Question

我设法编写了一个代码，使用BFS算法迭代所有图形节点。这很简单：）

不，我被困在获得两个节点之间的最短路径。

到目前为止，这是我的代码：

function table.contains(table, element)
  for _, value in pairs(table) do
    if value == element then
      return true
    end
  end
  return false
end

queue = {stack={}}

function queue:en(e)
    table.insert(self.stack, e)
end

function queue:de()
    local e = self.stack[1]   
    self.stack[1] = nil

    local new_stack = {}

    for _, v in pairs(self.stack) do
        table.insert(new_stack, v)
    end

    self.stack = new_stack

    return e
end

function queue:count()
    return #self.stack
end

function bfs()
    start = 1
    goal  = 10
    visited = {}
    path = {}

    graph = {}
    graph[1] = {2,3,4}
    graph[2] = {6,5,1}
    graph[3] = {1}
    graph[4] = {7,8}
    graph[5] = {9,10,2}
    graph[6] = {2}
    graph[7] = {4,11,12}
    graph[8] = {4}
    graph[9] = {5}
    graph[10] = {5}
    graph[11] = {7}
    graph[12] = {7}


    queue:en(start)
    table.insert(visited, start)
    depth = 1

    while queue:count() > 0 do
        node = queue:de()

        for _, exit in pairs(graph[node]) do
            if not table.contains(visited, exit) then
                table.insert(visited, exit)

                if exit == goal then 
                    print("GOAL : " .. exit)

                    do return end 
                end

                if graph[exit] then
                    print("Node: " .. exit .. ", Depth: " .. depth)

                    queue:en(exit)
                end
            end
        end

        depth = depth + 1
    end
end

bfs()

它生成的输出：

Node: 2, Depth: 1
Node: 3, Depth: 1
Node: 4, Depth: 1
Node: 6, Depth: 2
Node: 5, Depth: 2
Node: 7, Depth: 4
Node: 8, Depth: 4
Node: 9, Depth: 6
GOAL : 10

使用当前示例，我需要获得从1到10的最短路径。

Answer 1

这是如何做到的。

local queue = {} function queue:init() local q = {} q.stack = {} function q:push(e) table.insert(self.stack, e) end function q:pull() local e = self.stack[1] table.remove(self.stack, 1) return e end function q:count() return #self.stack end return q end return queue

tbl.lua

function table.contains(tbl, e) for _, v in pairs(tbl) do if v == e then return true end end return false end function table.copy(tbl) local t = {} for _, v in pairs(tbl) do table.insert(t, v) end return t end

bfs.lua

require "tbl" -- table.contains and table.copy queue = require "queue" local function bfs(graph, start, goal) if not graph[start] then return false end local visited = {} local queue = queue:init() queue:push({start}) table.insert(visited, start) while queue:count() > 0 do local path = queue:pull() local node = path[#path] if node == goal then return path end for _, exit in pairs(graph[node]) do if not table.contains(visited, exit) then table.insert(visited, exit) if graph[exit] then local new = table.copy(path) table.insert(new, exit) queue:push(new) end end end end return false end return bfs

&buffer[Int(index)]

用法：

将所有三个文件保存到您的lua路径。

bfs = require“bfs”

path = bfs（grap，start，goal）

使用Lua的BFS算法找到2个节点之间的最短路径

1 个答案: