2n皇后的两种解法 python（全面解析），第二种耗时大大缩短_黑皇后算法

题目

两种思路

思路一：同时放黑皇后和白皇后
思路二：放完黑皇后以后，再放白皇后

第一种解法（耗时较长）

同时考虑黑皇后和白皇后
考虑index行，分三步：

• 在index行放置黑皇后
• 在index行放置白皇后
• 递归，在index+1重复上述操作
• 递归结束条件 index == n
  代码

from collections import defaultdict

n = int(input())
board = []
for i in range(n):
    board.append(input().split())
col1 = defaultdict(bool)
dia11 = defaultdict(bool)
dia21 = defaultdict(bool)
col2 = defaultdict(bool)
dia12 = defaultdict(bool)
dia22 = defaultdict(bool)

count = 0


def dfs(index):
    global count
    if n <= 0:
        return 0
    if index == n:

        count += 1
        return
    # 在第index行放置黑皇后
    for j in range(n):
        if not col1[j] and not dia11[index + j] and not dia21[index - j] \
                and board[index][j] == "1":
            col1[j] = True
            dia11[index + j] = True
            dia21[index - j] = True
            board[index][j] = "0"
            for k in range(n):
                if not col2[k] and not dia12[index + k] and not dia22[index - k] \
                        and board[index][k] == "1":
                    col2[k] = True
                    dia12[index + k] = True
                    dia22[index - k] = True
                    board[index][k] = "0"
                    dfs(index + 1)
                    col2[k] = False
                    dia12[index + k] = False
                    dia22[index - k] = False
                    board[index][k] = "1"

            col1[j] = False
            dia11[index + j] = False
            dia21[index - j] = False
            board[index][j] = "1"

    return 0


dfs(0)
print(count)

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第二种解法（耗时大大缩短）

思路

• 先在index放置黑皇后
• 递归，在index+1行放置黑皇后
• 递归到底，index == n ，这个时候说明黑皇后已经全部放置完毕（一种情况），这个时候再开始第二次dfs2（0），这个时候是放置白皇后，白皇后也放置完毕以后，一个解就找到了。

代码：

from collections import defaultdict

n = int(input())
board = []
for i in range(n):
    board.append(input().split())
col1 = defaultdict(bool)
dia11 = defaultdict(bool)
dia21 = defaultdict(bool)
col2 = defaultdict(bool)
dia12 = defaultdict(bool)
dia22 = defaultdict(bool)
count = 0


def dfs2(index):
    global count
    if index == n:
        count += 1
        return
    # 在第index行放置黑皇后
    for j in range(n):
        if not col2[j] and not dia12[index + j] and not dia22[index - j] \
                and board[index][j] == "1":
            col2[j] = True
            dia12[index + j] = True
            dia22[index - j] = True
            board[index][j] = "0"
            dfs2(index + 1)
            col2[j] = False
            dia12[index + j] = False
            dia22[index - j] = False
            board[index][j] = "1"

    return 0

def dfs1(index):
    if n <= 0:
        return 0
    if index == n:
        dfs2(0)
        return
    # 在第index行放置黑皇后
    for j in range(n):
        if not col1[j] and not dia11[index + j] and not dia21[index - j] \
                and board[index][j] == "1":
            col1[j] = True
            dia11[index + j] = True
            dia21[index - j] = True
            board[index][j] = "0"
            dfs1(index + 1)
            col1[j] = False
            dia11[index + j] = False
            dia21[index - j] = False
            board[index][j] = "1"

    return 0


dfs1(0)
print(count)

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