207. Course Schedule -- 判断图是否有环

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本文深入探讨了课程安排问题的算法解决方案，通过图论中的有向图表示课程之间的先修关系，利用拓扑排序判断是否存在有效的课程完成顺序。文章详细解释了如何通过记录每个节点的入度和遍历过程来检测环的存在，从而确定所有课程是否可以顺利完成。

207. Course Schedule

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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.


class Solution {
public:
	bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
		int x, y;
		vector<int> cur;
		vector<bool>visited(numCourses, false);
		vector<vector<int> > v(numCourses, cur);
		vector<int> In_degree(numCourses, 0);
		for (int i = 0; i < prerequisites.size(); i++)
		{
			v[prerequisites[i][1]].push_back(prerequisites[i][0]);
			In_degree[prerequisites[i][0]]++;
		}
		int cnt = 0;
		//找入度 = 0 的节点
		bool has_0 = false;
		while (true) {
			has_0 = false;
			for (int i = 0; i < numCourses; i++)
			{
				if (!visited[i] && In_degree[i] == 0) {
					has_0 = true;
					visited[i] = true;
					cnt++;
					for (int j = 0; j < v[i].size(); j++)
					{
						In_degree[v[i][j]]--;
					}

				}
			}
			if (!has_0&&cnt<numCourses) {
				return false;
			}
			if (has_0&&cnt == numCourses)
				break;
		}

		return true;
	}
};