Number Pyramid (C++) kkmd66

difficulty:

Traverse from bottom to top to find the largest. The larger one of two adjacent numbers is selected and accumulated to the upper layer.

Description:

Consider the digital pyramid shown below. Write a program to calculate the maximum sum of numbers that a path starts from the highest point and ends anywhere at the bottom. Each step can go to the lower left point or the lower right point. 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5 In the example above, the path from 7 to 3 to 8 to 7 to 5 yields the maximum sum: 30

Input:

The first line enters the number of test data groups T; Next, each group of test data: The first line contains R(1<= R<=1000), which represents the number of lines. Each subsequent row contains the integer that a particular row of this number pyramid contains.

Output:

Output the maximum value for each set of data

Sample Input:

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

Sample Output:

30

#include<iostream>
#include "vector"

using namespace std;

/**
 * kkmd66
 * @return
 */

int main() {
    
    
    int t;
    cin >> t;

    //t个案例
    for (int i = 0; i < t; i++) {
    
    
        int r;
        cin >> r;

        //收集
        vector<vector<int>> matrix(r);
        for (int j = 0; j < r; j++) {
    
    
            for (int k = 0; k < j + 1; k++) {
    
    
                int temp;
                cin >> temp;
                matrix[j].push_back(temp);
            }
        }

        //累加找最大
        for (int j = r - 1; j > 0; j--) {
    
    
            for (int k = 0; k < j; k++) {
    
    
                matrix[j][k] > matrix[j][k + 1] ? matrix[j - 1][k] += matrix[j][k] : matrix[j - 1][k] += matrix[j][k +
                                                                                                                   1];
            }
        }

        //输出
        cout << matrix[0][0] << endl;
    }

    return 0;
}