2018-2-22 《啊哈，算法》再练习广度优先搜索，题：炸怪兽, 2-23改用深度优先搜索。宝岛探险（广度，深度，及地图着色）2-24水管工游戏，2-25测试水管工代码

2小时。

先是是纠错，通过对代码运行过程的测试。发现是变量打错。以及录入地图❌。

重构练习题，改使用while..end代替for in.

⚠️ ： 在while（k <= n）中如果用到next，为了不陷入死♻️，注意条件的判断, 一般要在next前k += 1。 

题：爆炸小子炸怪兽

代码：

1.广度优先搜索

设定一个点，先访问这个点的每个周边点，然后再扩展到每个周边点的还未访问的点。

代码：使用队列（栈）。 

# 地图

map = [

  ["#","#","#","#","#","#","#","#","#","#","#","#","#"],

  ["#","g","g",".","g","g","g","#","g","g","g",".","#"],

  ["#","#","#",".","#","g","#","g","#","g","#","g","#"],

  ["#",".",".",".",".",".",".",".","#",".",".","g","#"],

  ["#","g","#",".","#","#","#",".","#","g","#","g","#"],

  ["#","g","g",".","g","g","g",".","#",".","g","g","#"],

  ["#","g","#",".","#","g","#",".","#",".","#",".","#"],

  ["#","#","g",".",".",".","g",".",".",".",".",".","#"],

  ["#","g","#",".","#","g","#","#","#",".","#","g","#"],

  ["#",".",".",".","g","#","g","g","g",".","g","g","#"],

  ["#","g","#",".","#","g","#","g","#",".","#","g","#"],

  ["#","g","g",".","g","g","g","#","g",".","g","g","#"],

  ["#","#","#","#","#","#","#","#","#","#","#","#","#"]

]

# 地图副本book,用于标记走过的点。用compact复制出新的array。

# 因为不能嵌套复制。需要用map

book = map.map{|item| item.compact }

# 输出地图（对输出结果进行美化）

i = 0

n = map.size

n.times do |x|

  print "#{x} "

end

print " "

while i < n

  j= 0

  while j < n

    case map[i][j]

    when "#"

      print "◼︎ "

    when "g"

      print "☿ "

    when "."

      print "  "

    end

    j += 1

  end

  print " "

  i += 1

end

# 杀怪统计Method

def getnum(startx, starty, map)

# 四个方向各自统计杀怪数，加到sum中

  sum = 0

  x = y = 0

  # 向下，遇到“#”，停止。（爆炸受阻）

  x = startx

  y = starty

  while map[x][y] != "#"

    if map[x][y] == "g"

      sum += 1

    end

    x += 1

  end

  # 上

  x = startx

  y = starty

  while map[x][y] != "#"

    if map[x][y] == "g"

      sum += 1

    end

    x -= 1

  end

  # 右

  x = startx

  y = starty

  while map[x][y] != "#"

    if map[x][y] == "g"

      sum += 1

    end

    y += 1

  end

  # 左

  x = startx

  y = starty

  while map[x][y] != "#"

    if map[x][y] == "g"

      sum += 1

    end

    y -= 1

  end

  return sum

end

# 设定开始起点位置

print "请输入起点坐标x: "

startx = gets.to_i

print "请输入起点坐标y: "

starty = gets.to_i

# wide first search （核心）

x = []

y = []

# 指针head , tail(队列最后一位的后面空白处)

head = 0

tail = 1

x[head] = startx

y[head] = starty

# 坐标点周边的四个方向,目的是访问相邻的4个点

next_direction = [

  [0, 1],#右

  [1, 0],#下

  [0,-1],#左

  [-1,0] #上

]

# 统计最大击杀

kill_max = 0

max_x = max_y = 0

# 边界的判断

nx = map[0].size - 2

ny = map.size - 2

# 大循环，wide first search

while head < tail

  # 每个坐标点周边点的情况判定。

  k = 0

  while k <= 3

    tx = x[head] + next_direction[k][0]

    ty = y[head] + next_direction[k][1]

    # 界限

    if tx < 1 || tx > nx || ty <1 || ty > ny

      # k += 1必不可少，否则会死循环。用for语法的话可以避免。

      k += 1

      next

    end

    # 判断是否是平地，而且没有走过

    if map[tx][ty] == "." && book[tx][ty] == "."

