01背包问题是算法竞赛中经典的dp题目。文中总共包含五个版本的代码。从最朴素的枚举法开始，在不断的改进下，最终变成了dp解法。

问题定义

有若干个物品，每件物品的有重量weight和价值value：

struct Item {
  
Int
weight : 
Int
Int
  
Int
value : 
Int
Int
}

现在，给定一个物品列表items，和背包的容量capacity。从中选出若干件物品，使得这些物品的总重量不超过背包的容量，且物品的总价值最大。

typealias 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
@list.T as List

let 
@list.List[Item]
items_1 : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Item {
  weight: Int
  value: Int
}
Item] = 
(arr : FixedArray[Item]) -> @list.List[Item]
Create a list from a FixedArray.
Converts a FixedArray into a list with the same elements in the same order.
Example
let ls = @list.of([1, 2, 3, 4, 5])
assert_eq(ls.to_array(), [1, 2, 3, 4, 5])
@list.of([
  { 
Int
weight: 7, 
Int
value: 20 },
  { 
Int
weight: 4, 
Int
value: 10 },
  { 
Int
weight: 5, 
Int
value: 11 },
])

以上面的items_1为例，假设背包容量是$10$，那么最优的方案是选取后两个物品，占用$4+5=9$的容量，总共有$10+11=21$点价值。

注意，由于我们不能把物品切割，因此优先挑选性价比最高的物品并非正解。例如，在上面的例子中，若选取了性价比最高的物品1，则只有$20$点价值，而此时背包已经放不下其他物品了。

问题建模

我们先定义一些基础的对象与操作。

//物品的组合，下文简称组合
struct Combination {
  
@list.List[Item]
items : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Item {
  weight: Int
  value: Int
}
Item]
  
Int
total_weight : 
Int
Int
  
Int
total_value : 
Int
Int
}

//空的组合
let 
Combination
empty_combination : 
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination = {
  
@list.List[Item]
items: 
() -> @list.List[Item]
Creates an empty list.
Example
let ls : List[Int] = @list.empty()
assert_eq(ls.length(), 0)
@list.empty(),
  
Int
total_weight: 0,
  
Int
total_value: 0,
}

//往组合中添加物品，得到新的组合
fn 
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination::
(self : Combination, item : Item) -> Combination
add(
Combination
self : 
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination, 
Item
item : 
struct Item {
  weight: Int
  value: Int
}
Item) -> 
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination {
  {
    
@list.List[Item]
items: 
Combination
self.
@list.List[Item]
items.
(self : @list.List[Item], head : Item) -> @list.List[Item]
Add an element to the front of the list.
This is an alias for prepend - it creates a new list with the given
element added to the beginning.
Example
let ls = @list.of([2, 3, 4]).add(1)
assert_eq(ls, @list.of([1, 2, 3, 4]))
add(
Item
item),
    
Int
total_weight: 
Combination
self.
Int
total_weight 
(self : Int, other : Int) -> Int
Adds two 32-bit signed integers. Performs two's complement arithmetic, which
means the operation will wrap around if the result exceeds the range of a
32-bit integer.
Parameters:

self : The first integer operand.
other : The second integer operand.
Returns a new integer that is the sum of the two operands. If the
mathematical sum exceeds the range of a 32-bit integer (-2,147,483,648 to
2,147,483,647), the result wraps around according to two's complement rules.
Example:
  inspect(42 + 1, content="43")
  inspect(2147483647 + 1, content="-2147483648") // Overflow wraps around to minimum value
+ 
Item
item.
Int
weight,
    
Int
total_value: 
Combination
self.
Int
total_value 
(self : Int, other : Int) -> Int
Adds two 32-bit signed integers. Performs two's complement arithmetic, which
means the operation will wrap around if the result exceeds the range of a
32-bit integer.
Parameters:

self : The first integer operand.
other : The second integer operand.
Returns a new integer that is the sum of the two operands. If the
mathematical sum exceeds the range of a 32-bit integer (-2,147,483,648 to
2,147,483,647), the result wraps around according to two's complement rules.
Example:
  inspect(42 + 1, content="43")
  inspect(2147483647 + 1, content="-2147483648") // Overflow wraps around to minimum value
+ 
Item
item.
Int
value,
  }
}

//两个组合等效，意思是它们总价值一样
impl 
trait Eq {
  op_equal(Self, Self) -> Bool
}
Trait for types whose elements can test for equality
Eq for 
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination with 
(self : Combination, other : Combination) -> Bool
op_equal(
Combination
self, 
Combination
other) {
  
Combination
self.
Int
total_value 
(self : Int, other : Int) -> Bool
Compares two integers for equality.
Parameters:

self : The first integer to compare.
other : The second integer to compare.
Returns true if both integers have the same value, false otherwise.
Example:
  inspect(42 == 42, content="true")
  inspect(42 == -42, content="false")
== 
Combination
other.
Int
total_value
}

