207 Course Schedule （课程清单）

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:

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 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. 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.

这道课程清单的问题对于我们学生来说应该不陌生，因为我们在选课的时候经常会遇到想选某一门课程，发现选它之前必须先上了哪些课程，这道题给了很多提示，第一条就告诉我们了这道题的本质就是在有向图中检测环。 LeetCode中关于图的题很少，有向图的仅此一道，还有一道关于无向图的题是 Clone Graph 无向图的复制。个人认为图这种数据结构相比于树啊，链表啊什么的要更为复杂一些，尤其是有向图，很麻烦。第二条提示是在讲如何来表示一个有向图，可以用边来表示，边是由两个端点组成的，用两个点来表示边。第三第四条提示揭示了此题有两种解法，DFS和BFS都可以解此题。我们先来看BFS的解法，我们定义二维数组graph来表示这个有向图，一位数组in来表示每个顶点的入度。我们开始先根据输入来建立这个有向图，并将入度数组也初始化好。然后我们定义一个queue变量，将所有入度为0的点放入队列中，然后开始遍历队列，从graph里遍历其连接的点，每到达一个新节点，将其入度减一，如果此时该点入度为0，则放入队列末尾。直到遍历完队列中所有的值，若此时还有节点的入度不为0，则说明环存在，返回false，反之则返回true。代码如下：

class Solution {public:    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {        vector<vector<int>> graph(numCourses, vector<int>(0));        vector<int> in(numCourses, 0);        for(auto a : prerequisites) {            graph[a.second].push_back(a.first);            ++in[a.first];        }        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;    }};

下面我们来看DFS的解法，也需要建立有向图，还是用二维数组来建立，和BFS不同的是，我们像现在需要一个一维数组visit来记录访问状态，大体思路是，先建立好有向图，然后从第一个门课开始，找其可构成哪门课，暂时将当前课程标记为已访问，然后对新得到的课程调用DFS递归，直到出现新的课程已经访问过了，则返回false，没有冲突的话返回true，然后把标记为已访问的课程改为未访问。代码如下：

class Solution {public:    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {        vector<vector<int>> graph(numCourses, vector<int>(0));        vector<int> vis(numCourses, 0);        for(auto a : prerequisites) {            graph[a.second].push_back(a.first);        }        for(int i = 0; i < numCourses; ++i) {            if(!helper(graph, vis, i)) return false;        }        return true;    }    bool helper(vector<vector<int>> &graph, vector<int> vis, int i) { //DFS提交超时        if(vis[i] == -1) return false;        if(vis[i] == 1 ) return true;        vis[i] = -1;        for(auto a : graph[i]) {            if(!helper(graph, vis, a)) return false;        }        vis[i] = 1;        return true;    }};