LeetCode Course Schedule

最新推荐文章于 2024-01-21 17:16:08 发布

原创最新推荐文章于 2024-01-21 17:16:08 发布 · 1.1k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LeetCode Course Sche #LeetCode #Course Schedule

LeetCode 同时被 2 个专栏收录

44 篇文章

订阅专栏

LeetCode 算法实战

7 篇文章

订阅专栏

本文探讨了LeetCode上的一道经典图论题目——课程安排问题。通过两种方法——BFS（拓扑排序）和DFS（深搜判断是否存在环）来解决课程是否有可行的学习顺序的问题。

LeetCode 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.

此题用基本的图论搜索方法可以解决（BFS或者DFS），判断是否存在环。

解法一：BFS（拓扑排序）

bool canFinish(int numCourses, vector<pair<int, int> >& prerequisites)
{
	vector<set<int> > matrix(numCourses);//用矩阵(二维数组)的方式保存图
	for (int i = 0; i < prerequisites.size(); i++)
		matrix[prerequisites[i].second].insert(prerequisites[i].first);

	vector<int> degree(numCourses, 0);
	for (int i = 0; i < numCourses; i++)
	{
		for (set<int>::iterator it = matrix[i].begin(); it != matrix[i].end(); ++it)
		{
			++degree[*it];//入度增一
		}
	}

	for (int i = 0; i < numCourses; i++)
	{
		int j;
		for (j = 0; j < numCourses && degree[j] != 0; j++){}//找出入度为0的顶点

		if (j == numCourses)	//没有入度为0的点，则不能完成
			return false;
		
		degree[j] = -1;
		//将所有以该顶点为起始顶点的入度减一
		for (set<int>::iterator it = matrix[j].begin(); it != matrix[j].end(); ++it)
		{
			--degree[*it];
		}
	}

	return true;
}

解法二：DFS（深搜判断是否存在环）

bool DFS(vector<set<int> >& matrix, set<int>& visited, int v, vector<bool>& flag)
{
	flag[v] = true;//访问标记设为true
	visited.insert(v);//将顶点v插入已访问的顶点集合
	
	for (set<int>::iterator it = matrix[v].begin(); it != matrix[v].end(); ++it)
	{
		if (visited.find(*it) != visited.end() || DFS(matrix, visited, *it, flag))//如果已经访问过，则存在环
		{
			return true;
		}
	}
	visited.erase(v);//回溯
	return false;
}

bool canFinish(int numCourses, vector<pair<int, int> >& prerequisites) 
{
	vector<set<int> > matrix(numCourses);//用矩阵(二维数组)的方式保存图
	for (int i = 0; i < prerequisites.size(); i++)
		matrix[prerequisites[i].second].insert(prerequisites[i].first);

	set<int> visited;//已访问的顶点组成的集合
	vector<bool> flag(numCourses, false);//顶点的访问标记，false表示未访问
	for (int i = 0; i < numCourses; i++)
	{
		if (!flag[i])//该顶点为访问，则以该顶点为起始点进行深搜
		{
			if (DFS(matrix, visited, i, flag))//如果遍历到相同的顶点，则存在环，返回false
				return false;
		}
	}
	return true;
}