//比较两个组合的大小，就是比较它们总价值的大小
impl 
trait Compare {
  compare(Self, Self) -> Int
}
Trait for types whose elements are ordered
The return value of [compare] is:

zero, if the two arguments are equal
negative, if the first argument is smaller
positive, if the first argument is greater
Compare for 
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination with 
(self : Combination, other : Combination) -> Int
compare(
Combination
self, 
Combination
other) {
  
Combination
self.
Int
total_value.
(self : Int, other : Int) -> Int
Compares two integers and returns their relative order.
Parameters:

self : The first integer to compare.
other : The second integer to compare against.
Returns an integer indicating the relative order:

A negative value if self is less than other
Zero if self equals other
A positive value if self is greater than other
Example:
  let a = 42
  let b = 24
  inspect(a.compare(b), content="1") // 42 > 24
  inspect(b.compare(a), content="-1") // 24 < 42
  inspect(a.compare(a), content="0") // 42 = 42
compare(
Combination
other.
Int
total_value)
}

然后，我们就可以开始思考如何解决问题了。

一、朴素的枚举

枚举法是最朴素的方案，我们依照问题的定义，一步一步执行，就能得到答案：

1. 枚举出所有的组合；
2. 过滤出其中有效的组合，也就是那些能装入背包的；
3. 答案是其中总价值最大的那个。

得益于标准库提供的两个函数，我们可以将上面三行文字一比一地翻译为MoonBit代码。其中all_combinations是我们后续需要实现的函数，它的类型是(List[Item]) -> List[Combination]。

fn 
(items : @list.List[Item], capacity : Int) -> Combination
solve_v1(
@list.List[Item]
items : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Item {
  weight: Int
  value: Int
}
Item], 
Int
capacity : 
Int
Int) -> 
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination {
  
(items : @list.List[Item]) -> @list.List[Combination]
all_combinations(
@list.List[Item]
items)
  .
(self : @list.List[Combination], f : (Combination) -> Bool) -> @list.List[Combination]
Filter the list.
Example
assert_eq(@list.of([1, 2, 3, 4, 5]).filter(x => x % 2 == 0), @list.of([2, 4]))
filter(fn(
Combination
comb) { 
Combination
comb.
Int
total_weight 
(self_ : Int, other : Int) -> Bool
<= 
Int
capacity })
  .
(self : @list.List[Combination]) -> Combination
Get the maximum element of the list.
Warning: This function panics if the list is empty.
Use maximum() for a safe alternative that returns Option.
Example
let ls = @list.of([1, 3, 2, 5, 4])
assert_eq(ls.unsafe_maximum(), 5)
Panics
Panics if the list is empty.
unsafe_maximum()
}

注意这里使用的是unsafe_maximum而不是maximum。这是因为空列表列表中没有最大值，maximum在这种情况下会返回一个None。但我们知道，题目保证答案存在（只要capacity不是负数），所以我们可以改用unsafe_maximum。它在输入空列表的情况下直接中断程序，其它情况返回列表中的最大值。

接下来我们去实现枚举的过程。函数all_combinations接受一个物品的列表，返回一个组合的列表，其中包含所有能由这些物品构造出的组合。也许你现在没有头绪，这时我们可以先查看一下列表的定义。它大概长这样：

enum List[A] {
  Empty
  More(A, tail~ : List[A])
}

也就是说，列表分为两种：

1. 第一种是空的列表，叫Empty；
2. 第二种是非空的列表，叫More，其中包含了第一个元素（A）和剩余的部分（tail~ : T[A]），剩余部分也是一个列表。

这启示我们按物品列表是否为空来分情况讨论：

• 如果物品列表为空，那么唯一的一种组合方式就是空的组合；
• 否则，一定存在第一个物品item1和剩余部分items_tail。这种情况下，我们可以：
  1. 先求出不含item1的那些组合。这其实就是items_tail能凑出的那些组合，可以递归地求出。
  2. 再求出包含item1的那些组合。它们与不含item1的组合一一对应，只差把item1加入其中。
  3. 将这两者合并起来，就是所有items能凑出的组合。

例如，当物品列表包含a,b,c三个元素时，答案分为以下两个部分：

不含a的部分	包含a的部分
{ }	{ a }
{ b }	{ a, b }
{ c }	{ a, c }
{ b, c }	{ a, b, c }

fn 
(items : @list.List[Item]) -> @list.List[Combination]
all_combinations(
@list.List[Item]
items : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Item {
  weight: Int
  value: Int
}
Item]) -> 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination] {
  match 
@list.List[Item]
items {
    
@list.List[Item]
Empty => 
(x : Combination) -> @list.List[Combination]
Create a list with a single element.
Returns a list containing only the given element.
Example
  let ls = @list.singleton(42)
  assert_eq(ls, @list.of([42]))
  assert_eq(ls.length(), 1)
@list.singleton(
Combination
empty_combination)
    
