系，实现课程的合理安排。具体示例包括两门课程和四门课程的调度方案，展示了算法的应用过程。

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题目：

https://leetcode.com/problems/course-schedule-ii/

There are a total of n courses you have to take, labeled from 0 to n - 1.

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.
For example:

2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

思路：

就是利用队列bfs进行拓扑排序。如果能拓扑排序，输出一种合法的拓扑排序结果。

代码：

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<vector<int>> graph(numCourses,vector<int>());
        vector<int> inDegree(numCourses,0);
        vector<int> res;
        int len=prerequisites.size();
        for(int i=0;i<len;i++){
            graph[prerequisites[i].second].push_back(prerequisites[i].first);
            inDegree[prerequisites[i].first]++;
        }
        queue<int> q;
        for(int i=0;i<numCourses;i++){
            if(inDegree[i]==0){
                q.push(i);
            }
        }
        int cur;
        while(!q.empty()){
            cur=q.front();
            res.push_back(cur);
            q.pop();
            for(int i=0;i<graph[cur].size();i++){
                inDegree[graph[cur][i]]--;
                if(inDegree[graph[cur][i]]==0){
                    q.push(graph[cur][i]);
                }
            }
        }
        for(int i=0;i<numCourses;i++){
            if(inDegree[i]!=0){
                return vector<int>();
            }
        }
        return res;
    }
};