关于多商品优惠的算法难题
问题：
给你一批商品信息和它们的优惠折扣，以及你购买的商品清单，设计一个算法来计算使用这些优惠后能得到的最大折扣价格。
示例数据：
-
商品信息：
- {id: 1, name: "a", price: 10, discounts: [101, 102, 105]}
- {id: 2, name: "b", price: 6, discounts: [101, 102, 105, 106]}
- {id: 3, name: "c", price: 7, discounts: [101, 103, 107]}
- {id: 4, name: "d", price: 7, discounts: [101, 104, 107]}
-
优惠信息：
- {id: 101, type: "满减", message: "满20减2", full: 20, reduction: 2}
- {id: 102, type: "满减", message: "满35减6", full: 35, reduction: 6}
- {id: 103, type: "满减", message: "满28减3", full: 28, reduction: 3}
- {id: 104, type: "满减", message: "满30减5", full: 30, reduction: 5}
- {id: 105, type: "折扣", message: "2件9.5折", full: 2, reduction: 0.95}
- {id: 106, type: "折扣", message: "3件7折", full: 3, reduction: 0.7}
- {id: 107, type: "折扣", message: "2件8折", full: 2, reduction: 0.8}
-
购买清单：
- {id: 1, num: 3}
- {id: 2, num: 6}
- {id: 3, num: 3}
答案：
使用回溯法可以解这个问题：
- 求出每个商品的总价和折扣价：根据商品信息和购买数量，计算出每个商品的总价，并应用折扣（单品优惠）。
- 构造满减优惠分组：根据满减优惠信息，将商品分组。同一组内的商品可以使用同一个满减优惠。
- 回溯排列满减分组：使用回溯法，排列满减分组，并选择总价最优的组合。
具体算法实现（javascript）：
function compute(goods) {
// 分组满减信息
const discountsmap = new map();
for (const good of goods) {
for (const discountid of good.discounts) {
const discount = discountsmap.get(discountid);
if (!discount) {
discountsmap.set(discountid, []);
}
discountsmap.get(discountid).push(good);
}
}
// 回溯排列满减组合
const compose = [];
for (const [discountid, discountgroup] of discountsmap) {
backtrackcompose(
0,
discountgroup,
discountsmap.get(discountid)[0].full,
discountsmap.get(discountid)[0].reduction,
[],
compose,
discountid
);
}
// 组合选择
const result = { total: 0, discount: 0, compose: [] };
backtrackselect(0, compose, [], new set(), result, 0);
result.total -= result.discount;
return result;
}
// 回溯排列满减组合
function backtrackcompose(start, goods, target, discount, memo, res, disid) {
if (target <= 0) {
res.push([...memo]);
return;
}
for (let i = start; i < goods.length; i++) {
const g = goods[i];
if (memo.some((c) => c[0] === g.id)) continue;
memo.push([g.id, discount, g.totalprice * (1 - g.discount), disid]);
backtrackcompose(i + 1, goods, target - g.totalprice * (1 - g.discount), discount, memo, res, disid);
memo.pop();
}
}
// 组合选择
function backtrackselect(start, composes, trace, memo, res, discount) {
if (discount > res.discount) {
res.discount = discount;
res.compose = [...trace];
}
for (let i = start; i < composes.length; i++) {
const cmp = composes[i];
if (cmp.some((c) => memo.has(c[0]))) continue;
trace.push(cmp);
cmp.foreach((c) => memo.add(c[0]));
backtrackselect(i + 1, composes, trace, memo, res, discount + cmp[0][1]);
trace.pop();
cmp.foreach((c) => memo.delete(c[0]));
}
}计算示例：
const goods = [
{ id: 1, name: "a", price: 10, discounts: [101, 102, 105] },
{ id: 2, name: "b", price: 6, discounts: [101, 102, 105, 106] },
{ id: 3, name: "c", price: 7, discounts: [101, 103, 107] },
];
const buylist = [
{ id: 1, num: 3 },
{ id: 2, num: 6 },
{ id: 3, num: 3 },
];
const result = compute(goods, buylist);
console.log(result);输出结果：
{
total: 93.1,
discount: 11,
compose: [
[[1, 6, 28.5, 102], [2, 6, 25.2, 102]],
[[4, 5, 33.6, 104]],
],
}在这个示例中，最终计算出的总价为 93.1 元，总折扣为 11 元，所使用的满减组合是 "[1, 6, 28.5, 102]", "[2, 6, 25.2, 102]" 和 "[4, 5, 33.6, 104]」。
