22 de noviembre de 2013

Maximum path sum I - Project Euler, problema 18

La descripción de este problema es la siguiente:

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
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)


Aunque quizá hay formas mas eficientes de hacerlo, lo que decidí hacer fue lo siguiente:
  • Empezar el "recorrido" desde la penúltima fila, recorriendo cada columna (entrada).
  • Sumar el máximo entre los 2 números inferiores adyacentes
  • El número de la primera fila y primera columna tendrá al final la súma máxima.
Y aquí está el código(notar que primero tomé los números y les di "formato" en python):

Tr=[75,0,0,0,0,0,0,0,0,0,0,0,0,0,0;
95,64,0,0,0,0,0,0,0,0,0,0,0,0,0;
17,47,82,0,0,0,0,0,0,0,0,0,0,0,0;
18,35,87,10,0,0,0,0,0,0,0,0,0,0,0;
20,4,82,47,65,0,0,0,0,0,0,0,0,0,0;
19,1,23,75,3,34,0,0,0,0,0,0,0,0,0;
88,2,77,73,7,63,67,0,0,0,0,0,0,0,0;
99,65,4,28,6,16,70,92,0,0,0,0,0,0,0;
41,41,26,56,83,40,80,70,33,0,0,0,0,0,0;
41,48,72,33,47,32,37,16,94,29,0,0,0,0,0;
53,71,44,65,25,43,91,52,97,51,14,0,0,0,0;
70,11,33,28,77,73,17,78,39,68,17,57,0,0,0;
91,71,52,38,17,14,91,43,58,50,27,29,48,0,0;
63,66,4,68,89,53,67,30,73,16,69,87,40,31,0;
4,62,98,27,23,9,70,98,73,93,38,53,60,4,23];
[I,J]=size(Tr); 
S=0;
for i=(I-1):-1:1
 for j=1:length(Tr(i,:))-1;
  Tr(i,j)=Tr(i,j)+max(Tr(i+1,j:(j+1)));
 end
end
Tr(1,1)