如果有人帮助我了解如何将广度优先搜索更改为深度优先搜索，或者我需要遵循哪些步骤，我将不胜感激。
该算法基本上属于下一个功能：
CODE：
import sys
import copy
from os import system
#Set the console title
system("title Michael Fiford - Breadth First N-Queen Solver")
class QueenSolver:
#Store for the amount of queens we're placing, or table size
tableSize = 0
#The alphabet, for nice cell referencing on the output
alphabet = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']
#The queue of possible moves that we will create and loop through
queue = []
#Whether or not the solver can be ran
canRun = False
def setup(self, queenNumber):
#Set the number of queens/table size
self.tableSize = queenNumber
#Can run, so long as there are no errors
self.canRun = True
#Show error if there is no solution, or would take too long
if queenNumber < 4:
print "ERROR: A solution is not available for this few number of queens"
self.canRun = False
elif queenNumber > 13:
print "ERROR: This solution would take too long to calculate, and your computer would probably run out of memory first!"
self.canRun = False
#Create an empty table
def blankTable(self):
table = []
for row in xrange(self.tableSize):
new = []
for col in xrange(self.tableSize):
new.append(0);
table.append(new)
return table
#Place a queen in a table
def placeQueen(self, table, row, col):
#Copy the table, as python is annoying and will change both
t2 = copy.deepcopy(table)
t2[row][col] = 1
return t2
#The main program loop
def loopBoard(self):
#The column we are currently looking at
col = 1
#Loop while the queue isn't empty, and there are still move possibilities to explore
while len(self.queue):
#Create a new empty queue
queue2 = []
#Update status
print col, "Queens Placed"
#Loop the queue, looking for positions from each status
#s is an array containing the current queen positions for this status
for s in self.queue:
#Get what moves are available from this status
availableMoves = self.getPositions(s, col)
#If we are placing the last queen, and there are solutions available, finish
if col == self.tableSize -1 and len(availableMoves):
#Clear queue
self.queue = []
#Get the solution (or one of them, we only care about the first one)
s = availableMoves[0]
break;
#Add the possible moves to the new queue
#This a table containing all queens now placed
if len(availableMoves):
queue2 += availableMoves
#Replace the old queue with the new one
self.queue = queue2
#Increase Queen/col counter
col += 1
self.finish(s, col)
#Get an array of moves that are available, given info about the current table, and the current column
def getPositions(self, table, col):
#Create a row counter, and array to contain our position info
row = 0
possiblePositions = []
#Loop all rows on the board
while row < self.tableSize:
#If we can place in this space
if self.canPlace(table, row, col):
#Add the table with the newly added queen to the list of possible moves
possiblePositions.append(self.placeQueen(table, row, col))
row += 1
return possiblePositions
#Check whether or not we can place a queen in a position, given the table and the row and col of the desired position
#Return True if space is available
def canPlace(self, table, row, col):
# - Direction
# Check left/right
x = 0
#Loop across the table
while x < self.tableSize:
if table[x][col]:
return False
x += 1
# | Direction
#Check up/down
y = 0
#Loop down the table
while y < self.tableSize:
if table[row][y]:
return False
y += 1
# / Direction
#Check up right Diagonal
#We can start in the cell 1 up and 1 right of the cell in question, as we have already checked the actual cell in the above 2 checks
x = row + 1
y = col + 1
#Loop up/right through the table
while x < self.tableSize and y < self.tableSize:
if table[x][y]:
return False
x += 1
y += 1
#Check down left Diagonal
#Again, not starting in the cell specified
x = row - 1
y = col - 1
#Loop down/left through the table
while x >= 0 and y >= 0:
if table[x][y]:
return False
x -= 1
y -= 1
# \ Direction
#Check up left diagonal
#Again, not starting in the cell specified
x = row - 1
y = col + 1
#Loop up left through the table
while x >= 0 and y < self.tableSize:
if table[x][y]:
return False
x -= 1
y += 1
#Check down right diagonal
#Again, not starting in the cell specified
x = row + 1
y = col - 1
#Loop down right through the table
while x < self.tableSize and y >= 0:
if table[x][y]:
return False
x += 1
y -= 1
return True
#Output a table to a user, looking all pretty
def display(self, table):
#Max Number Length, so we can indent our table nicely later
mnl = len(str(len(table)))
#New Line
print ""
#Top of the table, E.g " A B C D"
print " "*mnl, " ",
for x in range(self.tableSize):
print self.alphabet[x],
#New Line
print ""
#Row spacer, E.g " * - - - - *
print " " * mnl, " *",
for x in range(self.tableSize):
print "-",
print "*"
#Row Counter
#Print the actual table, with the Queens as 1's, empty space as 0
#Also prefixed by the row number, E.g " 3 | 0 1 0 0 |
x = 1
for row in table:
#If numbers are shorter than the largest number, give them extra spaces so the rows still line up
extraPadding = mnl - len(str(x))
#Show the number prefix, spaces, and | symbol, E.g " 6 | "
print "", x, " "*int(extraPadding) + "|",
#Show the value of the cell (1 or 0)
for col in row:
print col,
#End of the row
print "|"
#Next Row
x += 1
#Show the same row spacer as at the top of the table, E.g " * - - - - *
print " " * mnl, " *",
for x in range(self.tableSize):
print "-",
print "*"
#We're done! Show output to the user
def finish(self, table, col):
#If we found the right number of queens
if col == self.tableSize:
print ""
print "Total of", self.tableSize, "Queens placed!"
