# 回溯算法-八皇后问题

8x8棋盘摆8个皇后，使得不能相互攻击（不在同一行、列、斜线上）。求多少种摆法

```python
n=8
x=[] #当前尝试方案
X=[] #所有可行方案
def conflict(k):
    global x
    for i in range(k):#遍历前0 - k-1
        # 判断是否与x[k]冲突
        if x[i]==x[k] or abs(x[i]-x[k])==abs(i-k):
            return True
    return False

def queens(k):#到达第k行，进行判断
    global n,x,X
    if k>=n:#如果到底部，那就是一个解，添加进去，因为到底部，意味着不与面前所有皇后冲突
#         print(x)
        X.append(x[:])
    else:#否则继续往下走，走的地方有n个，依次尝试
        for i in range(n):
            x.append(i)
            if not conflict(k):#不冲突，就继续往下走
                queens(k+1)#到k+1行进行判断
            #无论怎样，回溯，退回上一步尝试的位置，尝试其它位置
            x.pop()

def show(x):
    global n
    for i in range(n):
        print('. '*(x[i])+'X '+'. '*(n-x[i]-1))

queens(0)
print(X[-1],'\n')
show(X[-1])
print('-'*50)
print(X[-2],'\n')
show(X[-2])
```

```
[7, 3, 0, 2, 5, 1, 6, 4] 

. . . . . . . X 
. . . X . . . . 
X . . . . . . . 
. . X . . . . . 
. . . . . X . . 
. X . . . . . . 
. . . . . . X . 
. . . . X . . . 
--------------------------------------------------
[7, 2, 0, 5, 1, 4, 6, 3] 

. . . . . . . X 
. . X . . . . . 
X . . . . . . . 
. . . . . X . . 
. X . . . . . . 
. . . . X . . . 
. . . . . . X . 
. . . X . . . .
```


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