Project Euler：Problem 18 Maximum path sum I

标签：project euler   c++   

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)

安装这题目的说法，可以暴力搜索来做

自顶向下：

#include <iostream>
using namespace std;

int main()
{
	int s[15][15];
	for (int i = 0; i < 15; i++)
	{
		for (int j = 0; j <= i; j++)
			cin >> s[i][j];
	}
	for (int i = 1; i < 15; i++)
	{
		for (int j = 1; j <= i; j++)
		{
			if (j == 0)
			{
				s[i][j] += s[i - 1][j];
			}
			else if (i == j)
				s[i][j] += s[i - 1][j - 1];
			else
			{
				int tp = s[i - 1][j - 1] > s[i - 1][j] ? s[i - 1][j - 1] : s[i - 1][j];
				s[i][j] += tp;
			}
		}
	}
	int res = 0;
	for (int i = 0; i < 15; i++)
	{
		if (res < s[14][i])
			res = s[14][i];
	}
	cout << res << endl;
	system("pause");
	return 0;
}

自底向上：

#include <iostream>
using namespace std;

int main()
{
	int s[15][15];
	for (int i = 0; i < 15; i++)
	{
		for (int j = 0; j <= i; j++)
			cin >> s[i][j];
	}
	for (int i = 14; i >0; i--)
	{
		for (int j = 0; j < i; j++)
		{
			int tp = s[i][j]>s[i][j + 1] ? s[i][j] : s[i][j + 1];
			s[i - 1][j] += tp;
		}
	}
	cout << s[0][0] << endl;
	system("pause");
	return 0;
}

Project Euler：Problem 18 Maximum path sum I

标签：project euler   c++   

原文地址：http://blog.csdn.net/youb11/article/details/46286355