中山大学算法课程题目详解（第四周）

问题描述：

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.

问题解决：

根据老师上课讲过的DFS算法，任意选择一个course，也就是图的任意一个点，然后进行dfs遍历，在遍历过程中如果发现有回边，那么则存在环，返回false，否则，返回true。具体的做法是，给每一个点设置一个状态，状态包括三种：未访问，正在访问，以及已访问。在dfs遍历过程中，如果发现一个正在访问的点访问另外一个访问的点，则可以判断图中存在环，则可返回false。

具体代码如下：

class Solution {public:Solution() {}vector<vector<int> > graph;vector<int> visit;bool dfs(int u) {visit[u] = 1;for (auto v : graph[u]) {if (visit[v] == 1) {return false;}if (dfs(v) == false) {return false;}}visit[u] = 2;return true;}bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {if (numCourses == 0 || prerequisites.empty()) {return true;}graph = vector<vector<int> >(numCourses);visit = vector<int>(numCourses, 0);for (auto i : prerequisites) {graph[i.second].push_back(i.first);}for (int u = 0; u < numCourses; u++) {if (visit[u] == 0 && !dfs(u)) {return false;}}return true;}};