      # 在地图上标记已经走过

      book[tx][ty] = "X"

      x << tx

      y << ty

      # 队列尾巴的后一位，用于判断是否结束大循环

      tail += 1

      # 统计杀怪数量,调用Method

      sum_kill = getnum(tx, ty, map)

      if sum_kill > kill_max

        kill_max = sum_kill

        max_x = tx

        max_y = ty

      end

    end

    k += 1

  end

# 头部指针前移动

  head += 1

end

p "最大击杀#{kill_max}"

p "坐标：（#{max_x},#{max_y}）"

2.深度优先搜索

核心，设定一个边界（目标），延一个方向走到满足条件的极限，然后回朔换方向走，一旦不满足条件就回朔。只关注当下

代码： dfs()嵌套 + 界限

# 地图

$map = [

  ["#","#","#","#","#","#","#","#","#","#","#","#","#"],

  ["#","g","g",".","g","g","g","#","g","g","g",".","#"],

  ["#","#","#",".","#","g","#","g","#","g","#","g","#"],

  ["#",".",".",".",".",".",".",".","#",".",".","g","#"],

  ["#","g","#",".","#","#","#",".","#","g","#","g","#"],

  ["#","g","g",".","g","g","g",".","#",".","g","g","#"],

  ["#","g","#",".","#","g","#",".","#",".","#",".","#"],

  ["#","#","g",".",".",".","g",".",".",".",".",".","#"],

  ["#","g","#",".","#","g","#","#","#",".","#","g","#"],

  ["#",".",".",".","g","#","g","g","g",".","g","g","#"],

  ["#","g","#",".","#","g","#","g","#",".","#","g","#"],

  ["#","g","g",".","g","g","g","#","g",".","g","g","#"],

  ["#","#","#","#","#","#","#","#","#","#","#","#","#"]

]

# 地图副本book,用于标记走过的点。用compact复制出新的array。

# 因为不能嵌套复制。需要用map

$book = $map.map{|i| i.compact }

# 输出地图（对输出结果进行美化）

i = 0

nx = $map.size

ny = $map.first.size

nx.times do |n|

  if n >=10

    print "#{n}"

  else

    print "#{n} "

  end

end

print " "

while i < nx

  j = 0

  while j < ny

    case $map[i][j]

    when "#"

      print "◼︎ "

    when "g"

      print "☿ "

    when "."

      print "  "

    end

    j += 1

  end

  print "#{i} "

  i += 1

end

# 杀怪统计Method

def getnum(startx, starty)

# 四个方向各自统计杀怪数，加到sum中

  sum = 0

  x = y = 0

  # 向下，遇到“#”，停止。（爆炸受阻）

  x = startx

  y = starty

  while $map[x][y] != "#"

    if $map[x][y] == "g"

      sum += 1

    end

    x += 1

  end

  # 上

  x = startx

  y = starty

  while $map[x][y] != "#"

    if $map[x][y] == "g"

      sum += 1

    end

    x -= 1

  end

  # 右

  x = startx

  y = starty

  while $map[x][y] != "#"

    if $map[x][y] == "g"

      sum += 1

    end

    y += 1

  end

  # 左

  x = startx

  y = starty

  while $map[x][y] != "#"

    if $map[x][y] == "g"

      sum += 1

    end

    y -= 1

  end

  return sum

end

#深度优先搜索。----------------------------

# 设定开始起点位置

print "请输入起点坐标x: "

startx = gets.to_i

print "请输入起点坐标y: "

starty = gets.to_i

# 统计最大击杀

$kill_max = 0

$max_x = $max_y = 0

def dfs(x,y)

