638. Shopping Offers

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: 14Explanation: 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 0BIn 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: 11Explanation: 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:

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

好长的题目，看了一会才明白意思。。。。。

第一个参数price[2,5],代表a和b的原价格，第二个参数special [[3,0,5] （3*a+ 0*b => 5）,[1,2,10] (1*a + 2*b => 10)] 这里面=>代表折扣价，求[3 *a, 2*b] 使用special的套餐最低价格是多少。

解题思路，DFS，就是一个一个尝试，过程如下：

一个special[x]能否被needs容纳进去？

若可以，needs中减去一个sepcial[x]，

--------这里开始递归，也就是开始DFS。

继续尝试容纳。

容纳结束，恢复needs的内容。

若出现needs容纳不了sepcial[x]，尝试下一个，继续深度优先容纳，

直到所有的都不能容纳，此时依次推出尝试之前每一步没有尝试的容纳，就是DFS的过程。

其中还要计算每一步的价格。取最小。

简单点就是DFS搜索出所有可能，然后取最小值。

class Solution {public:    int result = 0;    int shoppingOffers(const vector<int>& price, const vector<vector<int>>& special, const vector<int>& needs) {        int len_s = special.size();        int len_p = price.size();        for(int i = 0;i<len_p;i++){            result += price[i]*needs[i];        }        vector<int> need(needs);        helper(price, special, need, 0, len_s, len_p, result);                return result;            }private:    void helper(const vector<int>& price, const vector<vector<int>>& special,  vector<int> &need, int index, int len_s, int len_p, int total)    {        for(int i = index; i < len_s; i++){            bool valid = true;            for(int j=0; j<len_p; j++){                if(need[j] < special[i][j]){                    valid = false;                    break;                }            }            if(valid){                int nextTotal  = total;                for(int j=0; j<len_p; j++){                    need[j] -= special[i][j];                    nextTotal  -= price[j]*special[i][j];                }                nextTotal  += special[i][len_p];                result = min(result, nextTotal );                                helper(price, special, need, i, len_s, len_p, nextTotal );                                for(int j=0; j<len_p; j++){                    need[j] += special[i][j];                }            }        }    }};