Spree (dynamic programming)

In LeetCode shop, there are many items in the sale.

However, there are some spree, each spree for a price bundling a group of articles.

Given the current list price of each item, each gift package contains items, as well as a list of your purchases to be. Please complete the minimum output of the exact cost of the wish list.

Each package is a large array is described by a set of data, the last number represents the price of a large package, the number of other types of other digital items are contained FIG.

Any spree can buy unlimited.

thought:分为3步：

  1. 每种物品都单独购买，需要money1
  2. 用大礼包进行替换， 需要money2
  3. 取最小值min(money1, money2)
 
  减枝：
      1. 当礼包中的物品的数量  >  所需物品的数量， 要进行减枝
      2. 为了避免像，礼包1，2 和 礼包2， 1这种情况重复计算两次，可以使用pos来指向当前的位置
         允许获取的礼包的索引只能大于等于pos（这种减枝比较隐蔽）

public int shoppingOffers(List<Integer> price, List<List<Integer>> special, List<Integer> needs) {
    return shoppingOffers(price,special, needs, 0);
}

    private int shoppingOffers(List<Integer> price, List<List<Integer>> special, List<Integer> needs, int pos){
        int local_min = directParchase(price, needs);
        for(int i = pos; i < special.size() ; i ++){
            List<Integer> tmp = new ArrayList<>();
            List<Integer> offer = special.get(i);
            for(int j = 0 ; j < needs.size(); j ++){
                if(offer.get(j) > needs.get(j)){
                    tmp = null;
                    break;
                }
                tmp.add(needs.get(j) - offer.get(j));
            }
            if(tmp != null){
                local_min = Math.min(local_min, offer.get(offer.size() - 1) +            shoppingOffers(price, special, tmp, i));
            }
        }
        return local_min;
}

    private int directParchase(List<Integer>price, List<Integer> needs){
        int sum = 0;
        for(int i = 0 ; i < needs.size() ; i ++){
            sum += price.get(i) * needs.get(i);
        }
        return sum;
    }