[LeetCode] Knapsack Problem背包问题

1. 动态规划

1. 三要素：定义状态，分解子问题（找到迭代公式），设置边界条件
2. 三种解题思路：
   1. Brutal Force Search
   2. Top Down（回溯+剪枝）
   3. Bottom Up（memorization）

2. 解题模板

2.1 题目定义

1. \(v[i]\) 第i件物品的价值
2. \(w[i]\)第i件物品的重量
3. \(C\)背包的总容量
4. 是否可重复

2.2 状态转移方程

设\(dp[i, j]\)，有i个物品，背包剩余容量为j。则动态转移方程：

2.1 【存在性】是否能够恰好装满容量为C的背包

\[f(i, j) = \begin{cases} f(i-1, j)|f(i-1, j-w[i])& \text{without repetition}\\ f(i-1, j)|f(i-1, j-w[i])|\cdots f(i-1, j-k\times w[i])& \text{with repetition} \end{cases}\]

2.2 【计数】容量为C的背包最多/少可以装多少东西

\[f(i, j) = \begin{cases} max\{f(i-1, j-w[i])+1, f(i-1, j)\}& \text{without repetition}\\ max\{f(i-1, j-k\times w[i]) + k \mid k \in [0, \frac{C}{v[i]}]\}& \text{with repetition} \end{cases}\]

2.3 【计数】装满容量为C的背包一共有多少方案

\[f(i, j) = \begin{cases} f(i-1, j-w[i]) + f(i-1, j)& \text{without repetition}\\ \sum_0^{C//v[i]} f(i-1, j-k\times w[i]) \mid k \in [0, \frac{C}{v[i]}]\}& \text{with repetition} \end{cases}\]

3. 例题

3.1 LeetCode 474.一和零

class Solution:
	def findMaxForm_TopDown(self, strs: list, m: int, n: int) -> int:

		# knapsack problem without repitition
		# recursion equation (take it or leave it):
		#   f(i, m, n) = max{f(i-1, m-numZeros[i], n-numOnes[i]) + 1, f(i-1, m, n)}
		# f(i, m, n) implies the maximum amount of strings that m 0s and n 1s can spell out.

		# TOP-DOWN

		def getMaxAmount(strs, m, n):
		
			if strs and m >= 0 and n >= 0 and not (m == 0 and n == 0):
				num_TakeIt = getMaxAmount(strs[1:], m - strs[0].count('0'), n - strs[0].count('1')) + 1
				num_LeaveIt = getMaxAmount(strs[1:], m, n)
				return max(num_TakeIt, num_LeaveIt)
			elif not strs or m <= 0 or n <= 0:
				if m >= 0 and n >= 0: return 0
				else: return -1  # if m < 0 or n < 0 or not strs:
		
		return getMaxAmount(strs, m, n)

	def findMaxForm_BottomUp(self, strs: list, m: int, n: int) -> int:

		import numpy as np
		# BOTTOM-UP
		dp = np.zeros((len(strs)+1, m+1, n+1), dtype=np.int32)

		# Initialization
		for i in range(m+1):
			for j in range(n+1):
				dp[0, i, j] = 0
		for i in range(1, len(strs)+1):
			dp[i, 0, 0] = 1

		for i in range(1, len(strs)+1):
			for j in range(m+1):
				for k in range(n+1):
					if j - strs[i - 1].count('0') >= 0 and k - strs[i - 1].count('1') >= 0:
						dp[i, j, k] = max(dp[i-1, j-strs[i-1].count('0'), k-strs[i-1].count('1')] + 1, dp[i-1, j, k])
					else:
						dp[i, j, k] = dp[i-1, j, k]

		return int(dp[len(strs), m, n])

两种方案都超时了，但是重要的是思路给掌握了。代码以后还要继续优化，使之通过OJ。超时的数据：

["0","0","1","1","1","0","1","0","0","1","1","0","1","0","1","0","1","0","0","1","0","1","0","0","1","1","1","0","1","1","0","0","1","1","1","0","1","0","0","0","1","0","1","0","0","1","0","0","1","1","1","1","1","0","0","1","0","1","0","1","1","0","0","0","1","1","1","1","1","1","0","1","1","1","0","0","1","1","0","0","1","1","0","1","0","0","1"]
93
91


["0","11","1000","01","0","101","1","1","1","0","0","0","0","1","0","0110101","0","11","01","00","01111","0011","1","1000","0","11101","1","0","10","0111"]
9
80

3.2 LeetCode518.零钱兑换2

class Solution:
    def change(self, amount: int, coins: List[int]) -> int:
        import numpy as np

        dp = np.zeros([len(coins) + 1, amount + 1])
        for i in range(0, len(coins) + 1): dp[i, 0] = 1
        for j in range(1, amount + 1): dp[0, j] = 0

        for i in range(1, len(coins) + 1):
            for j in range(1, amount + 1):
                dp[i, j] = dp[i - 1, j]
                if j-coins[i-1] >= 0: dp[i][j] += dp[i][j-coins[i-1]];

        return int(dp[-1, -1])