三角形最长路径

最新推荐文章于 2023-02-23 18:15:35 发布

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最新推荐文章于 2023-02-23 18:15:35 发布

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原文链接：http://www.cnblogs.com/xiaoyi115/p/3176520.html

本文介绍了一种计算三角形网格中从顶部到底部的最大路径和的算法。通过动态规划的方法，计算到达每个节点的最大路径和，最终找出整体的最大路径值。

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

解题思路：

在解本题的时候，我习惯性的想到利用最长递增子序列等问题的求解方式，即要求i到j行的距离A[i,j]，需要求到A[i,j-1]即i到j-1行的路径最长，但是通过这种思路，并不能够

找到一个对应的等式。后来通过借鉴别人的思路，可以求出到达每个点的最长路径，再求出其最长路径，再由某点i,j，能够到达该点的上一行一定是F[i-1,j-1]或者F[i-1,j]。比较其大值加上该点的长度即可。

设某点的路径数值为M[i,j]

F[i,j]=Max(F[i-1,j-1],F[i-1,j])+M[i,j]

 1 public static int BiggestSum(int[,] M, int len) 
 2         {
 3             int[,] F = new int[len+1, len+1];
 4             int biggest = 0;
 5             for (int i = 0; i <= len; i++)
 6                 for (int j = 0; j <= len; j++)
 7                     F[i, j] = 0;
 8             for (int i = 1; i <= len; i++)
 9                 for (int j = 1; j <= i; j++)
10                 {
11                     F[i, j] = Math.Max(F[i - 1, j - 1], F[i - 1, j]) + M[i, j];
12                     if (F[i, j] > biggest)
13                         biggest = F[i, j];
14                 }
15             return biggest;
16         }