LeetCode Course Schedule(Topological Sorting)

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.

解法一(BFS):

class Solution {
public:
    bool canFinish(int numCourses, vector>& prerequisites) {
        vector> graph = make_graph(numCourses, prerequisites);
        vector indegree = compute_indegree(graph);
        for(int i = 0; i> make_graph(int numCourses, vector>& prerequisites) {
        vector> graph(numCourses);
        for(auto pre: prerequisites)
            graph[pre.first].insert(pre.second);
        return graph;
    }
    vector compute_indegree(vector> graph) {
        vector indegree(graph.size(), 0);
        for(auto neighbors: graph)
            for(int neigh: neighbors)
                indegree[neigh]++;
        return indegree;
    }
};

入度为0的结点也可以用栈或队列维护，用一个计数变量判断到底有多少个结点可以出队。

解法二(DFS):

class Solution {
public:
    bool canFinish(int numCourses, vector>& prerequisites) {
        vector> graph = make_graph(numCourses, prerequisites);
        vector visited(numCourses, false), onpath(numCourses, false);
        for(int i = 0; i> make_graph(int numCourses, vector>& prerequisites) {
        vector> graph(numCourses);
        for(auto pre: prerequisites)
            graph[pre.first].insert(pre.second);
        return graph;
    }
    bool dfs_cycle(vector>& graph, int node, vector& visited, vector& onpath) {
        if(visited[node]) return false;
        onpath[node] = visited[node] = true;
        for(int neigh: graph[node])
            if(onpath[neigh] || dfs_cycle(graph, neigh, visited, onpath))
                return true;
        onpath[node] = false;
        return false;
    }
};

维护了两个标志数组，一个是全局的，一个是本路径的。

详细解析参考：
https://leetcode.com/problems/course-schedule/discuss/58509/18-22-lines-C++-BFSDFS-Solutions

附：
拓扑排序的算法（DFS）:
https://www.geeksforgeeks.org/topological-sorting/