(Item, @list.List[Item]) -> @list.List[Item]
More(
Item
item1, 
@list.List[Item]
tail=
@list.List[Item]
items_tail) => {
      let 
@list.List[Combination]
combs_without_item1 = 
(items : @list.List[Item]) -> @list.List[Combination]
all_combinations(
@list.List[Item]
items_tail)
      let 
@list.List[Combination]
combs_with_item1 = 
@list.List[Combination]
combs_without_item1.
(self : @list.List[Combination], f : (Combination) -> Combination) -> @list.List[Combination]
Maps the list.
Example
assert_eq(@list.of([1, 2, 3, 4, 5]).map(x => x * 2), @list.of([2, 4, 6, 8, 10]))
map(_.
(self : Combination, item : Item) -> Combination
add(
Item
item1))
      
@list.List[Combination]
combs_with_item1 
(self : @list.List[Combination], other : @list.List[Combination]) -> @list.List[Combination]
Add implementation for List - concatenates two lists.
The + operator for lists performs concatenation.
a + b is equivalent to a.concat(b).
Example
let a = @list.of([1, 2, 3])
let b = @list.of([4, 5, 6])
let result = a + b
assert_eq(result, @list.of([1, 2, 3, 4, 5, 6]))
+ 
@list.List[Combination]
combs_without_item1
    }
  }
}

通过使用模式匹配（match），我们再一次将上面的五行文字一比一地翻译成了MoonBit代码。

二、提前过滤，仅枚举有效的组合

在第一个版本中，枚举所有组合和过滤出能放入背包的组合是不相干的两个过程。在枚举的过程中，出现了很多无效的组合。这些组合早已放不进背包中，却还在后续的过程中被添加物品。不如早一点过滤它们，避免在它之上不断产生新的无效组合。观察代码，发现无效的组合只会在.map(_.add(item1))这一步产生。于是我们可以做出改进：仅向能再装下item1的组合添加item1。

我们将all_combinations改为all_combinations_valid，仅返回能装入这个背包的组合。现在枚举和过滤将交替进行。

fn 
(items : @list.List[Item], capacity : Int) -> @list.List[Combination]
all_combinations_valid(
  
@list.List[Item]
items : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Item {
  weight: Int
  value: Int
}
Item],
  
Int
capacity : 
Int
Int // 添加一个参数，因为过滤需要知道背包的容量
) -> 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination] {
  match 
@list.List[Item]
items {
    
@list.List[Item]
Empty => 
(x : Combination) -> @list.List[Combination]
Create a list with a single element.
Returns a list containing only the given element.
Example
  let ls = @list.singleton(42)
  assert_eq(ls, @list.of([42]))
  assert_eq(ls.length(), 1)
@list.singleton(
Combination
empty_combination) // 空的组合自然是有效的
    
(Item, @list.List[Item]) -> @list.List[Item]
More(
Item
item1, 
@list.List[Item]
tail=
@list.List[Item]
items_tail) => {
      // 我们假设 all_combinations_valid 返回的组合都是有效的（归纳假设）
      let 
@list.List[Combination]
valid_combs_without_item1 = 
(items : @list.List[Item], capacity : Int) -> @list.List[Combination]
all_combinations_valid(
        
@list.List[Item]
items_tail, 
Int
capacity,
      )
      // 由于添加了过滤，所以它里面的组合都是有效的
      let 
@list.List[Combination]
valid_combs_with_item1 = 
@list.List[Combination]
valid_combs_without_item1
        .
(self : @list.List[Combination], f : (Combination) -> Bool) -> @list.List[Combination]
Filter the list.
Example
assert_eq(@list.of([1, 2, 3, 4, 5]).filter(x => x % 2 == 0), @list.of([2, 4]))
filter(fn(
Combination
comb) { 
Combination
comb.
Int
total_weight 
(self : Int, other : Int) -> Int
Adds two 32-bit signed integers. Performs two's complement arithmetic, which
means the operation will wrap around if the result exceeds the range of a
32-bit integer.
Parameters:

self : The first integer operand.
other : The second integer operand.
Returns a new integer that is the sum of the two operands. If the
mathematical sum exceeds the range of a 32-bit integer (-2,147,483,648 to
2,147,483,647), the result wraps around according to two's complement rules.
Example:
  inspect(42 + 1, content="43")
  inspect(2147483647 + 1, content="-2147483648") // Overflow wraps around to minimum value
+ 
Item
item1.
Int
weight 
(self_ : Int, other : Int) -> Bool
<= 
Int
capacity })
        .
(self : @list.List[Combination], f : (Combination) -> Combination) -> @list.List[Combination]
Maps the list.
Example
assert_eq(@list.of([1, 2, 3, 4, 5]).map(x => x * 2), @list.of([2, 4, 6, 8, 10]))
map(_.
(self : Combination, item : Item) -> Combination
add(
Item
item1))
      // 两个部分都仅包含有效组合，所以合并后也仅包含有效组合
      
@list.List[Combination]
valid_combs_with_item1 
(self : @list.List[Combination], other : @list.List[Combination]) -> @list.List[Combination]
Add implementation for List - concatenates two lists.
The + operator for lists performs concatenation.
a + b is equivalent to a.concat(b).
Example
let a = @list.of([1, 2, 3])
let b = @list.of([4, 5, 6])
let result = a + b
assert_eq(result, @list.of([1, 2, 3, 4, 5, 6]))
+ 
@list.List[Combination]
valid_combs_without_item1
    }
  }
}

