ACWING60. The maximum value of the gift (to prove safety offer)

In each cell a m × n board are placed a gift, each gift has a certain value (value greater than 0).

You can begin to get a gift from the upper left corner of the grid the board, and each time to the right or down until you reach the bottom right corner of a checkerboard grid.

Given a chessboard and above gift, calculate how much you can get the most valuable gift?

note:

m, n> 0
Sample:

Input:
[
[2,3,1],
[1,7,1],
[4, 6,]
]

Output: 19

Explanation: get the maximum value of the gift can be obtained along path 2 → 3 → 7 → 6 → 1.

DP does not explain naked

class Solution {
public:
    int getMaxValue(vector<vector<int>>& grid) {
        int n = 0,m = 0;n = grid.size();
        if(n != 0) m = grid[0].size();
        int dp[n + 1][m + 1];memset(dp,0,sizeof(dp));
        for(int i = 1;i <= n;i++) {
            for(int j = 1;j <= m;j++) {
                dp[i][j] = grid[i - 1][j - 1] + max(dp[i - 1][j],dp[i][j - 1]);
            }
        }
        return dp[n][m];
    }
};