207. Course Schedule

#week17

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.

分析

这一题，很明显，可以看是否能够得到一个拓扑序列，就可以看能够完成

其实也就是检验是否是一个有环的图，如果有环就没法完成课程

如果无环则可以

所以检验有环无环，就可以通过求拓扑或者DFS等方式完成

题解

这个是摘要别人的BFS代码，我这个写的有点多有点丑：

1 class Solution {
2 public:
3     bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
4         vector<unordered_set<int>> graph = make_graph(numCourses, prerequisites);
5         vector<int> degrees = compute_indegree(graph);
6         for (int i = 0; i < numCourses; i++) {
7             int j = 0;
8             for (; j < numCourses; j++)
9                 if (!degrees[j]) break;
10             if (j == numCourses) return false;
11             degrees[j] = -1;
12             for (int neigh : graph[j])
13                 degrees[neigh]--;
14         }
15         return true;
16     }
17 private:
18     vector<unordered_set<int>> make_graph(int numCourses, vector<pair<int, int>>& prerequisites) {
19         vector<unordered_set<int>> graph(numCourses);
20         for (auto pre : prerequisites)
21             graph[pre.second].insert(pre.first);
22         return graph;
23     }
24     vector<int> compute_indegree(vector<unordered_set<int>>& graph) {
25         vector<int> degrees(graph.size(), 0);
26         for (auto neighbors : graph)
27             for (int neigh : neighbors)
28                 degrees[neigh]++;
29         return degrees;
30     }
31 };