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

DFS, detect cycle. For each node, we have 3 states during DFS: 1) un-visited, 2) visiting, 3) visited.

class Solution {public:    bool canFinish(int n, vector
     
      
       >& p) {        vector
       
         visited(n, -1);     // un-visited        vector
        
         
          > v(n, set
          
           ()); for (int i = 0; i < p.size(); i++) { v[p[i].second].insert(p[i].first); } for (int i = 0; i < n; i++) if (visited[i] == -1 && dfs(v, visited, i)) return false; return true; } bool dfs(vector
           
            
             >& v, vector
             
              & visited, int n) { visited[n] = 1; // visiting for (auto nb : v[n]) { if (visited[nb] == 1) return true; else if (visited[nb] == 2) continue; else { if (dfs(v, visited, nb)) return true; } } visited[n] = 2; // visited return false; }};
             
            
           
          
         
        
       
      
     

 is another way.