  # 计算杀敌总数。（深度优先搜索的边界）

  sum = getnum(x,y)

  if sum > $kill_max

    $kill_max = sum

    $max_x = x

    $max_y = y

  end

  # 坐标点周边的四个方向,目的是访问相邻的4个点

  next_direction = [

    [0, 1],#右

    [1, 0],#下

    [0,-1],#左

    [-1,0] #上

  ]

  # 四个方向枚举

  k = 0

  while k <= 3

    tx = x + next_direction[k][0]

    ty = y + next_direction[k][1]

    # 判断是否越界

    if tx < 1 || tx > 11 || ty <1 || ty > 11

      # k += 1必不可少，否则会死循环。用for语法的话可以避免。

      k += 1

      next

    end

    #判断是否围墙，或走过。

    if $map[tx][ty] == "." && $book[tx][ty] == "."

      $book[tx][ty] = "X"

      dfs(tx,ty)   #通过不断的深度嵌套，再通过回朔改变方向，来完成遍历每个点。

    end

    k += 1

  end

end

dfs(startx,starty)

print "最大击杀:#{$kill_max} "

p "坐标（#{$max_x}，#{$max_y}）"

---

 宝岛探险：

广度，深度，和上色。

给所有岛着色（深度）：嵌套循环，遍历所有岛（大于0）并标记，着色（num,不同的岛用不同的数表示。）

# 地图，0代表海洋，1-9代表陆地及其海拔。

$map = [

  [1,2,1,0,0,0,0,0,2,3],

  [3,0,2,0,1,2,1,0,1,2],

  [4,0,1,0,1,2,3,2,0,1],

  [3,2,0,0,0,1,2,4,0,0],

  [0,0,0,0,0,0,1,5,3,0],

  [0,1,2,1,0,1,5,4,3,0],

  [0,1,2,3,1,3,6,2,1,0],

  [0,0,3,4,8,9,7,5,0,0],

  [0,0,0,3,7,8,6,0,1,2],

  [0,0,0,0,0,0,0,0,1,0]

]

# 飞机降落位置坐标。

startx = 0

starty = 0

# 复制一个全新的地图用于在图上做标记。

$book = $map.map { |e| e.compact }

# 用于给不同的到上不同的颜色

$map_color = $map.map { |e| e.compact  }

# 记录岛的面积

$sum = 0

# dfs方法 深度优先搜索。

def dfs(startx, starty,num)

  # 方向坐标，按照右，下，左，上顺序排列。

  next_direction = [

    [0, 1],

    [1, 0],

    [0,-1],

    [-1,0]

  ]

  # 边界,本题不需要

  i = 0

  for i in 0..3 do

    tx = startx + next_direction[i][0]

    ty = starty + next_direction[i][1]

    # 判断是否越界

    if tx < 0 || tx > 9 || ty < 0 || ty > 9

      next

    end

    # 判断是否为没发现的陆地

    if $map[tx][ty] > 0 && $book[tx][ty] != "x"

      $book[tx][ty] = "x"

      $sum += 1

      $map_color[tx][ty] = num #陆地染色为num

      dfs(tx,ty,num)

    end

  end

end

num = 0 #不同小岛的编号

# 遍历所有陆地(>0),每块陆地，一旦走过就$book标记“x”

i = j = 0

for i in 0..9 do

  for j in 0..9 do

    if $map[i][j] > 0 && $book[i][j] != "x"

      num -= 1

      $map_color[i][j] = num  #陆地染色为num

      $book[i][j] = "x"

      $sum += 1 

      #调用函数，相邻的陆地都会染色相同num

      dfs(i,j,num)

    end

  end

end

p "岛的面积#{$sum}"

$map_color.each do |x|

  x.each do |y|

    printf("%3s",y)

  end

  print " "

end

$book.each do |x|

  x.each do |y|

    if y == "x"

      print "❌ "

    else

      print "#{y} "

    end

  end

  print " "

end

---

 水管工

 深度优先搜索实现。

⚠️ ：判断每个管子的进水口inflow,用dfs方法时，除了x,y坐标参数，还有inflow进水口参数。每一次调用dfs(x,y,inflow)都是下一个管道，此时的出水口，是下一个的进水口，如果用东南西北表示，要注意改成进水口方向。例子：如果当前管子的出水口是东，那么下一个管子的进水口就是西。

