#week19
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：
这一题在之前有用拓扑排序做过
然后再DFS里面也看到了它
之前也分析过，实际上就是判断是否有环
所以DFS也可以进行判断是否有环
题解：
1 class Solution { 2 public: 3 bool DFS(vector<unordered_set<int>> &matrix, unordered_set<int> &visited, int idx, vector<bool> &flag) { 4 flag[idx] = true; // 标记该结点访问过 5 visited.insert(idx); 6 7 // 找出该结点的所有邻居结点，如果存在访问过的结点或者递归，则返回true 8 for (auto it = matrix[idx].begin(); it != matrix[idx].end(); ++it) { 9 if (visited.find(*it) != visited.end() || DFS(matrix, visited, *it, flag)) { 10 return true; 11 } 12 } 13 14 visited.erase(idx); 15 return false; 16 } 17 bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) { 18 vector<unordered_set<int>> matrix(numCourses); 19 // 要想完成第一门课程，则先完成第二门课程(后面的是先要完成的课程) 20 // 构建图 21 for (int i = 0; i < prerequisites.size(); ++i) { 22 matrix[prerequisites[i].second].insert(prerequisites[i].first); 23 } 24 25 unordered_set<int> visited; // 记录一个递归访问过的结点 26 vector<bool> flag(numCourses, false); // 记录是否访问过结点 27 28 /** 29 * 遍历所有课程，也就是结点 30 * 判断是否标记过结点，如果没有则进行DFS判断是否存在回路，存在回路则返回false 31 */ 32 for (int i = 0; i < numCourses; ++i) { 33 if (!flag[i]) 34 // 如果递归中存在访问过的结点，则该拓扑排序是不存在的，也就无法完成课程 35 if (DFS(matrix, visited, i, flag)) 36 return false; 37 } 38 return true; 39 } 40 };