遵循代码的结构进行分类讨论，很容易证明all_combinations_valid的正确性——它返回的所有组合确实都是有效的。

由于all_combinations_valid返回的那些组合都是有效的，就不再需要在solve中过滤了。我们将solve中的filter删去。

fn 
(items : @list.List[Item], capacity : Int) -> Combination
solve_v2(
@list.List[Item]
items : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Item {
  weight: Int
  value: Int
}
Item], 
Int
capacity : 
Int
Int) -> 
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination {
  
(items : @list.List[Item], capacity : Int) -> @list.List[Combination]
all_combinations_valid(
@list.List[Item]
items, 
Int
capacity).
(self : @list.List[Combination]) -> Combination
Get the maximum element of the list.
Warning: This function panics if the list is empty.
Use maximum() for a safe alternative that returns Option.
Example
let ls = @list.of([1, 3, 2, 5, 4])
assert_eq(ls.unsafe_maximum(), 5)
Panics
Panics if the list is empty.
unsafe_maximum()
}

三、维护升序性质，提前结束过滤

在上个版本中，为了过滤出那些能装下item1的组合，我们必须遍历valid_combs_without_item1中的每一个组合。

但我们可以发现：如果item1没法放入一个组合，那么item1一定都无法放入比这个组合总重量更大的那些组合。

这也就是说，如果valid_combs_without_item1能按总重量升序排列，那么过滤时就不需要完整地遍历它了。在过滤的过程中，一旦碰到一个放不下item1的组合，就可以立刻舍去后续的所有组合。由于这种逻辑很常见，标准库提供了一个叫take_while的函数，我们用它替换掉filter。

要想让valid_combs_without_item1升序排列，可以用排序算法，但这却要遍历整个列表，违背了初衷。因此，我们得采用另一种方案：想办法让all_combinations_valid返回的列表是升序的。这需要一次递归的信仰之跃：

fn 
(items : @list.List[Item], capacity : Int) -> @list.List[Combination]
all_combinations_valid_ordered(
  
@list.List[Item]
items : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Item {
  weight: Int
  value: Int
}
Item],
  
Int
capacity : 
Int
Int
) -> 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination] {
  match 
@list.List[Item]
items {
    
@list.List[Item]
Empty => 
(x : Combination) -> @list.List[Combination]
Create a list with a single element.
Returns a list containing only the given element.
Example
  let ls = @list.singleton(42)
  assert_eq(ls, @list.of([42]))
  assert_eq(ls.length(), 1)
@list.singleton(
Combination
empty_combination) // 单元素的列表，自然是升序的
    
(Item, @list.List[Item]) -> @list.List[Item]
More(
Item
item1, 
@list.List[Item]
tail=
@list.List[Item]
items_tail) => {
      // 我们假设 all_combinations_valid_ordered 返回的列表是升序的（归纳假设）
      let 
@list.List[Combination]
valid_combs_without_item1 = 
(items : @list.List[Item], capacity : Int) -> @list.List[Combination]
all_combinations_valid_ordered(
        
@list.List[Item]
items_tail, 
Int
capacity,
      )
      // 那么它也是升序的，因为一个升序的列表先截取一部分，再往每个元素加上同样的重量，它们的总重量还是升序的
      let 
@list.List[Combination]
valid_combs_with_item1 = 
@list.List[Combination]
valid_combs_without_item1
        .
(self : @list.List[Combination], p : (Combination) -> Bool) -> @list.List[Combination]
Take the longest prefix of a list of elements that satisfies a given predicate.
Example
  let ls = @list.from_array([1, 2, 3, 4])
  let r = ls.take_while(x => x < 3)
  assert_eq(r, @list.of([1, 2]))
take_while(fn(
Combination
comb) { 
Combination
comb.
Int
total_weight 
(self : Int, other : Int) -> Int
Adds two 32-bit signed integers. Performs two's complement arithmetic, which
means the operation will wrap around if the result exceeds the range of a
32-bit integer.
Parameters:

self : The first integer operand.
other : The second integer operand.
Returns a new integer that is the sum of the two operands. If the
mathematical sum exceeds the range of a 32-bit integer (-2,147,483,648 to
2,147,483,647), the result wraps around according to two's complement rules.
Example:
  inspect(42 + 1, content="43")
  inspect(2147483647 + 1, content="-2147483648") // Overflow wraps around to minimum value
+ 
Item
item1.
Int
weight 
(self_ : Int, other : Int) -> Bool
<= 
Int
capacity })
        .
(self : @list.List[Combination], f : (Combination) -> Combination) -> @list.List[Combination]
Maps the list.
Example
assert_eq(@list.of([1, 2, 3, 4, 5]).map(x => x * 2), @list.of([2, 4, 6, 8, 10]))
map(_.
(self : Combination, item : Item) -> Combination
add(
Item
item1))
      // 现在我们只需要确保合并后也升序，就能衔接上最开始的假设
      
