Week 9 - 638.Shopping Offers - Medium

638.Shopping Offers - Medium

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.

my solution:

#include<vector>
#include<iostream>
#include<algorithm>
using namespace std;
class Solution {
public:

  int shoppingOffers(vector<int>& price, vector<vector<int>>& special, vector<int>& needs) {
    vector<int> costlist; // 价格列表
    costlist.push_back(noSpecCost(price, needs));
    for (int i = 0; i < special.size(); i++) {
      vector<int> needs1 = needs; //需求的拷贝
      if (canUseSpec(special[i], needs1)) { //判断能否使用套餐
        for (int j = 0; j < needs1.size(); j++) {
          needs1[j] -= special[i][j]; //使用套餐，需求发生变化
        }
        costlist.push_back(shoppingOffers(price, special, needs1) + *special[i].rbegin()); // 计算使用该套餐后的价格
      }
    }
    return *min_element(costlist.begin(), costlist.end()); // 求最便宜的价格
  }

  // 单买价格
  int noSpecCost(vector<int>& price, vector<int>& needs) {
    int sum = 0;
    for (int i = 0; i < needs.size(); i++) {
      sum += needs[i] * price[i];
    }
    return sum;
  }

  // 判断当前特价套餐能否使用
  bool canUseSpec(vector<int> spec, vector<int> needs) {
    for (int i = 0; i < needs.size(); i++) {
      if (needs[i] < spec[i]) {
        return false;
      }
    }
    return true;
  }
};

这道题的大意是这样：给定需要买的商品的数量和价格，再给出几个特价套餐，求恰好满足数量要求的最便宜的买法。那么这道题用动态规划的思想来解的话，可以写出这样的状态式：

Cost（n） = min（Cost（n-1）+套餐1，Cost（n-1）+套餐2，...，Cost（n-1）+套餐k，）

在此基础上使用递归容易得出答案。

而我认为这道题的难点在于如何处理套餐和单买的关系。这一点是我自己没有想到，看了答案才恍然大悟的：先考虑单买的花费，再考虑能否使用套餐，如果不能使用套餐的话，costlist中自然就只有单买的选项。这样单买的情况就被合并到一起了。