【DFS】207. Course Schedule

#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:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. 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：

class Solution {
public:
    bool DFS(vector<unordered_set<int>> &matrix, unordered_set<int> &visited, int idx, vector<bool> &flag) {
        flag[idx] = true; // 标记该结点访问过
        visited.insert(idx);
        
        // 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
        for (auto it = matrix[idx].begin(); it != matrix[idx].end(); ++it) {
            if (visited.find(*it) != visited.end() || DFS(matrix, visited, *it, flag)) {
                return true;
            }
        }
        
        visited.erase(idx);
        return false;
    }
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<unordered_set<int>> matrix(numCourses);
        // 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)
        // 构建图
        for (int i = 0; i < prerequisites.size(); ++i) {
            matrix[prerequisites[i].second].insert(prerequisites[i].first);
        }
        
        unordered_set<int> visited; // 记录一个递归访问过的结点
        vector<bool> flag(numCourses, false); // 记录是否访问过结点
        
        /**
         *  遍历所有课程，也就是结点
         *  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
         */
        for (int i = 0; i < numCourses; ++i) {
            if (!flag[i])
                // 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
                if (DFS(matrix, visited, i, flag))
                    return false;
        }
        return true;
    }
};