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[pull] main from itcharge:main #72

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Merged
pull merged 7 commits into AlgorithmAndLeetCode:main from itcharge:main
Mar 25, 2023
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View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -67,14 +67,15 @@ $dp[i][w] = \begin{cases} dp[i - 1][w] & w < weight[i - 1] \cr max \lbrace dp[i

```Python
class Solution:
# 思路 1:动态规划 + 二维基本思路
def zeroOnePackMethod1(self, weight: [int], value: [int], W: int):
size = len(weight)
dp = [[0 for _ in range(W + 1)] for _ in range(size + 1)]

# 枚举前 i 种物品
for i in range(1, size + 1):
# 枚举背包装载重量
for w in range(1, W + 1):
for w in range(W + 1):
# 第 i - 1 件物品装不下
if w < weight[i - 1]:
# dp[i][w] 取「前 i - 1 件物品装入载重为 w 的背包中的最大价值」
Expand Down Expand Up @@ -137,6 +138,7 @@ $dp[w] = \begin{cases} dp[w] & w < weight[i - 1] \cr max \lbrace dp[w], dp[w - w

```Python
class Solution:
# 思路 2:动态规划 + 滚动数组优化
def zeroOnePackMethod2(self, weight: [int], value: [int], W: int):
size = len(weight)
dp = [0 for _ in range(W + 1)]
Expand Down Expand Up @@ -232,14 +234,18 @@ $dp[w] = \begin{cases} dp[w] & w < nums[i - 1] \cr max \lbrace dp[w], \quad dp[w

```Python
class Solution:
# 思路 2:动态规划 + 滚动数组优化
def zeroOnePackMethod2(self, weight: [int], value: [int], W: int):
size = len(weight)
dp = [0 for _ in range(W + 1)]

# 枚举前 i 种物品
for i in range(1, size + 1):
# 逆序枚举背包装载重量(避免状态值错误)
for w in range(W, weight[i - 1] - 1, -1):
# dp[w] 取「前 i - 1 件物品装入载重为 w 的背包中的最大价值」与「前 i - 1 件物品装入载重为 w - weight[i - 1] 的背包中,再装入第 i - 1 物品所得的最大价值」两者中的最大值
dp[w] = max(dp[w], dp[w - weight[i - 1]] + value[i - 1])

return dp[W]

def canPartition(self, nums: List[int]) -> bool:
Expand All @@ -248,7 +254,7 @@ class Solution:
return False

target = sum_nums // 2
return self.zeroOnePackOptimization(nums, nums, target) == target
return self.zeroOnePackMethod2(nums, nums, target) == target
```

##### 思路 1:复杂度分析
Expand Down
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -51,14 +51,15 @@ $dp[i][w] = max \lbrace dp[i - 1][w - k \times weight[i - 1]] + k \times value[i

```Python
class Solution:
# 思路 1:动态规划 + 二维基本思路
def completePackMethod1(self, weight: [int], value: [int], W: int):
size = len(weight)
dp = [[0 for _ in range(W + 1)] for _ in range(size + 1)]

# 枚举前 i 种物品
for i in range(1, size + 1):
# 枚举背包装载重量
for w in range(1, W + 1):
for w in range(W + 1):
# 枚举第 i 种物品能取个数
for k in range(w // weight[i - 1] + 1):
# dp[i][w] 取所有 dp[i - 1][w - k * weight[i - 1] + k * value[i - 1] 中最大值
Expand Down Expand Up @@ -142,15 +143,16 @@ $\quad dp[i][w] = \begin{cases} dp[i - 1][w] & w < weight[i - 1] \cr max \lbrac
#### 思路 2:代码

```Python
class Solution:
class Solution:
# 思路 2:动态规划 + 状态转移方程优化
def completePackMethod2(self, weight: [int], value: [int], W: int):
size = len(weight)
dp = [[0 for _ in range(W + 1)] for _ in range(size + 1)]

# 枚举前 i 种物品
for i in range(1, size + 1):
# 枚举背包装载重量
for w in range(1, W + 1):
for w in range(W + 1):
# 第 i - 1 件物品装不下
if w < weight[i - 1]:
# dp[i][w] 取「前 i - 1 种物品装入载重为 w 的背包中的最大价值」
Expand Down Expand Up @@ -209,6 +211,7 @@ $dp[w] = \begin{cases} dp[w] & w < weight[i - 1] \cr max \lbrace dp[w], \quad d

```Python
class Solution:
# 思路 3:动态规划 + 滚动数组优化
def completePackMethod3(self, weight: [int], value: [int], W: int):
size = len(weight)
dp = [0 for _ in range(W + 1)]
Expand Down
9 changes: 6 additions & 3 deletions Solutions/0416. 分割等和子集.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -73,14 +73,17 @@ $dp[w] = \begin{cases} dp[w] & w < nums[i - 1] \cr max \lbrace dp[w], dp[w - num

```Python
class Solution:
def zeroOnePackOptimization(self, weight: [int], value: [int], W: int):
def zeroOnePack(self, weight: [int], value: [int], W: int):
size = len(weight)
dp = [0 for _ in range(W + 1)]

