Problem 18

Project Euler

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