> For the complete documentation index, see [llms.txt](https://hao-fu-1.gitbook.io/oj/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hao-fu-1.gitbook.io/oj/search/topological-sort/course-schedule.md).

# Course Schedule

<https://leetcode.com/problems/course-schedule/description/>

> 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, is it possible for you to finish all courses?
>
> 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 it is possible.
>
> 2, \[\[1,0],\[0,1]]
>
> There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.
>
> Note:
>
> The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
>
> You may assume that there are no duplicate edges in the input prerequisites.

## Thoughts

看成是有向图, 每个课是一个节点, 每条边由prerequisite -> ready两个节点相连. 依次从入度为0的结点开始遍历, 每遍历完一个节点, 把它从图中删去, 相当于把以它作为入边的节点的入度减1, 当一个节点入度为0时代表它可以被上(遍历)了.

## Code

```
class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<unordered_set<int>> edges(numCourses);
        vector<int> in_degrees(numCourses, 0);
        for (const auto pre : prerequisites) {
            edges[pre.second].insert(pre.first);
            ++in_degrees[pre.first];
        }

        queue<int> q;
        for (int i = 0; i < numCourses; ++i) {
            if (in_degrees[i] == 0) q.push(i);
        }

        int count = 0;
        while (!q.empty()) {
            const auto& u = q.front();
            q.pop();
            ++count;
            for (const auto v : edges[u]) {
                if (--in_degrees[v] == 0) {
                    q.push(v);
                }
            }
        }

        return count == numCourses;
    }
};
```

## Analysis

时间复杂度为边+节点数, O(M+N).