Course Schedule

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.

Analyse: Using BFS. 

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        if(prerequisites.empty()) return true;
        
        int n = prerequisites.size();
        vector<vector<bool> > graph(numCourses, vector<bool>(numCourses, false));
        vector<int> indegree(numCourses, 0);
        for(int i = 0; i < n; i++){
            if(graph[prerequisites[i].second][prerequisites[i].first] == false){//avoid duplicate pairs
                graph[prerequisites[i].second][prerequisites[i].first] = true;
                indegree[prerequisites[i].first]++;
            }
        }
        stack<int> stk;
        int count = 0;
        for(int i = 0; i < numCourses; i++){
            if(indegree[i] == 0)
                stk.push(i);
        }
        while(!stk.empty()){
            int source = stk.top();
            stk.pop();
            count++;
            for(int i = 0; i < numCourses; i++){
                if(graph[source][i]){
                    indegree[i]--;
                    if(indegree[i] == 0)
                        stk.push(i);
                }
            }
        }
        return count == numCourses;
    }
    //1. construct a graph and compute each node's indegree
    //2. push all nodes with 0 indegree to a stack
    //3. pop out the first node, decrease its neighbour's degree by 1
    //   push all nodes with 0 indegree to a stack and continue
};