lavos/python/cap5/work6.py

# 等分液体
# 在一个瓶子中装有8（一般情形下为偶数N）升汽油，要平均分成两份，
# 但只有一个装3（一般情形下为N/2-1）升和装5（一般情形下为N/2+1）升的量杯（都没有刻度）。
# 若有解，则打印出所有把汽油分成两等分的操作过程。若无解，则打印“No”。

from collections import deque


def pour(source, target, target_capacity):
    amount = min(source, target_capacity - target)
    return source - amount, target + amount


def get_next_states(state, capacities):
    next_states = []
    bottle, cup3, cup5 = state
    max_bottle, max_cup3, max_cup5 = capacities

    # bottle -> cup3
    next_states.append(pour(bottle, cup3, max_cup3) + (cup5,))
    # bottle -> cup5
    next_states.append(pour(bottle, cup5, max_cup5) + (cup3,))
    # cup3 -> bottle
    next_states.append(
        (pour(cup3, bottle, max_bottle)[1],) + pour(cup3, bottle, max_bottle)
    )
    # cup5 -> bottle
    next_states.append(
        (pour(cup5, bottle, max_bottle)[1], cup3, pour(cup5, bottle, max_bottle)[0])
    )
    # cup3 -> cup5
    next_states.append((bottle,) + pour(cup3, cup5, max_cup5))
    # cup5 -> cup3
    next_states.append((bottle,) + pour(cup5, cup3, max_cup3)[::-1])

    return next_states


def bfs(initial_state, capacities, target):
    queue = deque([(initial_state, [])])
    visited = set()

    while queue:
        (state, path) = queue.popleft()

        if (state[0] == target and state[1] == target) or (
            state[0] == target and state[2] == target
        ):
            return path

        if state in visited:
            continue

        visited.add(state)

        for next_state in get_next_states(state, capacities):
            if next_state not in visited:
                queue.append((next_state, path + [next_state]))

    return "No"


N = 8
capacities = (N, N // 2 - 1, N // 2 + 1)
initial_state = (N, 0, 0)
target = N // 2

result = bfs(initial_state, capacities, target)

if result == "No":
    print("No")
else:
    print("状态步骤：(汽油瓶, 5L杯, 3L杯)")
    for step in result:
        print(step)