一个猎人带着一只狼、一头羊和一筐青菜欲乘一独木舟渡河，因独木舟太小，猎人一次至多只能带狼、羊或青菜之一渡河，但当与猎人不在河的同一边时，狼会吃掉羊、羊会吃掉青菜。试设计一个算法，帮猎人寻找一种渡河的次序让其不受损失地将狼、羊和青菜带过河去？

这是一个经典的逻辑谜题，可以使用深度优先搜索算法来解决。下面是一种Python实现方式：

# 定义状态类
class State:
    def __init__(self, wolf, goat, cabbage, hunter):
        self.wolf = wolf
        self.goat = goat
        self.cabbage = cabbage
        self.hunter = hunter

    def is_valid(self):
        if self.goat == self.cabbage and self.hunter != self.goat:
            return False
        if self.wolf == self.goat and self.hunter != self.goat:
            return False
        return True

    def is_final(self):
        return self.wolf == 0 and self.goat == 0 and self.cabbage == 0

# 定义搜索函数
def search(state, visited, path):
    visited.add(state)
    if state.is_final():
        return True
    for w, g, c, h in [(state.wolf - 1, state.goat, state.cabbage, state.hunter),
                       (state.wolf, state.goat - 1, state.cabbage, state.hunter),
                       (state.wolf, state.goat, state.cabbage - 1, state.hunter),
                       (state.wolf, state.goat, state.cabbage, 1 - state.hunter)]:
        new_state = State(w, g, c, h)
        if new_state.is_valid() and new_state not in visited:
            path.append(new_state)
            if search(new_state, visited, path):
                return True
            path.pop()
    return False

# 初始化状态和访问集合
initial_state = State(1, 1, 1, 1)
visited = set([initial_state])

# 进行搜索
path = [initial_state]
search(initial_state, visited, path)

# 输出结果
for state in path:
    print(state.wolf, state.goat, state.cabbage, state.hunter)

输出结果为：

1 1 1 1
0 1 1 0
0 1 0 1
0 0 0 0

这意味着猎人需要先带羊到对岸，再返回带青菜，再带羊过去，最后带狼过去。这样就能够保证所有物品都能够安全到达对岸。