NYOJ 18 The Triangle(经典动态规划)_nyoj-18-the triangle-优快云博客

本文介绍了如何使用动态规划解决NYOJ 18题——The Triangle，通过自底向上的计算方法，从三角形的最后一行开始，逐行计算每个位置的最大路径值，最终找到路径最大值。

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The Triangle

时间限制：1000 ms | 内存限制：65535 KB

难度：4

描述

7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
(Figure 1)
Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.

输入

Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.

输出

Your program is to write to standard output. The highest sum is written as an integer.

样例输入

样例输出

自底向上进行计算。

利用一个新的数组先存储三角形最后一行的值，然后上一行存储最后一行的各值与上一行左右相加的较大值，

以此类推，最后在三角形的定点的得到路径最大值。

#include<iostream>
#include<algorithm>
using namespace std;

int tri[105][105], cal[105][105];

int main()
{
	int n;
	cin >> n;
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= i; j++)
			cin >> tri[i][j];
	for(int j = 1; j <= n; j++)
		cal[n][j] = tri[n][j];
	for(int i = n - 1; i >= 1; i--)
		for(int j = 1; j <= i; j++)
			cal[i][j] = tri[i][j] + max(cal[i + 1][j], cal[i + 1][j + 1]);
	cout << cal[1][1] << endl;
	return 0;
}