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Question

我正在尝试计算两个节点之间的所有简单路径。我试过DFS，看起来DFS不起作用。

下图使我的目标更加明确。

给出了两个节点'a'和'f'，并且必须找到它们之间的路径。对于上面的例子，有两个简单的路径：

a - ＆gt; b - ＆gt; c - ＆gt; e - ＆gt; f

a - ＆gt; b - ＆gt; d - ＆gt; e - ＆gt; f

我检查了一些帖子，但他们似乎没有处理周期（没有我的DFS工作的周期）。

图表是无向的并且有周期。我相信有一个解决方案。谁能指出我一个可靠的算法？如果这样的算法带有伪代码会很棒。

提前致谢最好，

Answer 1

这是一个使用加权的adj矩阵并避免循环的Ruby解决方案。它有点乱，但接缝工作。我正在测试连接循环（如环）中的节点a，b，c，d，e，f，但是有两个额外的连接（b＆lt; - ＆gt; f和c - f ）生成循环以测试算法。

# Generate all paths from A to E
@grapascii =
'
   C-----D
  / \    |
 B---F---E
  \ /     
   A
'

@nodes = %w{A B C D E F}
n = -1 # represents no connection

#weighted adj matrix
@adjmat = [
    #a  b  c  d  e  f
    [n, 1, n, n, n, 1], # a
    [1, n, 3, n, n, 1], # b
    [n, 4, n, 2, n, 1], # c
    [n, n, 4, n, 1, n], # d
    [n, n, n, 1, n, 1], # e
    [1, 6, 1, n, 2, n]  # f
]

# generate a neighbors hash for easy and fast future acccess
def gen_neighbors(nodes, adjmat)
    len = nodes.length
    neighbors = {}
    for i in 0..(len - 1)
        for j in 0..(len - 1)
            if adjmat[i][j] >= 0 && i != j
                neighbors[nodes[i]] ||= []
                neighbors[nodes[i]] << {:node => nodes[j], :cost => adjmat[i][j]}
            end
        end
    end
    return neighbors
end

@neighbors = gen_neighbors(@nodes, @adjmat)

def all_paths(currpath, destnode, currcost, paths)
    #the current node is always tha last one on the current evaluated path
    currnode = currpath.last
    # just the neighbors that is nor on the current evaluated path, to avoid cycles
    current_neighbors = @neighbors[currnode].select{|n| !currpath.include?(n[:node])}

    #each neighbor is a hash with :node and :cost
    current_neighbors.each do |neighbor|
        # yes. we have to duplicate that. maybe there is a better solution...
        new_path = currpath + [neighbor[:node]]
        cost = currcost + neighbor[:cost]

        if neighbor[:node] == destnode
            #FOUND PATH
            paths << {:path => new_path, :cost => cost}
        else
            all_paths(new_path, destnode, cost, paths)
        end
    end
end

puts @grapascii

puts "NEIGHBORS HASH:"
@neighbors.each do |node, neighbors|
    neighbors_and_cost = neighbors.map{|n| "#{n[:node]}(#{n[:cost]})"}
    puts "#{node} => #{neighbors_and_cost.join(', ')}"
end


# start path with the start node
startpath = ['A']
# paths variable will hold all possible paths without cycles
paths = []
all_paths(startpath, 'E', 0, paths)

puts "PATHS:"
# sort paths by total cost(optional)
paths.sort!{|a, b| a[:cost] <=> b[:cost]}
paths.each do |path|
    puts "#{path[:path]} => #{path[:cost]}"
end

输出：

   C-----D
  / \    |
 B---F---E
  \ /     
   A
NEIGHBORS HASH:
A => B(1), F(1)
B => A(1), C(3), F(1)
C => B(4), D(2), F(1)
D => C(4), E(1)
E => D(1), F(1)
F => A(1), B(6), C(1), E(2)
PATHS:
["A", "F", "E"] => 3
["A", "B", "F", "E"] => 4
["A", "F", "C", "D", "E"] => 5
["A", "B", "F", "C", "D", "E"] => 6
["A", "B", "C", "D", "E"] => 7
["A", "B", "C", "F", "E"] => 7
["A", "F", "B", "C", "D", "E"] => 13

查找具有周期

1 个答案:

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