给定一个二维网格，其每个像元都包含整数成本，它代表穿越该像元的成本。任务是找到从左下角到右上角的最大成本路径。
注意：仅使用向上和向右移动
例子：
Input : mat[][] = {{20, -10, 0},
{1, 5, 10},
{1, 2, 3}}
Output : 18
(2, 0) ==> (2, 1) ==> (1, 1) ==> (1, 2) ==> (0, 2)
cost for this path is (1+2+5+10+0) = 18
Input : mat[][] = {{1, -2, -3},
{1, 15, 10},
{1, -2, 3}}
Output : 24
先决条件：最小成本路径，允许向左，向右，向下和向上移动
方法：想法是维护一个单独的阵列，以使用队列存储所有单元的最大成本。对于每个单元格，检查到达该单元格的当前成本是否大于先前的成本。如果先前的成本最低，则使用当前成本更新单元格。
下面是上述方法的实现：
C++
// C++ program to find maximum cost to reach
// top right corner from bottom left corner
#include
using namespace std;
#define ROW 3
#define COL 3
// To store matrix cell coordinates
struct Point
{
int x;
int y;
};
// Check whether given cell (row, col)
// is a valid cell or not.
bool isValid(Point p)
{
// Return true if row number and column number
// is in range
return (p.x >=0) && (p.y q;
q.push(src); // Enqueue source cell
// Do a BFS starting from source cell
// on the allowed direction
while (!q.empty())
{
Point curr = q.front();
q.pop();
// Find up point
Point up = {curr.x-1, curr.y};
// if adjacent cell is valid, enqueue it.
if (isValid(up))
{
max_val[up.x][up.y] = max(max_val[up.x][up.y],
mat[up.x][up.y] + max_val[curr.x][curr.y]);
q.push(up);
}
// Find right point
Point right = {curr.x, curr.y+1};
if(isValid(right))
{
max_val[right.x][right.y] = max(max_val[right.x][right.y],
mat[right.x][right.y] + max_val[curr.x][curr.y]);
q.push(right);
}
}
// Return the required answer
return max_val[0][COL-1];
}
// Driver code
int main()
{
int mat[ROW][COL] = { {20, -10, 0},
{1, 5, 10},
{1, 2, 3},};
std::cout<<"Given matrix is "< Java
// Java program to find maximum cost to reach
// top right corner from bottom left corner
import java.util.*;
class GFG
{
static int ROW = 3;
static int COL = 3;
// To store matrix cell coordinates
static class Point
{
int x;
int y;
public Point(int x, int y)
{
this.x = x;
this.y = y;
}
}
// Check whether given cell (row, col)
// is a valid cell or not.
static boolean isValid(Point p)
{
// Return true if row number and column number
// is in range
return (p.x >= 0) && (p.y < COL);
}
// Function to find maximum cost to reach
// top right corner from bottom left corner
static int find_max_cost(int mat[][])
{
int [][]max_val = new int[ROW][COL];
max_val[ROW - 1][0] = mat[ROW - 1][0];
// Starting point
Point src = new Point(ROW - 1, 0);
// Create a queue for traversal
Queue q = new LinkedList<>();
q.add(src); // Enqueue source cell
// Do a BFS starting from source cell
// on the allowed direction
while (!q.isEmpty())
{
Point curr = q.peek();
q.remove();
// Find up point
Point up = new Point(curr.x - 1, curr.y);
// if adjacent cell is valid, enqueue it.
if (isValid(up))
{
max_val[up.x][up.y] = Math.max(max_val[up.x][up.y],
mat[up.x][up.y] + max_val[curr.x][curr.y]);
q.add(up);
}
// Find right point
Point right = new Point(curr.x, curr.y + 1);
if(isValid(right))
{
max_val[right.x][right.y] = Math.max(max_val[right.x][right.y],
mat[right.x][right.y] + max_val[curr.x][curr.y]);
q.add(right);
}
}
// Return the required answer
return max_val[0][COL - 1];
}
// Driver code
public static void main(String[] args)
{
int mat[][] = {{20, -10, 0},
{1, 5, 10},
{1, 2, 3}};
System.out.println("Given matrix is ");
for(int i = 0 ; i < ROW; ++i)
{
for(int j = 0; j < COL; ++j)
System.out.print(mat[i][j] + " ");
System.out.println();
}
System.out.print("Maximum cost is " +
find_max_cost(mat));