(a : @list.List[Combination], b : @list.List[Combination]) -> @list.List[Combination]
merge_keep_order(
@list.List[Combination]
valid_combs_with_item1, 
@list.List[Combination]
valid_combs_without_item1)
    }
  }
}

最后的任务是完成函数merge_keep_order，它将两个升序的列表合并为一个升序的列表：

fn 
(a : @list.List[Combination], b : @list.List[Combination]) -> @list.List[Combination]
merge_keep_order(
  
@list.List[Combination]
a : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination],
  
@list.List[Combination]
b : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination]
) -> 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination] {
  match (
@list.List[Combination]
a, 
@list.List[Combination]
b) {
    (
@list.List[Combination]
Empty, 
@list.List[Combination]
another) | (
@list.List[Combination]
another, 
@list.List[Combination]
Empty) => 
@list.List[Combination]
another
    (
(Combination, @list.List[Combination]) -> @list.List[Combination]
More(
Combination
a1, 
@list.List[Combination]
tail=
@list.List[Combination]
a_tail), 
(Combination, @list.List[Combination]) -> @list.List[Combination]
More(
Combination
b1, 
@list.List[Combination]
tail=
@list.List[Combination]
b_tail)) =>
      // 如果 a1 比 b1 更轻，而 b 又是升序的，说明
      //   a1 比 b 里所有组合都轻
      // 由于 a 是升序的，所以
      //   a1 比 a_tail 里所有组合都轻
      // 所以 a1 是 a 和 b 中最小的那一个
      if 
Combination
a1.
Int
total_weight 
(self_ : Int, other : Int) -> Bool
< 
Combination
b1.
Int
total_weight {
        // 我们先递归地合并出答案的剩余部分，再把 a1 加到开头
        
(a : @list.List[Combination], b : @list.List[Combination]) -> @list.List[Combination]
merge_keep_order(
@list.List[Combination]
a_tail, 
@list.List[Combination]
b).
(self : @list.List[Combination], head : Combination) -> @list.List[Combination]
Add an element to the front of the list.
This is an alias for prepend - it creates a new list with the given
element added to the beginning.
Example
let ls = @list.of([2, 3, 4]).add(1)
assert_eq(ls, @list.of([1, 2, 3, 4]))
add(
Combination
a1)
      } else { // 同理
        
(a : @list.List[Combination], b : @list.List[Combination]) -> @list.List[Combination]
merge_keep_order(
@list.List[Combination]
a, 
@list.List[Combination]
b_tail).
(self : @list.List[Combination], head : Combination) -> @list.List[Combination]
Add an element to the front of the list.
This is an alias for prepend - it creates a new list with the given
element added to the beginning.
Example
let ls = @list.of([2, 3, 4]).add(1)
assert_eq(ls, @list.of([1, 2, 3, 4]))
add(
Combination
b1)
      }
  }
}

虽然看起来有点啰嗦，但我还是想提一句：通过遵循代码结构的分类讨论，很容易证明all_combinations_valid_ordered和merge_keep_order的正确性——它确实返回的一个升序的列表。

对于一个升序的列表，它的最大值就是最后一个。于是我们将unsafe_maximum替换成unsafe_last。

fn 
(items : @list.List[Item], capacity : Int) -> Combination
solve_v3(
@list.List[Item]
items : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Item {
  weight: Int
  value: Int
}
Item], 
Int
capacity : 
Int
Int) -> 
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination {
  
(items : @list.List[Item], capacity : Int) -> @list.List[Combination]
all_combinations_valid_ordered(
@list.List[Item]
items, 
Int
capacity).
(self : @list.List[Combination]) -> Combination
Get the last element of the list.
Warning: This function panics if the list is empty.
Use last() for a safe alternative that returns Option.
Example
let ls = @list.of([1, 2, 3, 4, 5])
assert_eq(ls.unsafe_last(), 5)
Panics
Panics if the list is empty.
unsafe_last()
}

回过头来看，在这一版的改进中，我们似乎并没有得到什么太大的好处，毕竟在合并列表的过程中，我们仍然需要遍历整个列表。最初我也是这么想的，但后来意外地发现merge_keep_order的真正作用在下一个版本。

四、去除等同重量的冗余组合，达到最优时间复杂度

目前为止，我们进行的都不是时间复杂度层面的优化，但这些优化恰恰为接下来的步骤铺平了道路。现在让我们来考察一下时间复杂度。

在最差情况下（背包很大，全都放得下），组合列表（all_combinations的返回值）将最多包含 $2^{物品数量}$ 个元素。这导致整个算法的时间复杂度也是指数级的，因为all_combinations会被调用 $物品数量$ 次，而每次都会遍历组合列表。

为了降低时间复杂度，我们就需要降低组合列表的长度。这基于一个观察：如果有两个组合，它们总重量相同，那么总价值更高的那个组合总是比另一个更好。因此，我们不需要在列表中同时保留两者。

