[Project Euler] 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 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)

 

按照下面的提示，如果把所有的路线的和求出了，会有16384种情况。 

我们从三角形顶向下计算无法剪枝了，我们换一种思路来计算。从下往上计算，把小的结果直接舍去。

   3              3            3         23
  7 4            7 4         20 19
 2 4 6   ->    10 13 15  ->           ->
8 5 9 3

 

可以看出，只要简单的阅历一遍

我们用同样的思路来解决那个15层的三角形

#include < iostream >
using namespace std;

int main(){
int item[ 120 ] = {
75 ,
95 , 64 ,
17 , 47 , 82 ,
18 , 35 , 87 , 10 ,
20 , 4 , 82 , 47 , 65 ,
19 , 1 , 23 , 75 , 3 , 34 ,
88 , 2 , 77 , 73 , 7 , 63 , 67 ,
99 , 65 , 4 , 28 , 6 , 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 , 4 , 68 , 89 , 53 , 67 , 30 , 73 , 16 , 69 , 87 , 40 , 31 ,
4 , 62 , 98 , 27 , 23 , 9 , 70 , 98 , 73 , 93 , 38 , 53 , 60 , 4 , 23
};
int step = 15 ;
int tmp1,tmp2;
for ( int i = 14 ; i > 0 ; i -- ){
tmp1 = i * (i - 1 ) / 2 - 1 ;
tmp2 = i * (i + 1 ) / 2 - 1 ;
for ( int j = i; j > 0 ; j -- ){
item[tmp1 + j] += item[tmp2 + j] > item[tmp2 + j + 1 ] ? item[tmp2 + j]:item[tmp2 + j + 1 ];
}
}
cout << item[ 0 ] << endl;
}