638. Shopping Offers-优快云博客

本文链接：https://blog.youkuaiyun.com/lianggx3/article/details/109711530

该博客介绍了一个关于购物优惠的问题，其中涉及到不同商品的价格、特殊优惠组合及购买需求。博主通过实现一个深度优先搜索（DFS）算法，找出购买指定商品的最低价格。在示例中，算法考虑了各种优惠组合，如购买一定数量的商品获得折扣，以找到最优解。最终，博主展示了如何应用DFS来计算最小花费，并给出了具体代码实现。

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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:
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:
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.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/shopping-offers
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

DFS

class Solution {
    int ans = Integer.MAX_VALUE;
    public int shoppingOffers(List<Integer> price, List<List<Integer>> special, List<Integer> needs) {
        dfs(price, special, needs, 0);
        return ans;
    }

    private void dfs(List<Integer> price, List<List<Integer>> special, List<Integer> needs, int tmpPrice) {
        // needs中都是0，迭代结束
        int count = 0;
        for (int i = 0; i < needs.size(); i++) {
            count += needs.get(i);
        }
        if (count == 0) {
            ans = Math.min(ans, tmpPrice);
            return;
        }
        int nonOfferPrice = 0;
        for(int i = 0; i < needs.size(); i++) {
            nonOfferPrice += price.get(i) * needs.get(i);
        }
        // 算出剩下的商品，都用单价买的价格。
        ans = Math.min(ans, nonOfferPrice + tmpPrice);
        for (int i = 0; i < special.size(); i++) {
            boolean isValid = true;
            for (int j = 0; j < special.get(i).size() - 1; j++) {
                if (needs.get(j) < special.get(i).get(j)) {
                    isValid = false;
                    break;
                }
            }
            if (!isValid) continue;
            for (int j = 0; j < special.get(i).size() - 1; j++) {
                needs.set(j, needs.get(j) - special.get(i).get(j));
            }
            dfs(price, special, needs, tmpPrice + special.get(i).get(special.get(i).size() - 1));
            for (int j = 0; j < special.get(i).size() - 1; j++) {
                needs.set(j, needs.get(j) + special.get(i).get(j));
            }
        }
    }
}