Gena Playing Hanoi
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Gena Playing Hanoi
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思路:首先思路是BFS,要把結果都放到第一個柱子上,所以后面3個的狀態順序無所謂,可以可以快6倍
from collections import dequedef 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=1q,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 stepif a3 in vis: continuevis.add(a3)qq.append(a2)q,qq=qq,qstep+=1if __name__ == '__main__':N = int(input())a = list(map(int, input().rstrip().split()))print(helper(N,a))?
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