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.

1. This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
2. There are several ways to represent a graph. For example, the input prerequisites is a graph represented by a list of edges. Is this graph representation appropriate?
3. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
4. Topological sort could also be done via BFS.

給定n個課程，上課的順序有先後要求，用pair表示，判斷是否能完成所有課程。

對於每一對課程的順序關系，把它看做是一個有向邊，邊是由兩個端點組成的，用兩個點來表示邊，所有的課程關系即構成一個有向圖，問題相當於判斷有向圖中是否有環。判斷有向圖是否有環的方法是拓撲排序。

拓撲排序：維護一張表記錄所有點的入度，移出入度為0的點並更新其他點的入度，重復此過程直到沒有點的入度為0。如果原有向圖有環的話，此時會有剩余的點且其入度不為0；否則沒有剩余的點。

圖的拓撲排序可以DFS或者BFS。遍歷所有邊，計算點的入度；將入度為0的點移出點集，並更新剩余點的入度；重復步驟2，直至沒有剩余點或剩余點的入度均大於0。

這裏不能使用鄰接矩陣，應該使用鄰接表來存儲有向圖的信息。鄰接表可以使用結構體來實現，每個結構體存儲一個值以及一個指向下一個節點的指針，同時維護一個存儲多個頭結點的數組即可。除此之外，在數據結構簡單的情況下，還可以使用數組來模擬簡單的鄰接表。

Java：BFS

public class Solution {  
    public boolean canFinish(int numCourses, int[][] prerequisites) {  
        int[] pre = new int[numCourses];  
        List<Integer>[] satisfies = new List[numCourses];  
        for(int i=0; i<numCourses; i++) satisfies[i] = new ArrayList<>();  
        for(int i=0; i<prerequisites.length; i++) {  
            satisfies[prerequisites[i][1]].add(prerequisites[i][0]);  
            pre[prerequisites[i][0]] ++;  
        }  
        int finish = 0;  
        LinkedList<Integer> queue = new LinkedList<>();  
        for(int i=0; i<numCourses; i++) {  
            if (pre[i] == 0) queue.add(i);  
        }  
        while (!queue.isEmpty()) {  
            int course = queue.remove();  
            finish ++;  
            if (satisfies[course] == null) continue;  
            for(int c: satisfies[course]) {  
                pre[c] --;  
                if (pre[c] == 0) queue.add(c);  
            }  
        }  
        return finish == numCourses;  
    }  
} 

Java：DFS

public class Solution {  
    private boolean[] canFinish;  
    private boolean[] visited;  
    private List<Integer>[] depends;  
    private boolean canFinish(int course) {  
        if (visited[course]) return canFinish[course];  
        visited[course] = true;  
        for(int c: depends[course]) {  
            if (!canFinish(c)) return false;  
        }  
        canFinish[course] = true;  
        return canFinish[course];  
    }  
    public boolean canFinish(int numCourses, int[][] prerequisites) {  
        canFinish = new boolean[numCourses];  
        visited = new boolean[numCourses];  
        depends = new List[numCourses];  
        for(int i=0; i<numCourses; i++) depends[i] = new ArrayList<Integer>();   
        for(int i=0; i<prerequisites.length; i++) {  
            depends[prerequisites[i][0]].add(prerequisites[i][1]);  
        }  
        for(int i=0; i<numCourses; i++) {  
            if (!canFinish(i)) return false;  
        }  
        return true;  
    }  
}  

Python：

import collections

class Solution(object):
    def canFinish(self, numCourses, prerequisites):
        """
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: bool
        """
        zero_in_degree_queue, in_degree, out_degree = collections.deque(), {}, {}
        
        for i, j in prerequisites:
            if i not in in_degree:
                in_degree[i] = set()
            if j not in out_degree:
                out_degree[j] = set()
            in_degree[i].add(j)
            out_degree[j].add(i)
        
        for i in xrange(numCourses):
            if i not in in_degree:
                zero_in_degree_queue.append(i)
        
        while zero_in_degree_queue:
            prerequisite = zero_in_degree_queue.popleft()
            
            if prerequisite in out_degree:
                for course in out_degree[prerequisite]:
                    in_degree[course].discard(prerequisite)
                    if not in_degree[course]:
                        zero_in_degree_queue.append(course)
            
                del out_degree[prerequisite]
        
        if out_degree:
            return False
        
        return True

C++: BFS

class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int> > graph(numCourses, vector<int>(0));
        vector<int> in(numCourses, 0);
        for (auto a : prerequisites) {
            graph[a[1]].push_back(a[0]);
            ++in[a[0]];
        }
        queue<int> q;
        for (int i = 0; i < numCourses; ++i) {
            if (in[i] == 0) q.push(i);
        }
        while (!q.empty()) {
            int t = q.front();
            q.pop();
            for (auto a : graph[t]) {
                --in[a];
                if (in[a] == 0) q.push(a);
            }
        }
        for (int i = 0; i < numCourses; ++i) {
            if (in[i] != 0) return false;
        }
        return true;
    }
};

C++: DFS

class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int> >& prerequisites) {
        vector<vector<int> > graph(numCourses, vector<int>(0));
        vector<int> visit(numCourses, 0);
        for (auto a : prerequisites) {
            graph[a[1]].push_back(a[0]);
        }
        for (int i = 0; i < numCourses; ++i) {
            if (!canFinishDFS(graph, visit, i)) return false;
        }
        return true;
    }
    bool canFinishDFS(vector<vector<int> > &graph, vector<int> &visit, int i) {
        if (visit[i] == -1) return false;
        if (visit[i] == 1) return true;
        visit[i] = -1;
        for (auto a : graph[i]) {
            if (!canFinishDFS(graph, visit, a)) return false;
        }
        visit[i] = 1;
        return true;
    }
};

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