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:
-
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 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 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])); } }
登录后复制
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);
登录后复制
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]], ], }
登录后复制
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]」。
© 版权声明
文章版权归作者所有,未经允许请勿转载。
【小浪云服务商 - 服务器12元起 - 挂机宝5元起】
THE END
暂无评论内容