Problem 18

Maximum path sum I

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)

最大路径和 I

从如下数字三角形的顶端出发，不断移动到下一行与其相邻的数直至到达底部，所能得到的最大路径和是 23。

如上图，最大路径和为 3 + 7 + 4 + 9 = 23。

从如下数字三角形的顶端出发到达底部，求所能得到的最大路径和。

注意：在这个问题中，由于只有 16384 条路径，通过穷举所有的路径来解决问题是可行的。然而，对于第 67 题，虽然是一道相同类型的题目，但是其中的数字三角形将拥有一百行，就不再能够通过暴力枚举的方法来解决，而需要一个更加聪明的办法！;o)

经典的动态规划问题，不过这里我们首先还是给出穷举解法：

(defvar eu18-data
  [[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]])

(defun eu18-find (i j)
  (if (= i 0) (aref (aref eu18-data 0) 0)
    (cond
     ((= j 0) (+ (aref (aref eu18-data i) j)
		 (eu18-find (1- i) 0)))
     ((= j i) (+ (aref (aref eu18-data i) j)
		 (eu18-find (1- i) (1- j))))
     (t
      (+ (aref (aref eu18-data i) j)
	 (max (eu18-find (1- i) (1- j))
	      (eu18-find (1- i) j)))))))

(apply 'max (cl-loop for i from 0 below 15
		     collect (eu18-find 14 i)))
;; 1074

由于我比较懒，这里就直接在递归法的基础上做修改了，我就不写迭代法了：

(setq eu18-h (make-hash-table :test 'equal))

(defun eu18-find2 (i j)
  (if-let ((v (gethash (cons i j) eu18-h)))
      v
    (cond
     ((= i 0) (puthash '(0 0) (aref (aref eu18-data 0) 0) eu18-h))
     ((= j 0) (puthash `(,i ,j) (+ (aref (aref eu18-data i) j)
				   (eu18-find2 (1- i) 0))
		       eu18-h))
     ((= j i) (puthash `(,i ,j) (+ (aref (aref eu18-data i) j)
				   (eu18-find2 (1- i) (1- j)))
		       eu18-h))
     (t
      (puthash `(,i ,j) (+ (aref (aref eu18-data i) j)
			   (max (eu18-find2 (1- i) (1- j))
				(eu18-find2 (1- i) j)))
	       eu18-h)))))

老实说提升并不大，毕竟规模还没有上来。

Problem 18

1. Problem

2. Solution