62 lines
1.6 KiB
Python
62 lines
1.6 KiB
Python
|
from math import floor
|
||
|
|
||
|
def print_table(table):
|
||
|
print('\n-- TABLE --')
|
||
|
for row in range(len(table)):
|
||
|
x = ''
|
||
|
for col in range(len(table[row])):
|
||
|
x += ' ' + str(table[row][col])
|
||
|
print(x)
|
||
|
|
||
|
def leftover(capacity, kg):
|
||
|
n = floor(capacity / kg)
|
||
|
return n, capacity - (n * kg)
|
||
|
|
||
|
def init_table(num_rows, num_cols):
|
||
|
table = []
|
||
|
for _ in range(num_rows):
|
||
|
row = []
|
||
|
for _ in range(num_cols):
|
||
|
row.append(0)
|
||
|
table.append(row)
|
||
|
return table
|
||
|
|
||
|
def get(table, row, col):
|
||
|
if row < 0 or col < 0:
|
||
|
return 0
|
||
|
return table[row][col]
|
||
|
|
||
|
def max_haul(items, capacity):
|
||
|
table = init_table(len(items), capacity)
|
||
|
|
||
|
for row in range(len(table)):
|
||
|
for col in range(len(table[row])):
|
||
|
curr_capacity = col + 1
|
||
|
kg, val = items[row]
|
||
|
# A
|
||
|
a = get(table, row - 1, col)
|
||
|
# B
|
||
|
n, lo = leftover(curr_capacity, kg)
|
||
|
b = (val * n) + get(table, row - 1, lo - 1)
|
||
|
# commit
|
||
|
if kg > curr_capacity:
|
||
|
table[row][col] = a
|
||
|
else:
|
||
|
print(n, lo)
|
||
|
table[row][col] = max([a, b])
|
||
|
print_table(table)
|
||
|
return table[-1][-1]
|
||
|
|
||
|
# There are multiple variants of this problem:
|
||
|
# 1. We're allowed to take multiple of each item.
|
||
|
# 2. We can only take one of each item.
|
||
|
# 3. We can only take a fixed amount of each item.
|
||
|
|
||
|
items = [(7,160), (3,90), (2,15)]
|
||
|
capacity = 20
|
||
|
result = max_haul(items, capacity)
|
||
|
expected = None
|
||
|
print("Result: {} == Expected: {}".format(result, expected))
|
||
|
assert result == expected
|
||
|
print("Success!")
|