如果能排除那些冗余的组合，组合列表的长度将不会超过背包容量（抽屉原理），进而将整个算法的时间复杂度降低到 $\mathcal{O}(物品数量 \times 背包容量)$。观察代码，现在唯一有可能会向列表中引入冗余组合的地方是merge_keep_order的else分支。为了避免这种情况出现，我们只需要对这个地方进行一点改动：

fnalias 
(x : T, y : T) -> T
@math.maximum

fn 
(a : @list.List[Combination], b : @list.List[Combination]) -> @list.List[Combination]
merge_keep_order_and_dedup(
  
@list.List[Combination]
a : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination],
  
@list.List[Combination]
b : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination]
) -> 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination] {
  match (
@list.List[Combination]
a, 
@list.List[Combination]
b) {
    (
@list.List[Combination]
Empty, 
@list.List[Combination]
another) | (
@list.List[Combination]
another, 
@list.List[Combination]
Empty) => 
@list.List[Combination]
another
    (
(Combination, @list.List[Combination]) -> @list.List[Combination]
More(
Combination
a1, 
@list.List[Combination]
tail=
@list.List[Combination]
a_tail), 
(Combination, @list.List[Combination]) -> @list.List[Combination]
More(
Combination
b1, 
@list.List[Combination]
tail=
@list.List[Combination]
b_tail)) =>
      if 
Combination
a1.
Int
total_weight 
(self_ : Int, other : Int) -> Bool
< 
Combination
b1.
Int
total_weight {
        
(a : @list.List[Combination], b : @list.List[Combination]) -> @list.List[Combination]
merge_keep_order_and_dedup(
@list.List[Combination]
a_tail, 
@list.List[Combination]
b).
(self : @list.List[Combination], head : Combination) -> @list.List[Combination]
Add an element to the front of the list.
This is an alias for prepend - it creates a new list with the given
element added to the beginning.
Example
let ls = @list.of([2, 3, 4]).add(1)
assert_eq(ls, @list.of([1, 2, 3, 4]))
add(
Combination
a1)
      } else if 
Combination
a1.
Int
total_weight 
(self_ : Int, other : Int) -> Bool
> 
Combination
b1.
Int
total_weight {
        
(a : @list.List[Combination], b : @list.List[Combination]) -> @list.List[Combination]
merge_keep_order_and_dedup(
@list.List[Combination]
a, 
@list.List[Combination]
b_tail).
(self : @list.List[Combination], head : Combination) -> @list.List[Combination]
Add an element to the front of the list.
This is an alias for prepend - it creates a new list with the given
element added to the beginning.
Example
let ls = @list.of([2, 3, 4]).add(1)
assert_eq(ls, @list.of([1, 2, 3, 4]))
add(
Combination
b1)
      } else { // 此时 a1 和 b1 一样重，出现冗余，保留总价值更高的那个
        let 
Combination
better = 
(x : Combination, y : Combination) -> Combination
maximum(
Combination
a1, 
Combination
b1)
        
(a : @list.List[Combination], b : @list.List[Combination]) -> @list.List[Combination]
merge_keep_order_and_dedup(
@list.List[Combination]
a_tail, 
@list.List[Combination]
b_tail).
(self : @list.List[Combination], head : Combination) -> @list.List[Combination]
Add an element to the front of the list.
This is an alias for prepend - it creates a new list with the given
element added to the beginning.
Example
let ls = @list.of([2, 3, 4]).add(1)
assert_eq(ls, @list.of([1, 2, 3, 4]))
add(
Combination
better)
      }
  }
}

all_combinations_valid_ordered_nodup（这是我这辈子写过的名字最长的函数了）和solve_v4替换相应部分即可。

fn 
(items : @list.List[Item], capacity : Int) -> @list.List[Combination]
all_combinations_valid_ordered_nodup(
  
@list.List[Item]
items : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Item {
  weight: Int
  value: Int
}
Item],
  
Int
capacity : 
Int
Int
) -> 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination] {
  match 
@list.List[Item]
items {
    
@list.List[Item]
Empty => 
(x : Combination) -> @list.List[Combination]
Create a list with a single element.
Returns a list containing only the given element.
Example
  let ls = @list.singleton(42)
  assert_eq(ls, @list.of([42]))
  assert_eq(ls.length(), 1)
@list.singleton(
Combination
empty_combination)
    
(Item, @list.List[Item]) -> @list.List[Item]
More(
Item
item1, 
@list.List[Item]
tail=
@list.List[Item]
items_tail) => {
      let 
@list.List[Combination]
combs_without_item1 = 
(items : @list.List[Item], capacity : Int) -> @list.List[Combination]
all_combinations_valid_ordered_nodup(
        
