LeetCode-Course Schedule-解题报告

最新推荐文章于 2022-11-01 19:09:45 发布

原创最新推荐文章于 2022-11-01 19:09:45 发布 · 337 阅读

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本文针对课程安排问题进行了详细的解析，并通过使用深度优先搜索（DFS）算法实现了一个解决方案。该问题探讨了在存在先修课程约束的情况下，是否有可能完成所有课程的学习。

原题链接 https://leetcode.com/problems/course-schedule/

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.

其实就是拓扑排序。

我使用的dfs，当然我用一个数组保存了每个节点的入度。每次输出入度为0的节点。

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<vector<int> >graph(numCourses);
		vector<int>out(numCourses, 0);
	    int num = 0;
		vector<bool>vis(numCourses, 0);
		for (auto &p : prerequisites)
			graph[p.second].push_back(p.first), out[p.first]++;
		dfs(graph, num, vis, out);
		return num == numCourses;
	}
	void dfs(vector<vector<int> >& g, int& num, vector<bool>& vis, vector<int>& out)
	{
		if (num == g.size())return;
		for (int i = 0; i < g.size(); ++i)
		{
			if (!vis[i] && out[i] == 0)
			{
				num++;
				vis[i] = true;
				for (int j = 0; j < g[i].size(); ++j)
					--out[g[i][j]];
				dfs(g, num, vis, out);
			}
		}
	}
};