递推和记忆化搜索--The Triangle--poj1163

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.

Input

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.

Output

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

Sample Input

5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
Sample Output

30

递推代码

#include <iostream>
#include <vector>
#include <algorithm>

int maxSum(const std::vector<std::vector<int>>& triangle) {
   
    if (triangle.empty()) return 0;

    // 创建一个三角形的拷贝，以便我们可以在其中进行计算
    std::vector<std::vector<int>> dp = triangle;

    // 从倒数第二行开始往上。因为最后一行的数字已经代表了从那里出发到底部的最大路径和。
    for (int i = dp.size() - 2; i >= 0; --i) {
   
        for (int j = 0; j < dp[i].size(); ++j) {
   
            // 将当前数字加上其下方的两个数字中的较大者
            dp[i][j]