回路计数#回溯法#python

动态规划算法参考博主，前两种时间复杂度一样的，第三种时间复杂度相对较低。
1.回溯算法一：

#回路计数#回溯法
import math
def backroad(path):
    if len(path)==21:
        Length+=1
        return
    for num in range(2,22):
        if len(path)==1:
            path.append(num)
            backroad(path,ss)
            path.pop()
        else:
            if num in path or math.gcd(path[-1],num)!=1:
                continue
            path.append(num)
            backroad(path)
            path.pop()

if __name__=='__main__':
    Length=[]
    backroad([1],0)
    print(Length)

2.纯暴力

#回路计数的解法二#纯暴力
import itertools
num=[]
for i in range(2,22):
    num.append(i)
data=[]
SS=0
AA=[]
for i in itertools.permutations(num,len(num)):
    for j in range(len(num)-1):
        if math.gcd(num[j],num[j+1])!=1:
            break
    else:
        AA.append(i)
        SS+=1
print(SS)

3.回溯算法二

#哈希表求解
import math
Road={}
for i in range(2,22):
    for j in range(2,22):
        if i!=j and math.gcd(i,j)==1:
            if i not in Road.keys():
                Road[i]=[j]
            else:
                Road[i].append(j)
def ComBack(path):
    global GG
    if len(path)==20:
        GG+=1
        print(GG)
        return
    for j in Road[path[-1]]:
        if j not in path:
            path.append(j)
            ComBack(path)
            path.pop()

if __name__=='__main__':
    GG=0
    for i in range(2,22):
        ComBack([i])
    print(GG)