> For the complete documentation index, see [llms.txt](https://docs.1l.lol/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://docs.1l.lol/algorithms/knapsack.md).

# 简单背包问题

### 01 背包

有 n 件物品，每件有重量 w\[i] 和价值 v\[i]，背包容量为 C，每件**最多取一次**，求最大价值。

状态：`dp[j]` 表示容量为 j 时的最大价值。

```cpp
for (int i = 1; i <= n; i++)
    for (int j = C; j >= w[i]; j--)        // 逆序遍历，保证每件只用一次
        dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
cout << dp[C];
```

### 完全背包

每件物品可以取**无限次**，只需把内层循环改为**正序**：

```cpp
for (int i = 1; i <= n; i++)
    for (int j = w[i]; j <= C; j++)
        dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
```

### 为什么 01 背包要逆序

正序会让同一件物品被重复使用（变成完全背包）；逆序时 dp\[j-w\[i]] 还是“上一轮”的状态，保证只用一次。

### 空间优化

用一维数组滚动即可，无需二维 dp\[i]\[j]，空间从 O(nC) 降到 O(C)。


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