Problem 18

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 5

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)

動的計画法のようなもの(笑)。

Problem 67も数字を15->100に変えて解ける。

import Data.Array.IArray
getTr ::Int->IO (Array Int Int)
getTr n=getContents>>=return.listArray(1,n*(n+1)`div`2).map read.words
sumTr n tr = tr'!(1,1)
where tr'=listArray((1,1),(n,n))[a i j|(i,j)<-range((1,1),(n,n))]::Array(Int,Int)Int
idx i j = i*(i-1)`div`2+j
a k l |k==n =  tr!(idx k l)
|otherwise = (tr'!)(k+1,l)`max` (tr'!)(k+1,l+1)+tr!(idx k l)
main =getTr 15 >>=return.sumTr 15>>=putStrLn.show
More Reading
Newer// Problem 17
Older// Problem 19