GoldMine problem

Given a gold mine of n*m dimensions. Each field in this mine contains a positive integer which is the amount of gold in tons. Initially the miner is at first column but can be at any row. He can move only (right->,right up /,right down\) that is from a given cell, the miner can move to the cell diagonally up towards the right or right or diagonally down towards the right. Find out maximum amount of gold he can collect.

// C++ program to solve Gold Mine problem 
#include<bits/stdc++.h> 
using namespace std; 

const int MAX = 100; 

// Returns maximum amount of gold that can be collected 
// when journey started from first column and moves 
// allowed are right, right-up and right-down 
int getMaxGold(int gold[][MAX], int m, int n) 
{ 
	// Create a table for storing intermediate results 
	// and initialize all cells to 0. The first row of 
	// goldMineTable gives the maximum gold that the miner 
	// can collect when starts that row 
	int goldTable[m][n]; 
	memset(goldTable, 0, sizeof(goldTable)); 

	for (int col=n-1; col>=0; col--) 
	{ 
		for (int row=0; row<m; row++) 
		{ 
			// Gold collected on going to the cell on the right(->) 
			int right = (col==n-1)? 0: goldTable[row][col+1]; 

			// Gold collected on going to the cell to right up (/) 
			int right_up = (row==0 || col==n-1)? 0: 
							goldTable[row-1][col+1]; 

			// Gold collected on going to the cell to right down (\) 
			int right_down = (row==m-1 || col==n-1)? 0: 
							goldTable[row+1][col+1]; 

			// Max gold collected from taking either of the 
			// above 3 paths 
			goldTable[row][col] = gold[row][col] + 
							max(right, max(right_up, right_down)); 
													
		} 
	} 

	// The max amount of gold collected will be the max 
	// value in first column of all rows 
	int res = goldTable[0][0]; 
	for (int i=1; i<m; i++) 
		res = max(res, goldTable[i][0]); 
	return res; 
} 

// Driver Code 
int main() 
{ 
	int gold[MAX][MAX]= { {1, 3, 1, 5}, 
		{2, 2, 4, 1}, 
		{5, 0, 2, 3}, 
		{0, 6, 1, 2} 
	}; 
	int m = 4, n = 4; 
	cout << getMaxGold(gold, m, n); 
	return 0; 
}