70 lines
1.9 KiB
Python
70 lines
1.9 KiB
Python
from collections import deque
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def bfs(capacities, target):
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# 初始化队列和访问过的状态集合
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queue = deque([(0, 0, 0)]) # 三个桶的初始状态
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visited = set([(0, 0, 0)])
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# 操作步骤计数
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steps = 0
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# 开始BFS搜索
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while queue:
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for _ in range(len(queue)):
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current_state = queue.popleft()
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a, b, c = current_state
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# 如果找到目标状态
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if target in current_state:
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return steps
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# 获取当前桶的容量
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A, B, C = capacities
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# 对每一对桶进行操作
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next_states = [
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# 倒水操作：将水从一个桶倒入另一个桶
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# 从a桶倒到b桶
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(min(a + b, B), max(0, a + b - B), c),
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# 从a桶倒到c桶
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(min(a + c, C), b, max(0, a + c - C)),
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# 从b桶倒到a桶
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(min(a + b, A), max(0, a + b - A), c),
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# 从b桶倒到c桶
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(a, min(b + c, C), max(0, b + c - C)),
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# 从c桶倒到a桶
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(min(a + c, A), b, max(0, a + c - A)),
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# 从c桶倒到b桶
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(a, min(b + c, B), max(0, b + c - B)),
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# 清空每个桶
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(0, b, c),
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(a, 0, c),
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(a, b, 0),
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# 将每个桶填满
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(A, b, c),
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(a, B, c),
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(a, b, C)
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]
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# 遍历所有新状态
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for state in next_states:
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if state not in visited:
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visited.add(state)
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queue.append(state)
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# 增加操作次数
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steps += 1
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return -1 # 如果找不到目标
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# 容器的容量
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capacities = [160, 119, 77]
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# 目标水量
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target = 76
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# 求解最少操作次数
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result = bfs(capacities, target)
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print("最少操作次数:", result)
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