# 枚举前 i 种物品
for i in range(1, size + 1):
# 逆序枚举背包装载重量(避免状态值错误)
for w in range(W, weight[i - 1] - 1, -1):
# dp[w] 取「前 i - 1 件物品装入载重为 w 的背包中的最大价值」与「前 i - 1 件物品装入载重为 w - weight[i - 1] 的背包中,再装入第 i - 1 物品所得的最大价值」两者中的最大值
dp[w] = max(dp[w], dp[w - weight[i - 1]] + value[i - 1])

return dp[W]

def canPartition(self, nums: List[int]) -> bool:
Expand All @@ -89,7 +92,7 @@ class Solution:
return False

target = sum_nums // 2
return self.zeroOnePackOptimization(nums, nums, target) == target
return self.zeroOnePack(nums, nums, target) == target
```

### 思路 1:复杂度分析
Expand Down
4 changes: 2 additions & 2 deletions Templates/10.Dynamic-Programming/Pack-CompletePack.py
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -7,7 +7,7 @@ def completePackMethod1(self, weight: [int], value: [int], W: int):
# 枚举前 i 种物品
for i in range(1, size + 1):
# 枚举背包装载重量
for w in range(1, W + 1):
for w in range(W + 1):
# 枚举第 i 种物品能取个数
for k in range(w // weight[i - 1] + 1):
# dp[i][w] 取所有 dp[i - 1][w - k * weight[i - 1] + k * value[i - 1] 中最大值
Expand All @@ -23,7 +23,7 @@ def completePackMethod2(self, weight: [int], value: [int], W: int):
# 枚举前 i 种物品
for i in range(1, size + 1):
# 枚举背包装载重量
for w in range(1, W + 1):
for w in range(W + 1):
# 第 i - 1 件物品装不下
if w < weight[i - 1]:
# dp[i][w] 取「前 i - 1 种物品装入载重为 w 的背包中的最大价值」
Expand Down
37 changes: 37 additions & 0 deletions Templates/10.Dynamic-Programming/Pack-GroupPack.py
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,37 @@
class Solution:
# 思路 1:动态规划 + 二维基本思路
def groupPackMethod1(self, group_count: [int], weight: [[int]], value: [[int]], W: int):
size = len(group_count)
dp = [[0 for _ in range(W + 1)] for _ in range(size + 1)]

# 枚举前 i 组物品
for i in range(1, size + 1):
# 枚举背包装载重量
for w in range(W + 1):
# 枚举第 i 组物品能取个数
dp[i][w] = dp[i - 1][w]
for k in range(group_count[i - 1]):
if w >= weight[i - 1][k]:
# dp[i][w] 取所有 dp[i - 1][w - weight[i - 1][k]] + value[i - 1][k] 中最大值
dp[i][w] = max(dp[i][w], dp[i - 1][w - weight[i - 1][k]] + value[i - 1][k])

return dp[size][W]

# 思路 2:动态规划 + 滚动数组优化
def groupPackMethod2(self, group_count: [int], weight: [[int]], value: [[int]], W: int):
size = len(group_count)
dp = [0 for _ in range(W + 1)]

# 枚举前 i 组物品
for i in range(1, size + 1):
# 逆序枚举背包装载重量
for w in range(W, -1, -1):
# 枚举第 i 组物品能取个数
for k in range(group_count[i - 1]):
if w >= weight[i - 1][k]:
# dp[i][w] 取所有 dp[i - 1][w - weight[i - 1][k]] + value[i - 1][k] 中最大值
dp[w] = max(dp[w], dp[w - weight[i - 1][k]] + value[i - 1][k])

return dp[W]


3 changes: 2 additions & 1 deletion Templates/10.Dynamic-Programming/Pack-MultiplePack.py
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -7,7 +7,7 @@ def multiplePackMethod1(self, weight: [int], value: [int], count: [int], W: int)
# 枚举前 i 种物品
for i in range(1, size + 1):
# 枚举背包装载重量
for w in range(1, W + 1):
for w in range(W + 1):
# 枚举第 i 种物品能取个数
for k in range(min(count[i - 1], w // weight[i - 1]) + 1):
# dp[i][w] 取所有 dp[i - 1][w - k * weight[i - 1] + k * value[i - 1] 中最大值
Expand Down Expand Up @@ -35,6 +35,7 @@ def multiplePackMethod2(self, weight: [int], value: [int], count: [int], W: int)
def multiplePackMethod3(self, weight: [int], value: [int], count: [int], W: int):
weight_new, value_new = [], []

# 二进制优化
for i in range(len(weight)):
cnt = count[i]
k = 1
Expand Down
2 changes: 1 addition & 1 deletion Templates/10.Dynamic-Programming/Pack-ZeroOnePack.py
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -7,7 +7,7 @@ def zeroOnePackMethod1(self, weight: [int], value: [int], W: int):
# 枚举前 i 种物品
for i in range(1, size + 1):
# 枚举背包装载重量
for w in range(1, W + 1):
for w in range(W + 1):
# 第 i - 1 件物品装不下
if w < weight[i - 1]:
# dp[i][w] 取「前 i - 1 件物品装入载重为 w 的背包中的最大价值」
Expand Down

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