Determining whether there has circumferential drawing

1  // will constitute a Pre-relationship diagram, each number of the second number to the first number even edges.
2  // First of all penetration is zero point into the team, ready topological sorting.
3  // wide search process, the degree of the current node minus associated with node 1; if the new level of nodes 0, it is found into the team.
4  // Finally, if through all nodes, then meet the requirements; otherwise, there is a ring Pre relationship.
. 5  
. 6  // find whether there is a ring 
. 7  class Solution 
 . 8  {
 . 9  public :
 10      BOOL canFinish ( int numCourses, Vector <Vector < int >> & the Prerequisites) 
 . 11      {
 12 is          Vector <Vector < int >> Graph (numCourses);
 13 is         vector<int> in_degree(numCourses, 0);
14         for (int i = 0; i < prerequisites.size(); i++) 
15         {
16             in_degree[prerequisites[i][0]]++;
17             graph[prerequisites[i][1]].push_back(prerequisites[i][0]);
18         }
19 
20         queue<int> q;
21         vector<bool> vis(numCourses, false);
22 
23         for (int i = 0; i < numCourses; i++)
24             if (in_degree[i] == 0)
25                 q.push(i);
26         while (!q.empty()) 
27         {
28             int sta = q.front();
29             q.pop();
30             vis[sta] = true;
31             //有哪些邻边
32             for (int i = 0; i < graph[sta].size(); i++) 
33             {
34 is                  in_degree [Graph [STA] [I]] -; // penetration -1 
35                  IF (in_degree [Graph [STA] [I]] == 0 ) // into 0 if the degree of added queue 
36                      Q. Push (Graph [STA] [I]);
 37 [              }
 38 is          }
 39          
40          // 0-> l-> 2 
41 is          for ( int I = 0 ; I <numCourses; I ++ )
 42 is              IF (VIS [I] == to false )
 43 is                  return  to false ; // A ring 
44 is          return  to true ;// Acyclic 
45      }
 46 };