# 638. Shopping Offers 大礼包

@TOC

## # 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:

``````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:

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.

## # 解题方法

### # DFS

dfs的做法是这样：求出直接购买这些商品的价格，然后遍历所有的套餐，看能不能使用这个套餐（判断的方式是使用套餐之后仍然还有剩余物品），保存所有情况下的最小值返回即可。

``````class Solution(object):
def shoppingOffers(self, price, special, needs):
"""
:type price: List[int]
:type special: List[List[int]]
:type needs: List[int]
:rtype: int
"""
return self.dfs(price, special, needs)

def dfs(self, price, special, needs):
local_min = self.directPurchase(price, needs)
for spec in special:
remains = [needs[j] - spec[j] for j in range(len(needs))]
if min(remains) >= 0:
local_min = min(local_min, spec[-1] + self.dfs(price, special, remains))
return local_min

def directPurchase(self, price, needs):
total = 0
for i, need in enumerate(needs):
total += price[i] * need
return total
``````

``````class Solution(object):
def shoppingOffers(self, price, special, needs):
"""
:type price: List[int]
:type special: List[List[int]]
:type needs: List[int]
:rtype: int
"""
return self.dfs(price, special, needs, {})

def dfs(self, price, special, needs, d):
val = sum(price[i] * needs[i] for i in range(len(needs)))
for spec in special:
remains = [needs[j] - spec[j] for j in range(len(needs))]
if min(remains) >= 0:
val = min(val, d.get(tuple(needs), spec[-1] + self.dfs(price, special, remains, d)))
d[tuple(needs)] = val
return val
``````

``````class Solution(object):
def shoppingOffers(self, price, special, needs):
"""
:type price: List[int]
:type special: List[List[int]]
:type needs: List[int]
:rtype: int
"""
N = len(needs)
res = sum(p * n for p, n in zip(price, needs))
for sp in special:
if all(sp[i] <= needs[i] for i in range(N)):
remain = [needs[i] - sp[i] for i in range(N)]
if min(remain) >= 0:
res = min(res, sp[-1] + self.shoppingOffers(price, special, remain))
return res
``````

### # 回溯法

``````class Solution {
public:
int shoppingOffers(vector<int>& price, vector<vector<int>>& special, vector<int>& needs) {
return helper(price, special, needs, 0);
}
int helper(vector<int>& price, vector<vector<int>>& special, vector<int>& needs, int start) {
const int N = price.size();
int ans = 0;
for (int i = 0; i < N; i++) {
ans += price[i] * needs[i];
}
for (int i = start; i < special.size(); i++) {
auto offer = special[i];
int total = offer.back();
for (int j = 0; j < N; j ++) {
needs[j] -= offer[j];
}
if (*min_element(needs.begin(), needs.end()) >= 0) {
total += helper(price, special, needs, i);
ans = min(total, ans);
}
for (int j = 0; j < N; j++) {
needs[j] += offer[j];
}
}
return ans;
}
};
``````

https://leetcode.com/problems/shopping-offers/discuss/105212/Very-Easy-to-understand-JAVA-Solution-beats-95-with-explanation https://leetcode.com/problems/shopping-offers/discuss/105204/Python-dfs-with-memorization.

## # 日期

2018 年 9 月 7 日 —— 中午不睡，下午崩溃 2019 年 3 月 23 日 —— 今天也是元气满满的一天！