######################################################################
# By starting at the top of the triangle below and moving to adjacent numbers on the
# row below, the maximum total from top to bottom is 23.
#
# 3
# 7 4
# 2 4 6
# 8 5 9 3
#
# That is, 3 + 7 + 4 + 9 = 23.
#
# Find the maximum total from top to bottom of the triangle below:
#
# 75
# 95 64
# 17 47 82
# 18 35 87 10
# 20 04 82 47 65
# 19 01 23 75 03 34
# 88 02 77 73 07 63 67
# 99 65 04 28 06 16 70 92
# 41 41 26 56 83 40 80 70 33
# 41 48 72 33 47 32 37 16 94 29
# 53 71 44 65 25 43 91 52 97 51 14
# 70 11 33 28 77 73 17 78 39 68 17 57
# 91 71 52 38 17 14 91 43 58 50 27 29 48
# 63 66 04 68 89 53 67 30 73 16 69 87 40 31
# 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
#
# NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route.
# However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved
# by brute force, and requires a clever method! ;o)
######################################################################

triangle = "75 \n\
95 64 \n\
17 47 82 \n\
18 35 87 10 \n\
20 04 82 47 65 \n\
19 01 23 75 03 34 \n\
88 02 77 73 07 63 67 \n\
99 65 04 28 06 16 70 92 \n\
41 41 26 56 83 40 80 70 33 \n\
41 48 72 33 47 32 37 16 94 29 \n\
53 71 44 65 25 43 91 52 97 51 14 \n\
70 11 33 28 77 73 17 78 39 68 17 57 \n\
91 71 52 38 17 14 91 43 58 50 27 29 48 \n\
63 66 04 68 89 53 67 30 73 16 69 87 40 31 \n\
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23 \n"

tri = []
temp = []
tempstr = ''

for c in triangle:
    if c is " ":
        temp.append(int(tempstr))
        tempstr = ""
    elif c is "\n":
        tri.append(temp)
        temp = []
    else:
        tempstr += c

tri.reverse()

for row in range(1, len(tri)):
    for number in range(0, len(tri[row])):
        tri[row][number] += max(tri[row - 1][number], tri[row - 1][number + 1])

tri.reverse()

print(tri[0][0])
# Solution: 1074