代码随性训练营第四十六天（Python）｜ 139.单词拆分 、多重背包

139.单词拆分

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        dp = [False] * (len(s) + 1)
        dp[0] = True

        # 求排列先遍历背包再遍历物品
        for i in range(len(s) + 1):
            for j in range(i):
                if dp[j] and s[j:i] in wordDict:
                    dp[i] = True # dp[j] 可以被拆分单词，并且 s[j:i] 在字典中。s[0:i] 也可以被拆分单词
                    break
        return dp[len(s)]

[多重背包]
有N种物品和一个容量为V 的背包。第i种物品最多有Mi件可用，每件耗费的空间是Ci ，价值是Wi 。求解将哪些物品装入背包可使这些物品的耗费的空间 总和不超过背包容量，且价值总和最大。

def multi_bag(weight, value, count, bag_weight):
    dp = [0] * (bag_weight + 1)

    for i in range(len(weight)): # 先遍历物品
        for j in range(bag_weight, weight[i] - 1, -1): # 再遍历背包
            # 遍历个数
            for k in range(1, count[i] + 1):
                if j - k * weight[i] >= 0:
                    dp[j] = max(dp[j], dp[j - k * weight[i]] + k * value[i])

        # 打印 dp
        print(" ".join([str(dp[j]) for j in range(bag_weight + 1)]))

    return dp[bag_weight]


if __name__ == "__main__":
    weight = [1, 3, 4]
    value = [15, 20, 30]
    bag_weight = 4
    count = [2, 3, 2]
    res = multi_bag(weight, value, count, bag_weight)
    print(res)