用Python语言写一个完全背包算法,不导入库就可以使用
时间: 2024-10-13 21:15:05 浏览: 7
在Python中,我们可以手动编写一个完全背包算法来解决背包问题,这是一种动态规划方法。以下是不依赖于外部库的一个简单实现:
```python
def knapsack(weights, values, capacity):
n = len(weights)
dp = [[0] * (capacity + 1) for _ in range(n + 1)]
# 动态规划填充过程
for i in range(1, n + 1):
for w in range(1, capacity + 1):
if weights[i - 1] <= w:
dp[i][w] = max(dp[i - 1][w], dp[i - 1][w - weights[i - 1]] + values[i - 1])
else:
dp[i][w] = dp[i - 1][w]
# 返回最大价值
return dp[n][capacity]
# 示例数据
weights = [5, 4, 6]
values = [10, 20, 30]
capacity = 8
print("最大价值:", knapsack(weights, values, capacity))
相关问题
用Python语言写一个完全背包算法,不导入库就可以使用,并且输出所放物品和价值
在Python中,我们可以手动编写一个基本的完全背包算法,不依赖于内置库。这里我们创建一个简单的函数`knapsack()`来实现这个过程。完全背包是指每个物品可以取任意次数,而不仅仅是取一次。
```python
def knapsack(capacity, weights, values, n):
# 创建二维数组,用于存储动态规划的结果
dp = [[0 for _ in range(capacity + 1)] for _ in range(n + 1)]
# 动态规划遍历所有物品和背包容量
for i in range(1, n + 1):
for w in range(1, capacity + 1):
if weights[i - 1] <= w:
# 如果当前物品重量小于等于剩余背包容量,则可以选择放入,取最大价值
dp[i][w] = max(dp[i - 1][w], dp[i - 1][w - weights[i - 1]] + values[i - 1])
else:
# 否则,如果不选择该物品,价值不变
dp[i][w] = dp[i - 1][w]
# 返回最优解
return dp[n][capacity], get_items(dp, capacity, weights, values)
def get_items(dp, capacity, weights, values, current_weight=0, item_index=-1 or current_weight > capacity:
return []
elif dp[item_index][current_weight] == dp[item_index - 1][current_weight]:
return get_items(dp, capacity, weights, values, current_weight, item_index - 1)
else:
result = [item_index]
return get_items(dp, capacity, weights, values, current_weight - weights[item_index], item_index + 1) + result
# 使用示例
capacity = 50
weights = [10, 20, 30]
values = [60, 100, 120]
n = len(weights)
optimal_value, selected_items = knapsack(capacity, weights, values, n)
print(f"最优价值: {optimal_value}")
print("选出的物品及价值:")
for i in selected_items:
print(f"物品{i}: {weights[i]}, 价值{values[i]}")
用Python语言写一个完全背包算法,不导入库就可以使用,并且输出所放物品和价值,代码简洁
当然可以。完全背包算法(0-1 Knapsack Problem)是一个经典的动态规划问题,我们可以通过循环和条件判断来实现。这里是一个简单的Python代码示例:
```python
def knap_sack(weights, values, capacity):
n = len(weights)
dp = [[0 for _ in range(capacity + 1)] for _ in range(n + 1)]
# 构建动态规划表
for i in range(1, n+1):
for w in range(1, capacity + 1):
if weights[i - 1] <= w:
dp[i][w] = max(values[i - 1] + dp[i - 1][w - weights[i - 1]], dp[i - 1][w])
else:
dp[i][w] = dp[i - 1][w]
# 找出最优解并记录物品
selected_items = []
w = capacity
for i in reversed(range(n)):
if dp[i][w] != dp[i - 1][w]:
selected_items.append((i, weights[i], values[i]))
w -= weights[i]
return dp[n][capacity], selected_items
# 示例输入
weights = [2, 3, 4, 5]
values = [6, 9, 7, 10]
capacity = 8
max_value, items = knap_sack(weights, values, capacity)
print(f"最大价值:{max_value}")
print("选择的物品及价值:")
for item in items:
print(f"物品 {item[0]+1},重量 {item[1]},价值 {item[2]}")
#
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