210. Course Schedule II

Description:

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, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

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 the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].

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.

click to show more hints.

Hints:

1. This problem is equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
2. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
3. Topological sort could also be done via BFS.

簡要題解：

判斷是否可以完成全部課程採用與文章207. Course Schedule相同的做法。

完成課程的順序由post（dfs中一個結點的後訪問時刻）決定。由於邊的方向與邏輯上方向相反，因此，按post值遞增的順序。

代碼：

class Solution {
private:
    bool* visited;
    int* post;
    int clock;
    vector<int> scheduling;
public:
    void postvisit(int v) {
        post[v] = clock++;
        scheduling.push_back(v);
    }

    void explore(int v, vector<pair<int, int>>& prerequisites) {
        visited[v] = true;

        for (int i = 0; i < prerequisites.size(); i++)
            if (prerequisites[i].first == v && !visited[prerequisites[i].second])
                explore(prerequisites[i].second, prerequisites);

        postvisit(v);
    }

    void dfs(int numCourses, vector<pair<int, int>>& prerequisites) {
        if (visited != NULL)
            delete[] visited;
        visited = new bool[numCourses];
        for (int i = 0; i < numCourses; i++)
            visited[i] = false;

        if (post != NULL)
            delete[] post;
        post = new int[numCourses];
        clock = 0;

        for (int i = 0; i < numCourses; i++)
            if (!visited[i])
                explore(i, prerequisites);
    }

    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        dfs(numCourses, prerequisites);

        for (int i = 0; i < prerequisites.size(); i++)
            if (post[prerequisites[i].first] < post[prerequisites[i].second])
                return false;

        if (visited != NULL)
            delete[] visited;
        if (post != NULL)
            delete[] post;

        return true;
    }

    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        scheduling.clear();

        if (canFinish(numCourses, prerequisites))
            return scheduling;
        else
            return vector<int>();
    }
};