print "Solution:"
self.display(table)
else:
print ""
print "ERROR: Could not place all queens for some unknown reason =["
#Run the script
def run(self):
if not self.canRun:
print "ERROR: Can not run"
else:
print ""
print "Working..."
print ""
self.queue = self.getPositions(self.blankTable(), 0)
self.loopBoard()
#Ask the user how many Queens they want to use
def ask():
while True:
print ""
print "How many Queens would you like use? [8]"
input = raw_input()
#Check if the input given is an integer
if input.isdigit():
return int(input)
#If no input is given, use the standard 8
elif input == "":
return 8;
print "ERROR: Invalid Input"
#Run the program
def run():
#Instantiate the solver
qs = QueenSolver()
#While ask hasn't given a valid input
while(not qs.canRun):
qs.setup(ask())
print ""
#GO!
qs.run()
#Prompt the user if they want to run the program again
def prompt():
#Has valid input been received?
while True:
print ""
print "Would you like to run the script again? Please enter Y/N [N]"
input = raw_input()
#Check if the input given is Y or N
if input == "Y" or input == "y":
return True
#Also accept an empty string in place of N
elif input == "N" or input == "n" or input == "":
return False
print "ERROR: Invalid Input"
if __name__ == "__main__":
print ""
print ""
print " #######################################"
print " ## Breadth First Search Queen Solver ##"
print " ## By: Michael Fiford - COMF3 ##"
print " ## Date: 03/12/2013 ##"
print " #######################################"
#Run the script, and prompt them after if they want to run it again
shouldRun = True
while(shouldRun):
run()
shouldRun = prompt()
答案 0 :(得分：0)
DFS和BFS之间的区别在于您正在探索图形的节点。 在DFS算法中，您首先要探索添加到堆栈中的最后一个节点（后进先出），而在BFS算法中，您是在先进先出（队列）的基础上探索它。
由此产生的变化是代码应该很小：在BFS算法中，您可以使用列表来实现堆栈，在堆栈顶部附加新节点，然后探索它们：
l=[]
while len(l)>0:
new_node=l.pop()
l.extend(new_node.get_neighbors())
切换到BFS算法的变化非常小：您从堆栈切换到队列。这在集合模块中的Python中实现为deque（具有popleft的高效实现（获取列表的第一项并将其删除）并追加（在队列末尾附加项目）：
import collections
l=collections.deque()
while len(l)>0:
new_node=l.popleft()
l.extend(new_node.get_neighbors())
您的代码可以重写以符合之前的描述：
while len(self.queue):
#Create a new empty queue
#queue2 = []
#Update status
print col, "Queens Placed"
#Loop the queue, looking for positions from each status
#s is an array containing the current queen positions for this status
s=queue.pop()
#Get what moves are available from this status
availableMoves = self.getPositions(s, col)
#If we are placing the last queen, and there are solutions available, finish
if col == self.tableSize -1 and len(availableMoves):
#Clear queue
self.queue = []
#Get the solution (or one of them, we only care about the first one)
s = availableMoves[0]
break;
#Add the possible moves to the new queue
#This a table containing all queens now placed
#if len(availableMoves):
# queue2 += availableMoves
queue.extend(availableMoves)
#Replace the old queue with the new one
#self.queue = queue2
#Increase Queen/col counter
col += 1
如果您需要更多解释，请与我们联系。希望这会有所帮助。