Question

我正在编写一个贪婪算法（Python 3.x.x）用于'宝石抢劫'。鉴于一系列珠宝和价值，该计划抓住了它可以放入袋子里的最有价值的宝石，而不超过袋子重量限制。我在这里有三个测试用例，它适用于其中两个。

每个测试用例都以相同的方式编写：第一行是行李重量限制，后面的所有行都是元组（重量，值）。

示例案例1（正常）：

示例案例2（不起作用）：

代码：

def take_input(infile):
    f_open = open(infile, 'r')
    lines = []
    for line in f_open:
        lines.append(line.strip())
    f_open.close()
    return lines

def set_weight(weight):
    bag_weight = weight
    return bag_weight

def jewel_list(lines):
    jewels = []
    for item in lines:
        jewels.append(item.split())
    jewels = sorted(jewels, reverse= True)
    jewel_dict = {}
    for item in jewels:
        jewel_dict[item[1]] = item[0]
    return jewel_dict

def greedy_grab(weight_max, jewels):
    #first, we get a list of values
    values = []
    weights = []
    for keys in jewels:
        weights.append(jewels[keys])
    for item in jewels.keys():
        values.append(item)
    values = sorted(values, reverse= True)
    #then, we start working
    max = int(weight_max)
    running = 0
    i = 0
    grabbed_list = []
    string = ''
    total_haul = 0
    # pick the most valuable item first. Pick as many of them as you can.            
    # Then, the next, all the way through.
    while running < max:
        next_add = int(jewels[values[i]])
        if (running + next_add) > max:
            i += 1
        else:
            running += next_add
            grabbed_list.append(values[i])
    for item in grabbed_list:
        total_haul += int(item)
    string = "The greedy approach would steal $" + str(total_haul) + " of  
             jewels."
    return string

infile = "JT_test2.txt"
lines = take_input(infile)
#set the bag weight with the first line from the input
bag_max = set_weight(lines[0])
#once we set bag weight, we don't need it anymore
lines.pop(0)

#generate a list of jewels in a dictionary by weight, value
value_list = jewel_list(lines)
#run the greedy approach
print(greedy_grab(bag_max, value_list))

有没有人有任何线索为什么它不适用于案例2？非常感谢您的帮助。 编辑：案例2的预期结果是6130美元。我好像得到了6090美元。

Answer 1

您的字典键是字符串，而不是整数，因此当您尝试对它们进行排序时，它们会按字符串排序。所以你会得到：

['6000', '3000', '30', '100']

反而想要：

['6000', '3000', '100', '30']

将此函数更改为此类并具有整数键：

def jewel_list(lines):
    jewels = []
    for item in lines:
        jewels.append(item.split())
    jewels = sorted(jewels, reverse= True)
    jewel_dict = {}
    for item in jewels:
        jewel_dict[int(item[1])] = item[0]  # changed line
    return jewel_dict

当你改变它时它会给你：

The greedy approach would steal $6130 of jewels.

Answer 2

In [237]: %paste
def greedy(infilepath):
  with open(infilepath) as infile:
    capacity = int(infile.readline().strip())
    items = [map(int, line.strip().split()) for line in infile]

  bag = []
  items.sort(key=operator.itemgetter(0))
  while capacity and items:
    if items[-1][0] <= capacity:
      bag.append(items[-1])
      capacity -= items[-1][0]
    items.pop()
  return bag

## -- End pasted text --

In [238]: sum(map(operator.itemgetter(1), greedy("JT_test1.txt")))
Out[238]: 8

In [239]: sum(map(operator.itemgetter(1), greedy("JT_test2.txt")))
Out[239]: 6130

Answer 3

我认为在这段代码中i也必须在其他方面增加

while running < max:
  next_add = int(jewels[values[i]])
  if (running + next_add) > max:
    i += 1
  else:
    running += next_add
    grabbed_list.append(values[i])
    i += 1 #here

这和@ iblazevic的答案解释了为什么它的表现如此

Python贪心算法

3 个答案: