leetcode 638. Shopping Offers 最佳购物优惠+十分典型深度优先遍历DFS

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 2and 2 a n d 5 respectively.
In special offer 1, you can pay 5for3Aand0BInspecialoffer2,youcanpay 5 f o r 3 A a n d 0 B I n s p e c i a l o f f e r 2 , y o u c a n p a y 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 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 2 , a n d 3 for B, 4forC.Youmaypay 4 f o r C . Y o u m a y p a y 4 for 1A and 1B, and 9for2A,2Band1C.Youneedtobuy1A,2Band1C,soyoumaypay 9 f o r 2 A , 2 B a n d 1 C . Y o u n e e d t o b u y 1 A , 2 B a n d 1 C , s o y o u m a y p a y 4 for 1A and 1B (special offer #1), and 3for1B, 3 f o r 1 B , 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.

本题题意很简单，就是在一系列的购物优惠中选取最佳的策略，以获取少的花费

直接递归即可，很棒的做法

代码如下：

#include 
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using namespace std;


class Solution {
public:
    int shoppingOffers(vector<int>& p, vector<vector<int>>& sp, vector<int>& need) 
    {
        int minsum = 0;
        for (int i = 0; i < p.size(); i++)
            minsum += p[i] * need[i];

        for (auto s : sp)
        {
            bool can = true;
            for (int i = 0; i < need.size(); i++)
            {
                if (need[i] < s[i])
                {
                    can = false;
                    break;
                }
            }

            if (can)
            {
                for (int i = 0; i < need.size(); i++)
                    need[i] -= s[i];
                minsum = min(minsum, s[need.size()] + shoppingOffers(p, sp, need));
                for (int i = 0; i < need.size(); i++)
                    need[i] += s[i];
            }

        }
        return minsum;
    }
};