题目描述：

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:

1
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.

1
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: The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

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.
  • 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.

Hint中已经说的很明确了，就是判断一个图中有没有环。我用DFS来遍历每个节点，有两个出现环的情况：

  1. 出现了路径中已经出现的节点
  2. 有的节点没有被遍历到过，说明它们没有起始的端点
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class Solution {
public:
	bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
		vector<vector<int>> gg(numCourses);
		vector<int> prerequested(numCourses, false), allVisited(numCourses, false);
		for (int i = 0; i < prerequisites.size(); i++) {
			gg[prerequisites[i].first].push_back(prerequisites[i].second);
			prerequested[prerequisites[i].second] = true;
		}
		int prerequestedNum = 0;
		for (int i = 0; i < numCourses; i++) {
			if (!prerequested[i]) {
				vector<int> thisTimeVisited(numCourses, false);
				thisTimeVisited[i] = true;
				allVisited[i] = true;
				if (!DFS(gg, i, allVisited, thisTimeVisited)){
					return false;
				}
				thisTimeVisited[i] = false;
			}
			else {
				prerequestedNum++;
			}
		}
		if (prerequestedNum == numCourses) {
			//no start node
			return false;
		}
		for (int i = 0; i < numCourses; i++) {
			if (!allVisited[i]) {
				//some nodes cannot be visited
				return false;
			}
		}
		return true;
	}

	bool DFS(vector<vector<int>> &gg, int node, vector<int> &allVisited, vector<int> &thisTimeVisited) {
		vector<int> nextNodes = gg[node];
		for (int i = 0; i < nextNodes.size(); i++) {
			if (thisTimeVisited[nextNodes[i]]) {
				//find cycle
				return false;
			}
			allVisited[nextNodes[i]] = true;
			thisTimeVisited[nextNodes[i]] = true;
			if (!DFS(gg, nextNodes[i], allVisited, thisTimeVisited))
				return false;
			thisTimeVisited[nextNodes[i]] = false;
		}
		return true;
	}
};