leetcode Course Schedule

最新推荐文章于 2024-01-21 17:16:08 发布

转载最新推荐文章于 2024-01-21 17:16:08 发布 · 42 阅读

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原文链接：http://www.cnblogs.com/GadyPu/p/5028150.html

本文探讨了如何通过拓扑排序解决课程计划问题，确保所有课程可以按顺序完成，包括实例解析与代码实现。

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题目连接

https://leetcode.com/problems/course-schedule/

Course Schedule

Description

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

拓扑排序。。

class Solution {
public:
	bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
		if (numCourses && prerequisites.empty()) return true;
		tot = 0;
		int num = 0, m = prerequisites.size();
		inq = new int[numCourses + 10];
		memset(inq, 0, sizeof(int)* (numCourses + 10));
		head = new int[numCourses + 10];
		memset(head, -1, sizeof(int)* (numCourses + 10));
		G = new edge[m + 10];
		for (int i = 0; i < m; i++) {
			int u = prerequisites[i].first, v = prerequisites[i].second;
			inq[v]++;
			add_edge(u, v);
		}
		queue<int> q;
		for (int i = 0; i < numCourses; i++) { if (!inq[i])  q.push(i); }
		while (!q.empty()) {
			num++;
			int u = q.front(); q.pop();
			for (int i = head[u]; ~i; i = G[i].next) {
				if (--inq[G[i].to] == 0) q.push(G[i].to);
			}
		}
		delete []G; delete []inq, delete []head;
		return num == numCourses;
	}
private:
	int tot, *inq, *head;
	struct edge { int to, next; }*G;
	inline void add_edge(int u, int v) {
		G[tot].to = v, G[tot].next = head[u], head[u] = tot++;
	}
};

转载于:https://www.cnblogs.com/GadyPu/p/5028150.html