Gena Playing Hanoi

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思路：首先思路是BFS，要把结果都放到第一个柱子上，所以后面3个的状态顺序无所谓，可以可以快6倍

from collections import deque

def serialize(a):
#    return tuple(sorted([tuple(t) for t in a]))
    l=[tuple(t) for t in a]
    l[1:]=sorted(l[1:])
    return tuple(l)
def deserialize(a):
    return deque([deque(t) for t in a])
def copy(a):
    return deque([deque(t) for t in a])
    
def helper(n,aa):
    a=[deque() for _ in range(4)]
    for i,v in enumerate(aa): a[v-1].append(i+1)
    step=1
    q,qq=[a],[]
    vis=set()
    vis.add(serialize(a))
    target=serialize([deque(range(1,n+1)),deque(),deque(),deque()])
    while q:
        while q:
            a=q.pop()
            for i in range(4):
                for j in range(4):
                    if a[i] and (not a[j] or a[i][0]<a[j][0]):
                        # a2=deserialize(a)
                        a2=copy(a)
                        a2[j].appendleft(a2[i].popleft())
                        a3=serialize(a2)
                        if a3==target: return step
                        if a3 in vis: continue
                        vis.add(a3)
                        qq.append(a2)
        q,qq=qq,q
        step+=1
                        
    
if __name__ == '__main__':
    N = int(input())
    a = list(map(int, input().rstrip().split()))
    print(helper(N,a))