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.

拓撲排序。用一個佇列存入度為0的節點，依次出隊，將與出隊節點相連的節點的入度減1，如果入度減為0，將其放入佇列中，直到佇列為空。如裡最後還有入度不為0的節點的話，說明有環，否則無環。

 1 class Solution {
 2 public:
 3     bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
 4         vector<vector<int>> graph(numCourses, vector<int>(0
 
));
 5         vector<int> inDegree(numCourses, 0);
 6         for (auto u : prerequisites) {
 7             graph[u[1]].push_back(u[0]);
 8             ++inDegree[u[0]];
 9         }
10         queue<int> que;
11         for (int i = 0; i < numCourses; ++i) {
12             if (inDegree[i] == 0) que.push(i);
13         }
14         while (!que.empty()) {
15             int u = que.front();
16             que.pop();
17             for (auto v : graph[u]) {
18                 --inDegree[v];
19                 if (inDegree[v] == 0) que.push(v);
20             }
21         }
22         for (int i = 0; i < numCourses; ++i) {
23             if (inDegree[i] != 0) return false;
24         }
25         return true;
26     }
27 };