Leetcode 207 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.

拓扑排序，维护入度为0的集合。首先放入所有入度为0的点，逐个引出这个原点所能到达的点并删除原点，如果被到达的点在删除这条原点到其的路径后入度为0，则将这个点放入集合内。

伪代码

L ← Empty list that will contain the sorted elements

S ← Set of all nodes with no incoming edges

while S is non-empty do

    remove a node n from S

    add n to tail of L

    for each node m with an edge e from n to m do

        remove edge e from the graph

        if m has no other incoming edges then

            insert m into S

if graph has edges then

    return error (graph has at least one cycle)

else 

    return L (a topologically sorted order)

require 'set'

def can_finish(num_courses, prerequisites)

  graph, neighbour = Hash.new{|hsh,key| hsh[key] = Set.new}, Hash.new{|hsh,key| hsh[key] = Set.new}

  prerequisites.each {|x,y| graph[x] << y; neighbour[y] << x}

  zero_degree, count = [], 0

  num_courses.times {|x| zero_degree << x if graph[x].empty?}

  while not zero_degree.empty?

    node = zero_degree.pop

    count += 1

    neighbour[node].each do |x|

      graph[x] -= [node]

      zero_degree << x if graph[x].empty?

    end

  end

  count == num_courses

end