@list.List[Item]
items_tail, 
Int
capacity,
      )
      let 
@list.List[Combination]
combs_with_item1 = 
@list.List[Combination]
combs_without_item1
        .
(self : @list.List[Combination], p : (Combination) -> Bool) -> @list.List[Combination]
Take the longest prefix of a list of elements that satisfies a given predicate.
Example
  let ls = @list.from_array([1, 2, 3, 4])
  let r = ls.take_while(x => x < 3)
  assert_eq(r, @list.of([1, 2]))
take_while(fn(
Combination
comb) { 
Combination
comb.
Int
total_weight 
(self : Int, other : Int) -> Int
Adds two 32-bit signed integers. Performs two's complement arithmetic, which
means the operation will wrap around if the result exceeds the range of a
32-bit integer.
Parameters:

self : The first integer operand.
other : The second integer operand.
Returns a new integer that is the sum of the two operands. If the
mathematical sum exceeds the range of a 32-bit integer (-2,147,483,648 to
2,147,483,647), the result wraps around according to two's complement rules.
Example:
  inspect(42 + 1, content="43")
  inspect(2147483647 + 1, content="-2147483648") // Overflow wraps around to minimum value
+ 
Item
item1.
Int
weight 
(self_ : Int, other : Int) -> Bool
<= 
Int
capacity })
        .
(self : @list.List[Combination], f : (Combination) -> Combination) -> @list.List[Combination]
Maps the list.
Example
assert_eq(@list.of([1, 2, 3, 4, 5]).map(x => x * 2), @list.of([2, 4, 6, 8, 10]))
map(_.
(self : Combination, item : Item) -> Combination
add(
Item
item1))
      
(a : @list.List[Combination], b : @list.List[Combination]) -> @list.List[Combination]
merge_keep_order_and_dedup(
@list.List[Combination]
combs_with_item1, 
@list.List[Combination]
combs_without_item1)
    }
  }
}

fn 
(items : @list.List[Item], capacity : Int) -> Combination
solve_v4(
@list.List[Item]
items : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Item {
  weight: Int
  value: Int
}
Item], 
Int
capacity : 
Int
Int) -> 
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination {
  
(items : @list.List[Item], capacity : Int) -> @list.List[Combination]
all_combinations_valid_ordered_nodup(
@list.List[Item]
items, 
Int
capacity).
(self : @list.List[Combination]) -> Combination
Get the last element of the list.
Warning: This function panics if the list is empty.
Use last() for a safe alternative that returns Option.
Example
let ls = @list.of([1, 2, 3, 4, 5])
assert_eq(ls.unsafe_last(), 5)
Panics
Panics if the list is empty.
unsafe_last()
}

至此，我们重新发明了01背包问题的dp解法。

总结

这篇文章的内容是我某天早上躺在床上的突发奇想，从第一版到第四版代码完全在手机上写成，没有经过任何调试，但却能轻松地保证了正确性。相比传统算法竞赛题解中常见的写法，本文中使用的函数式写法带来了以下优势：

1. 告别循环，使用递归分情况讨论。要想从列表中获取元素，必须使用模式匹配（match），这提醒我考虑列表为空时的答案。它相比dp数组的初始值拥有更加明确的含义。
2. 依赖库函数进行遍历。标准库中提供的高阶函数（filter、take_while、map、maximum）能替换掉样板化的循环（for、while），便于读者一眼看出遍历的目的。
3. 声明式编程。第一版的代码是想法的一比一地翻译。与其说是在描述一个算法，更像是在描述这个问题本身，这保证了第一版的正确性。而随后每次改进都在不影响结果的前提下进行，于是继承了第一版的正确性。

当然，从来就没有银弹。我们需要可读性和效率之间做取舍。函数式的风格固然好理解，但还是有许多优化余地的。进一步的优化方向是将列表替换成数组，再替换成从头到尾只使用两个滚动数组，甚至是只使用一个数组。这可以将空间复杂度优化成 $\mathcal{O}(背包容量)$，但不在本文的讨论范围内。我相信初学者更希望看到的是一个易于理解的代码。

附录

更多细节优化

利用items中物品的顺序不影响结果的总价值这个性质。可以把all_combinations转化成尾递归。

另外，take_while产生的列表在map后马上就被丢弃了，我们可以改用迭代器来避免产生这个一次性列表。

fn 
(items : @list.List[Item], capacity : Int) -> @list.List[Combination]
all_combinations_loop(
  
@list.List[Item]
items : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Item {
  weight: Int
  value: Int
}
Item],
  
Int
capacity : 
Int
Int
) -> 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination] {
  loop 
@list.List[Item]
items, 
(x : Combination) -> @list.List[Combination]
Create a list with a single element.
Returns a list containing only the given element.
Example
  let ls = @list.singleton(42)
  assert_eq(ls, @list.of([42]))
  assert_eq(ls.length(), 1)
@list.singleton(
Combination
empty_combination) {
    
@list.List[Item]
Empty, 
@list.List[Combination]
combs_so_far => 
@list.List[Combination]
combs_so_far
    
(Item, @list.List[Item]) -> @list.List[Item]
More(
Item
item1, 
@list.List[Item]
tail=
@list.List[Item]
items_tail), 
@list.List[Combination]
combs_so_far => {
      let 
@list.List[Combination]
combs_without_item1 = 
@list.List[Combination]
combs_so_far
      let 
@list.List[Combination]
combs_with_item1 = 
@list.List[Combination]
combs_without_item1
        .
(self : @list.List[Combination]) -> Iter[Combination]
Create an iterator over the list elements.
Returns an iterator that yields each element of the list in order.
Example
let ls = @list.of([1, 2, 3, 4, 5])
let iter = ls.iter()
let sum = iter.fold(init=0, (acc, x) => acc + x)
inspect(sum, content="15")
iter()
        .
(self : Iter[Combination], f : (Combination) -> Bool) -> Iter[Combination]
Takes elements from the iterator as long as the predicate function returns true.
Type Parameters

