[Project Euler] 来做欧拉项目练习题吧: 题目018

                                      [Project Euler] 来做欧拉项目练习题吧: 题目018

                                                        周银辉 

 

问题描述:

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! 

 

 

问题分析:

题目中数据看上去是一个三角形,但由于题目要求的是从一个点出发沿着其相邻的两个点向下一行前进,点和其下一行相邻的两个点刚好构成一个二叉树。
设A[i,j]表示第i行的第j个数,那么A[i,j]相邻的两个点就是A[i+1,j]和A[i+1,j+1],也就是它的左右子树。
变化成二叉树就好办了.设best[A]表示以点A出发点并且满足题目要求的运算结果,节点A在数组中对应的数值为(A), 节点A的左右子树分别为B和C,
那么best[A] = (A) + max(best[B], best[C]), 呵呵,递归的
(参考代码中的buffer是为了避免重复递归运算而设置的缓冲区 )

#include <stdio.h>


#define SZ 15
#define END (SZ-1)

int data[SZ][SZ] = 
{
{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 buffer[SZ][SZ];

int test(int i, int j)
{
if(j>i) //j>i时是上面数组中没有显式标明数字的那些区域,不参与运算
{
return 0;
}

int sum, sumLeft=0, sumRight=0, ni, nj; // ni: next i
int currentNode;
currentNode = data[i][j];

if(i==END)
{
return currentNode;
}

ni = i+1;
nj = j+1;

if(ni<SZ)
{
if(buffer[ni][j] != 0)
{
sumLeft = buffer[ni][j];
}
else
{
sumLeft = test(ni, j);
buffer[ni][j] = sumLeft;
}
}

if(nj<SZ)
{
if(buffer[ni][nj] != 0)
{
sumRight = buffer[ni][nj];
}
else
{
sumRight = test(ni, nj);
buffer[ni][nj] = sumRight;
}
}

sum = currentNode + (sumLeft>sumRight ? sumLeft:sumRight);
buffer[i][j] = sum;

return sum;
}

int main()
{
printf("result is %d\n", test(0,0));
return 0;

} 

 

速度还不错(intel 双核2.4G):

real    0m0.004s
user    0m0.001s

sys     0m0.003s 

posted @ 2011-03-29 22:28  周银辉  阅读(2169)  评论(2编辑  收藏