$map = [

  [5,3,5,3],

  [1,5,3,0],

  [2,3,5,1],

  [6,1,1,5],

  [1,5,5,5]

]

$book = $map.map { |e| e.compact }

# 用1-6分别表示水管的6种排列形式。4个弯管，2个直管

# n,m+1时候停止

# 参数inflow为进水方向。W进水口在西，N进水口在上，E进水口在right，S进水口在下。

inflow = ""

# 判断是否成功出水

$flag = "no"

#用于记录路径,栈：后进的先出，ruby用pop方法，去掉数组最后一个，也可以加个指针

$x = []

$y = []

def dfs(x, y, inflow)

p "开始dfs"

  # 界限：停止dfs得出结果

  # 判断水流是否走出地图，看地图东南角的边界。

  if x == 4 && y == 4

    $flag = "yes"

    len = $x.size #队列的长度，然后输出连个数组x,y

    len.times do |n|

      printf("(%d,%d)->",$x[n],$y[n])

    end

    return

  end

  # 判断是否边界

  if x < 0 || x > 4 || y < 0 || y > 3

    return

  end

  # 判断是否已经标记了

  if $book[x][y] == "Y"

    return

  end

  $book[x][y] = "Y"

  # 当前坐标入栈

  $x << x

  $y << y

  print "#{$x} #{$y}. "

  # 当前水管是直管的情况 5表示东西走向，6表示南北走向。

  if $map[x][y] >= 5 && $map[x][y] <= 6

    # "直管"

    case inflow

    when "W" #水从西进，下一个管道也是从西进,第三个参数是进水方向

      dfs(x, y+1, "W")

    when "N"

      dfs(x+1, y , "N")

    when "E"

      dfs(x, y-1, "E")

    when "S"

      dfs(x-1, y ,"S")

    end

  end

  # 当前水管是弯管的情况 1表示东北口，2表示东南口，3表示西南口，4表示西北口

  if $map[x][y] >= 1 && $map[x][y] <= 4

    #  "弯管"

    case inflow

    when "N" #当水从北流入，弯管只有两种流出结构,但inflow参数是进水方向。

        dfs(x, y+1, "W") #下一个管，和进水方向，不要混淆出水方向。

        dfs(x, y-1, "E")

    when "E" #当水从东流入，弯管只会两种流出方向，N，S

        dfs(x-1, y, "S")

        dfs(x+1, y, "N")

    when "S"

        dfs(x, y+1, "W")

        dfs(x, y-1 ,"E")

    when "W"

        dfs(x-1, y, "S")

        dfs(x+1, y, "N")

    end

  end

  # 在dfs方法不再深度后，执行下面代码，取消标记 。

  # 目的：用于回溯，产生其他方案

  $book[x][y] = "N"

  # 栈后进先出，去掉不合格的

  $x.pop

  $y.pop

  return

end

dfs(0,0,"W")

if $flag == "yes"

  print "找到出水方案"

else

  print "没找到出水方案"

end

问题处理：

第一问：我在整体代码最后希望输出管道路径，代码如下：

10.times do |n|

  $map[$x[n]][$y[n]] = "X"

end

$map.each do |x|

  print "#{x} "

end

出错❌： no implicit conversion from nil to integer (TypeError)

试❌：加入代码p $x 发现得到[],竟然是空数组，难怪错误提示是nil to integer. 

因为，dfs方法结束后 $x,$y已经清空。变为空数组 。可以放到dfs方法中的判断水流是否走出地图的if..end中，注意单独复制一个地图用于储存路径，$route = $map.map{|e|e.compact}

第二问： 代码正确得到结果，但是改变地图结构，就不能得到正确结构。

出错❌ 代码：

  # 目的：用于回溯，产生其他方案

  $book[x][y] == "N"

分析：

    粗心多写了一个=。导致不能正确回朔。之前的代码正确也是瞎猫碰死耗子。