T: The type of the elements in the iterator.
Arguments

self - The input iterator.
f - The predicate function that determines whether an element should be taken.
Returns
A new iterator that contains the elements as long as the predicate function returns true.
take_while(fn(
Combination
comb) { 
Combination
comb.
Int
total_weight 
(self : Int, other : Int) -> Int
Adds two 32-bit signed integers. Performs two's complement arithmetic, which
means the operation will wrap around if the result exceeds the range of a
32-bit integer.
Parameters:

self : The first integer operand.
other : The second integer operand.
Returns a new integer that is the sum of the two operands. If the
mathematical sum exceeds the range of a 32-bit integer (-2,147,483,648 to
2,147,483,647), the result wraps around according to two's complement rules.
Example:
  inspect(42 + 1, content="43")
  inspect(2147483647 + 1, content="-2147483648") // Overflow wraps around to minimum value
+ 
Item
item1.
Int
weight 
(self_ : Int, other : Int) -> Bool
<= 
Int
capacity })
        .
(self : Iter[Combination], f : (Combination) -> Combination) -> Iter[Combination]
Transforms the elements of the iterator using a mapping function.
Type Parameters

T: The type of the elements in the iterator.
R: The type of the transformed elements.
Arguments

self - The input iterator.
f - The mapping function that transforms each element of the iterator.
Returns
A new iterator that contains the transformed elements.
map(_.
(self : Combination, item : Item) -> Combination
add(
Item
item1))
        |> 
(iter : Iter[Combination]) -> @list.List[Combination]
Convert an iterator into a list, preserving order of elements.
Creates a list from an iterator, maintaining the same order as the iterator.
If order is not important, consider using from_iter_rev for better performance.
Example
let arr = [1, 2, 3, 4, 5]
let iter = arr.iter()
let ls = @list.from_iter(iter)
assert_eq(ls, @list.of([1, 2, 3, 4, 5]))
@list.from_iter
      continue 
@list.List[Item]
items_tail,
        
(a : @list.List[Combination], b : @list.List[Combination]) -> @list.List[Combination]
merge_keep_order_and_dedup(
@list.List[Combination]
combs_with_item1, 
@list.List[Combination]
combs_without_item1)
    }
  }
}

fn 
(items : @list.List[Item], capacity : Int) -> Combination
solve_v5(
@list.List[Item]
items : 
enum @list.List[A] {
  Empty
  More(A, tail~ : @list.List[A])
}
List[
struct Item {
  weight: Int
  value: Int
}
Item], 
Int
capacity : 
Int
Int) -> 
struct Combination {
  items: @list.List[Item]
  total_weight: Int
  total_value: Int
}
Combination {
  
(items : @list.List[Item], capacity : Int) -> @list.List[Combination]
all_combinations_loop(
@list.List[Item]
items, 
Int
capacity).
(self : @list.List[Combination]) -> Combination
Get the last element of the list.
Warning: This function panics if the list is empty.
Use last() for a safe alternative that returns Option.
Example
let ls = @list.of([1, 2, 3, 4, 5])
assert_eq(ls.unsafe_last(), 5)
Panics
Panics if the list is empty.
unsafe_last()
}

题外话

1. 在第一版中，all_combinations(items)产生的Combination甚至比其中的More还多一个，堪称链表节点复用大师。
2. 升序还可以换成降序，对应的take_while要换成drop_while。而改用Array后可以通过binary_search来寻找下标直接切分。
3. 如果你感兴趣，可以考虑一下怎么把上面的做法拓展到各种其它的背包问题。
4. all_combinations_loop原名：generate_all_ordered_combination_that_fit_in_backpack_list_without_duplicates_using_loop。

测试

test {
  for 
(@list.List[Item], Int) -> Combination
solve in [
(items : @list.List[Item], capacity : Int) -> Combination
solve_v1, 
(items : @list.List[Item], capacity : Int) -> Combination
solve_v2, 
(items : @list.List[Item], capacity : Int) -> Combination
solve_v3, 
(items : @list.List[Item], capacity : Int) -> Combination
solve_v4, 
(items : @list.List[Item], capacity : Int) -> Combination
solve_v5] {
    
(a : Int, b : Int, msg? : String, loc~ : SourceLoc = _) -> Unit raise Error
Asserts that two values are equal. If they are not equal, raises a failure
with a message containing the source location and the values being compared.
Parameters:

a : First value to compare.
b : Second value to compare.
loc : Source location information to include in failure messages. This is
usually automatically provided by the compiler.
Throws a Failure error if the values are not equal, with a message showing
the location of the failing assertion and the actual values that were
compared.
Example:
  assert_eq(1, 1)
  assert_eq("hello", "hello")
assert_eq(
(@list.List[Item], Int) -> Combination
solve(
@list.List[Item]
items_1, 10).
Int
total_value, 21)
  }
}