}
}
// This code is contributed by PrinciRaj1992 Python
# Python3 program to find maximum cost to reach
# top right corner from bottom left corner
from collections import deque as queue
ROW = 3
COL = 3
# Check whether given cell (row, col)
# is a valid cell or not.
def isValid(p):
# Return true if row number and column number
# is in range
return (p[0] >= 0) and (p[1] < COL)
# Function to find maximum cost to reach
# top right corner from bottom left corner
def find_max_cost(mat):
max_val = [[0 for i in range(COL)] for i in range(ROW)]
max_val[ROW - 1][0] = mat[ROW - 1][0]
# Starting po
src = [ROW - 1, 0]
# Create a queue for traversal
q = queue()
q.appendleft(src) # Enqueue source cell
# Do a BFS starting from source cell
# on the allowed direction
while (len(q) > 0):
curr = q.pop()
# Find up point
up = [curr[0] - 1, curr[1]]
# if adjacent cell is valid, enqueue it.
if (isValid(up)):
max_val[up[0]][up[1]] = max(max_val[up[0]][up[1]],mat[up[0]][up[1]] + max_val[curr[0]][curr[1]])
q.appendleft(up)
# Find right po
right = [curr[0], curr[1] + 1]
if(isValid(right)):
max_val[right[0]][right[1]] = max(max_val[right[0]][right[1]],mat[right[0]][right[1]] + max_val[curr[0]][curr[1]])
q.appendleft(right)
# Return the required answer
return max_val[0][COL - 1]
# Driver code
mat = [[20, -10, 0],
[1, 5, 10],
[1, 2, 3]]
print("Given matrix is ")
for i in range(ROW):
for j in range(COL):
print(mat[i][j],end=" ")
print()
print("Maximum cost is ", find_max_cost(mat))
# This code is contributed by mohit kumar 29C#
// C# program to find maximum cost to reach
// top right corner from bottom left corner
using System;
using System.Collections.Generic;
class GFG
{
static int ROW = 3;
static int COL = 3;
// To store matrix cell coordinates
public class Point
{
public int x;
public int y;
public Point(int x, int y)
{
this.x = x;
this.y = y;
}
}
// Check whether given cell (row, col)
// is a valid cell or not.
static Boolean isValid(Point p)
{
// Return true if row number and
// column number is in range
return (p.x >= 0) && (p.y < COL);
}
// Function to find maximum cost to reach
// top right corner from bottom left corner
static int find_max_cost(int [,]mat)
{
int [,]max_val = new int[ROW,COL];
max_val[ROW - 1, 0] = mat[ROW - 1, 0];
// Starting point
Point src = new Point(ROW - 1, 0);
// Create a queue for traversal
Queue q = new Queue();
q.Enqueue(src); // Enqueue source cell
// Do a BFS starting from source cell
// on the allowed direction
while (q.Count != 0)
{
Point curr = q.Peek();
q.Dequeue();
// Find up point
Point up = new Point(curr.x - 1, curr.y);
// if adjacent cell is valid, enqueue it.
if (isValid(up))
{
max_val[up.x, up.y] = Math.Max(max_val[up.x, up.y],
mat[up.x, up.y] + max_val[curr.x, curr.y]);
q.Enqueue(up);
}
// Find right point
Point right = new Point(curr.x,
curr.y + 1);
if(isValid(right))
{
max_val[right.x, right.y] = Math.Max(max_val[right.x, right.y],
mat[right.x, right.y] + max_val[curr.x, curr.y]);
q.Enqueue(right);
}
}
// Return the required answer
return max_val[0, COL - 1];
}
// Driver code
public static void Main(String[] args)
{
int [,]mat = {{20, -10, 0},
{1, 5, 10},
{1, 2, 3}};
Console.WriteLine("Given matrix is ");
for(int i = 0 ; i < ROW; ++i)
{
for(int j = 0; j < COL; ++j)
Console.Write(mat[i, j] + " ");
Console.WriteLine();
}
Console.Write("Maximum cost is " +
find_max_cost(mat));
}
}
// This code is contributed by Princi Singh 输出：
Given matrix is
20 -10 0
1 5 10
1 2 3
Maximum cost is 18