207. Course Schedule #

题目 #

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:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about  how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

题目大意 #

现在你总共有 n 门课需要选，记为 0 到 n-1。在选修某些课程之前需要一些先修课程。 例如，想要学习课程 0 ，你需要先完成课程 1 ，我们用一个匹配来表示他们: [0,1]。给定课程总量以及它们的先决条件，判断是否可能完成所有课程的学习？

解题思路 #

• 给出 n 个任务，每两个任务之间有相互依赖关系，比如 A 任务一定要在 B 任务之前完成才行。问是否可以完成所有任务。

• 这一题就是标准的 AOV 网的拓扑排序问题。拓扑排序问题的解决办法是主要是循环执行以下两步，直到不存在入度为0的顶点为止。

  • 1. 选择一个入度为0的顶点并输出之；
  • 2. 从网中删除此顶点及所有出边。

  循环结束后，若输出的顶点数小于网中的顶点数，则输出“有回路”信息，即无法完成所有任务；否则输出的顶点序列就是一种拓扑序列，即可以完成所有任务。

代码 #

package leetcode

// AOV 网的拓扑排序
func canFinish(n int, pre [][]int) bool {
	in := make([]int, n)
	frees := make([][]int, n)
	next := make([]int, 0, n)
	for _, v := range pre {
		in[v[0]]++
		frees[v[1]] = append(frees[v[1]], v[0])
	}
	for i := 0; i < n; i++ {
		if in[i] == 0 {
			next = append(next, i)
		}
	}
	for i := 0; i != len(next); i++ {
		c := next[i]
		v := frees[c]
		for _, vv := range v {
			in[vv]--
			if in[vv] == 0 {
				next = append(next, vv)
			}
		}
	}
	return len(next) == n
}