638. Shopping Offers

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638. Shopping Offers

题目描述

In LeetCode Store, there are some kinds of items to sell. Each item has a price.

However, there are some special offers, and a special offer consists of one or more different kinds of items with a sale price.

You are given the each item’s price, a set of special offers, and the number we need to buy for each item. The job is to output the lowest price you have to pay for exactly certain items as given, where you could make optimal use of the special offers.

Each special offer is represented in the form of an array, the last number represents the price you need to pay for this special offer, other numbers represents how many specific items you could get if you buy this offer.

You could use any of special offers as many times as you want.

Example 1:

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Input: [2,5], [[3,0,5],[1,2,10]], [3,2]
Output: 14
Explanation: 
There are two kinds of items, A and B. Their prices are $2 and $5 respectively. 
In special offer 1, you can pay $5 for 3A and 0B
In special offer 2, you can pay $10 for 1A and 2B. 
You need to buy 3A and 2B, so you may pay $10 for 1A and 2B (special offer #2), and $4 for 2A.

Example 2:

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Input: [2,3,4], [[1,1,0,4],[2,2,1,9]], [1,2,1]
Output: 11
Explanation: 
The price of A is $2, and $3 for B, $4 for C. 
You may pay $4 for 1A and 1B, and $9 for 2A ,2B and 1C. 
You need to buy 1A ,2B and 1C, so you may pay $4 for 1A and 1B (special offer #1), and $3 for 1B, $4 for 1C. 
You cannot add more items, though only $9 for 2A ,2B and 1C.

Note:

  • There are at most 6 kinds of items, 100 special offers.
  • For each item, you need to buy at most 6 of them.
    You are not allowed to buy more items than you want, even if that would lower the overall price.

题目大意

给定一组商品价格 prices ,以及一些组合的优惠策略 special （包含商品数目及其对应的价格）, 以及要购买的商品数目 needs，求在最优策略下，恰好购买 needs 所需的商品（不能超过），需要的最小金额。

解题思路

DFS,通过辅助字典，记录已经遍历过的数目组合。

代码

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class Solution {
    func shoppingOffers(_ price: [Int], _ special: [[Int]], _ needs: [Int]) -> Int {
        var dict = [String: Int]()
        func dfs(needs: [Int]) -> Int {
            var key = ""
            var total = 0
            for (n, p) in zip(needs, price) {
                total += n * p
                key += "\(n)-"
            }
            if let cost = dict[key] { return cost }
            dict[key] = total
            for discounts in special {
                var finalNeeds = [Int]()
                for (d, n) in zip(discounts, needs) {
                    let surplus = n - d
                    if surplus < 0 { break }
                    finalNeeds.append(surplus)
                }
                if finalNeeds.count != needs.count { continue }
                dict[key] = min(dict[key]!, dfs(needs: finalNeeds) + discounts.last!)
            }
            return dict[key]!
        }
        return dfs(needs: needs)
    }
}
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