给定一个n * m尺寸的金矿。该矿场中的每个字段都包含一个正整数，该整数是黄金的吨数。最初，矿工位于第一列，但可以位于任何行。他只能从给定单元移动(右->，右向上/，右向下\)，矿工可以向右或向右斜向右或向右对角向下移至该单元。找出他可以收集的最大数量的黄金。
例子：
Input : mat[][] = {{1, 3, 3},
{2, 1, 4},
{0, 6, 4}};
Output : 12
{(1,0)->(2,1)->(2,2)}
Input: mat[][] = { {1, 3, 1, 5},
{2, 2, 4, 1},
{5, 0, 2, 3},
{0, 6, 1, 2}};
Output : 16
(2,0) -> (1,1) -> (1,2) -> (0,3) OR
(2,0) -> (3,1) -> (2,2) -> (2,3)
Input : mat[][] = {{10, 33, 13, 15},
{22, 21, 04, 1},
{5, 0, 2, 3},
{0, 6, 14, 2}};
Output : 83来源Flipkart访谈
创建与给定矩阵mat [] []相同的二维矩阵goldTable [] [])。如果我们仔细观察这个问题，我们会注意到以下情况。
- 金的数量是正数，因此我们希望在给定的约束下覆盖最大值的最大像元。
- 在每一步中，我们都向右侧移动了一步。因此，我们总是最后一列。如果我们在最后一列，那么我们将无法向右移动
如果我们在第一行或最后一列，那么我们将无法向右上移，因此只需分配0即可，否则将goldTable [row-1] [col + 1]的值分配给right_up。如果我们在最后一行或最后一列，那么我们将无法向下移动，因此只分配0即可，否则将goldTable [row + 1] [col + 1]的值分配给右侧。
现在找到right，right_up和right_down的最大值，然后将其与该mat [row] [col]相加。最后，找到所有行和第一列的最大值，然后将其返回。
C++14
// C++ program to solve Gold Mine problem
#include
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)
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 Java
// Java program to solve Gold Mine problem
import java.util.Arrays;
class GFG {
static final 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
static int getMaxGold(int gold[][],
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[][] = new int[m][n];
for(int[] rows:goldTable)
Arrays.fill(rows, 0);
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]
+ Math.max(right, Math.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 = Math.max(res, goldTable[i][0]);
return res;
}
//driver code
public static void main(String arg[])
{
int gold[][]= { {1, 3, 1, 5},
{2, 2, 4, 1},
{5, 0, 2, 3},
{0, 6, 1, 2} };
int m = 4, n = 4;
System.out.print(getMaxGold(gold, m, n));
}
}
// This code is contributed by Anant Agarwal.Python3
# Python program to solve
# Gold Mine problem
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
def getMaxGold(gold, m, 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
goldTable = [[0 for i in range(n)]
for j in range(m)]
for col in range(n-1, -1, -1):
for row in range(m):
# Gold collected on going to
# the cell on the rigth(->)
if (col == n-1):
right = 0
else:
right = goldTable[row][col+1]
# Gold collected on going to
# the cell to right up (/)
if (row == 0 or col == n-1):
right_up = 0
else:
right_up = goldTable[row-1][col+1]
# Gold collected on going to
# the cell to right down (\)
if (row == m-1 or col == n-1):
right_down = 0
else:
right_down = goldTable[row+1][col+1]
# Max gold collected from taking
# either of the above 3 paths
goldTable[row][col] = gold[row][col] + max(right, right_up, right_down)
# The max amount of gold
# collected will be the max
# value in first column of all rows
res = goldTable[0][0]
for i in range(1, m):
res = max(res, goldTable[i][0])
return res
# Driver code
gold = [[1, 3, 1, 5],
[2, 2, 4, 1],
[5, 0, 2, 3],
[0, 6, 1, 2]]
m = 4
n = 4
print(getMaxGold(gold, m, n))
# This code is contributed
# by Soumen Ghosh.C#
// C# program to solve Gold Mine problem
using System;
class GFG
{
static 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
static int getMaxGold(int[,] gold,
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 = new int[m, n];
for(int i = 0; i < m; i++)
for(int j = 0; j < n; j++)
goldTable[i, j] = 0;
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] +
Math.Max(right, Math.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 = Math.Max(res, goldTable[i, 0]);
return res;
}
// Driver Code
static void Main()
{
int[,] gold = new int[,]{{1, 3, 1, 5},
{2, 2, 4, 1},
{5, 0, 2, 3},
{0, 6, 1, 2}
};
int m = 4, n = 4;
Console.Write(getMaxGold(gold, m, n));
}
}
// This code is contributed by DrRoot_PHP
= 0 ; $col--)
{
for ($row = 0 ; $row < $m ; $row++)
{
// Gold collected on going to
// the cell on the rigth(->)
if ($col == $n - 1)
$right = 0 ;
else
$right = $goldTable[$row][$col + 1];
// Gold collected on going to
// the cell to right up (/)
if ($row == 0 or $col == $n - 1)
$right_up = 0 ;
else
$right_up = $goldTable[$row - 1][$col + 1];
// Gold collected on going to
// the cell to right down (\)
if ($row == $m - 1 or $col == $n - 1)
$right_down = 0 ;
else
$right_down = $goldTable[$row + 1][$col + 1];
// Max gold collected from taking
// either of the above 3 paths
$goldTable[$row][$col] = $gold[$row][$col] +
max($right, $right_up,
$right_down);
}
}
// The max amount of gold collected will be the
// max value in first column of all rows
$res = $goldTable[0][0] ;
for ($i = 0; $i < $m; $i++)
$res = max($res, $goldTable[$i][0]);
return $res;
}
// Driver code
$gold = array(array(1, 3, 1, 5),
array(2, 2, 4, 1),
array(5, 0, 2, 3),
array(0, 6, 1, 2));
$m = 4 ;
$n = 4 ;
echo getMaxGold($gold, $m, $n) ;
// This code is contributed by Ryuga
?>输出
16