硬着头皮，又做了一道递归题。思路有，但没经验。

cnmlgb9998 · 发表于 2024-11-13 15:29

本帖最后由 cnmlgb9998 于 2024-11-13 15:39 编辑

请大神多多指点。如何剪枝？减少冗余。

let
源 = Excel.CurrentWorkbook(),
Data = 源{[Name = "表5"]}[Content],
Filter = 源{[Name = "表6"]}[Content],
fx = (x, y) =>
[
a = Table.SelectRows(Data, each [始发站] = x[出发]),
b =
if Table.IsEmpty(a) then
y
else
List.Combine(
List.Transform(
Table.ToRecords(a),
each
if [终点站] = x[目的] then
y & {{[终点站], [价格]}}
else
@fx(
Record.TransformFields(
x,
{{"出发", (z) => [终点站]}}
),
y & {{[终点站], [价格]}}
)
)
)
][b],
Tab = List.Transform(Table.ToRecords(Filter), each Table.FromRows(fx(_, {{[出发], 0}}))),
Group = Table.Group(Tab{0}, "Column1", {"D", each _}, 0, (x, y) => Number.From(y = Filter[出发]{0})),
Sele = List.Select(
Group[D],
each List.ContainsAll([Column1], List.Skip(Table.ToRows(Filter){0}))
),
Sele2 = [
a = List.Min(List.Transform(Sele, each List.Sum([Column2]))),
b = List.Select(Sele, each List.Sum([Column2]) = a)
][b],
Res1 = Sele2{0},
Chang = Text.Combine(
List.Accumulate(
{1 .. Table.RowCount(Res1) - 1},
{},
(x, y) =>
x
& {
Text.Format(
"#{0}>>#{1},￥#{2}",
{
Text.From(Res1[Column1]{y - 1}),
Text.From(Res1[Column1]{y}),
Text.From(Res1[Column2]{y})
}
)
}
),
" | "
)
& " | 合计："
& Text.From(List.Sum(Res1[Column2])),
Res2 = #table(1, {{Chang}})
in
Res2

复制代码

cnmlgb9998 · 发表于 2024-11-13 16:01

这题，我是自降难度。行程只能单向。请注意。

cnmlgb9998 · 发表于 2024-11-13 17:35

代码更新了一下。

let
源 = Excel.CurrentWorkbook(),
Data = 源{[Name = "表5"]}[Content],
Filter = 源{[Name = "表6"]}[Content],
fx = (x, y) =>
[
a = Table.SelectRows(Data, each [始发站] = x[出发]),
b =
if Table.IsEmpty(a) then
{y}
else
List.Combine(
List.Transform(
Table.ToRecords(a),
each
if [终点站] = x[目的] then
{y & {{[终点站], [价格]}}}
else
@fx(
Record.TransformFields(
x,
{{"出发", (z) => [终点站]}}
),
y & {{[终点站], [价格]}}
)
)
)
][b],
Tab = List.Combine(List.Transform(Table.ToRecords(Filter), each fx(_, {{[出发], 0}}))),
Sele = List.Select(
Tab,
each List.ContainsAll(List.Zip(_){0}, List.LastN(Table.ToRows(Filter){0}, 2))
),
Min = List.Min(List.Transform(Sele, each List.Sum(List.Zip(_){1}))),
Sele2 = List.Combine(List.Select(Sele, each List.Sum(List.Zip(_){1}) = Min)),
Res = Text.Combine(
List.Accumulate(
{1 .. List.Count(Sele2) - 1},
{},
(x, y) =>
x
& {
Text.Format(
"#{0}>>#{1},￥#{2}",
{
Text.From(Sele2{y - 1}{0}),
Text.From(Sele2{y}{0}),
Text.From(Sele2{y}{1})
}
)
}
),
" | "
)
& " | 合计："
& Text.From(Min),
转换为表 = #table(1, {{Res}})
in
转换为表

复制代码

deadzlq · 发表于 2024-11-15 13:39

PQ参考解法(与Leecode原题,增加往返同价,城市从1开始而已,基本照搬),推荐dp动态规划或BellmanFord算法求

cnmlgb9998 · 发表于 2024-11-15 13:57

deadzlq 发表于 2024-11-15 13:39
PQ参考解法(与Leecode原题,增加往返同价,城市从1开始而已,基本照搬),推荐dp动态规划或BellmanFord算法求

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[求助] 硬着头皮，又做了一道递归题。思路有，但没经验。