[LeetCode] 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?

Example 1:

Input: 2, [[1,0]] 
Output: true
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0. So it is possible.

Example 2:

Input: 2, [[1,0],[0,1]]
Output: false
Explanation: 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.
You may assume that there are no duplicate edges in the input prerequisites.

Hints:

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.
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?
Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
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。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。先来看 BFS 的解法，定义二维数组 graph 来表示这个有向图，一维数组 in 来表示每个顶点的入度。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：

解法一：

class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> graph(numCourses, vector<int>());
        vector<int> in(numCourses);
        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;
    }
};

下面来看 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，像现在需要一个一维数组 visit 来记录访问状态，这里有三种状态，0表示还未访问过，1表示已经访问了，-1 表示有冲突。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：

解法二：

class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> graph(numCourses, vector<int>());
        vector<int> visit(numCourses);
        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;
    }
};