用Python实现回溯算法

Question

我正在尝试实现一个算法，它接受两个整数n和k，其中n是一行中的座位数，k是试图坐在一行中的学生的数量。问题是，每个学生必须在两侧至少有两个座位。我拥有的是一个生成所有子集的函数(一个由0或1组成的数组，1表示有人坐在那里)，然后我将其发送到一个函数，以检查它是否是有效的子集。这是我为该函数编写的代码

def process(a,num,n):
    c = a.count('1')
    #If the number of students sitting down (1s) is equal to the number k, check the subset
    if(c == num):
        printa = True
        for i in range(0,n):
            if(a[i] == '1'):
                if(i == 0):
                    if( (a[i+1] == '0') and (a[i+2] == '0') ):
                        break
                    else:
                        printa = False
                elif(i == 1):
                    if( (a[i-1] == '0') and (a[i+1] == '0') and (a[i+2] == '0') ):
                        break
                    else:
                        printa = False
                elif(i == (n-1)):
                    if( (a[i-2] == '0') and (a[i-1] == '0') and (a[i+1] == '0') ):
                        break
                    else:
                        printa = False
                elif(i == n):
                    if( (a[i-2] == '0') and (a[i-1] == '0') ):
                        break
                else:
                    printa = False                    
            else:
                if( (a[i-2] == '0') and (a[i-1] == '0') and (a[i+1] == '0') and (a[i+2] == '0') ):
                    break
                else:
                    printa = False
        if(printa):
            print a
    else:
        return

这段代码适用于k和n的小输入，但是如果我得到更高的值，我会得到一个列表错误的索引，因为某些原因，我不能确定。

任何帮助都是非常感谢的。

O输入a是如下所示的列表

['1','0','0','1','0'] # a valid subset for n=5 and k=2
['0','0','0','1','1'] # an invalid subset

编辑：

调用进程的代码：

'''
This function will recursivly call itself until it gets down to the leaves then sends that
subset to process function.  It appends
either a 0 or 1 then calls itself
'''
def seatrec(arr,i,n,k):
    if(i==n):
        process(arr,k,n)
        return
    else:
        arr.append("0")
        seatrec(arr,i+1,n,k)
        arr.pop()
        arr.append("1")
        seatrec(arr,i+1,n,k)
        arr.pop()
    return
'''
This is the starter function that sets up the recursive calls
'''
def seat(n,k):
    q=[]
    seat(q,0,n,k)

def main():
    n=7
    k=3
    seat(n,k)

if __name__ == "__main__":
    main()

如果我使用这些数字，我得到的错误是

if( (a[i-2] == '0') and (a[i-1] == '0') and (a[i+1] == '0') ):
IndexError: list index out of range

Answer 1

排除无效的座位安排就足够了，即当学生彼此相邻时，或者当他们之间只有一个座位时，所有其他具有正确数量的['1', '1']，['1', '0', '1'] '1'，'0'是有效的，example

def isvalid(a, n, k):
    if not isinstance(a, basestring):
       a = ''.join(a) # `a` is a list of '1', '0'
    return (len(a) == n and a.count('1') == k and a.count('0') == (n-k) and
            all(p not in a for p in ['11', '101']))

存在在不检查所有子集的情况下生成有效子集的更有效的算法，

def subsets(n, k):
    assert k >= 0 and n >= 0
    if k == 0: # no students, all seats are empty
        yield '0'*n
    elif k == 1 and (n == 1 or n == 2): # the last student at the end of the row
        yield '1' + '0'*(n-1) # either '1' or '10'
        if n == 2: yield '01'
    elif n > 3*(k-1): # there are enough empty seats left for k students
        for s in subsets(n-3, k-1):
            yield '100' + s # place a student
        for s in subsets(n-1, k):
            yield '0' + s   # add empty seat

Example

n, k = 5, 2
for s in subsets(n, k):
    assert isvalid(s, n, k)
    print(s)

输出

10010
10001
01001

Answer 2

长度为n的数组的索引是从0到n-1。因此，访问n不在列表中。

如果你没有注意到在较小的值上，生成列表的代码一定有一个bug。