给定一个整数矩阵，其中每个元素代表电池的重量。在矩阵[NXN]中找到权重最大的路径。路径遍历规则是：
- 它应该从左上角的元素开始。
- 路径可以在最后一行的任何元素处结束。
- 我们可以从一个单元格(i，j)移至以下两个单元格。
- 下移：(i + 1，j)
- 对角移动：(i + 1，j + 1)
例子：
Input : N = 5
mat[5][5] = {{ 4, 2 ,3 ,4 ,1 },
{ 2 , 9 ,1 ,10 ,5 },
{15, 1 ,3 , 0 ,20 },
{16 ,92, 41, 44 ,1},
{8, 142, 6, 4, 8} };
Output : 255
Path with max weight : 4 + 2 +15 + 92 + 142 = 255
可以递归定义以上问题。
Let maxCost(i, j) be the cost maximum cost to
reach mat[i][j]. Since end point can be any point
in last row, we finally return maximum of all values
maxCost(N-1, j) where j varies from 0 to N-1.
If i == 0 and j == 0
maxCost(0, 0) = mat[0][0]
// We can traverse through first column only by
// down move
Else if j = 0
maxCost(i, 0) = maxCost(i-1, 0) + mat[i][0]
// In other cases, a cell mat[i][j] can be reached
// through previous two cells ma[i-1][j] and
// mat[i-1][j-1]
Else
maxCost(i, j) = mat[i][j] + max(maxCost(i-1, j),
maxCost(i-1, j-1)),
如果我们绘制上述递归解的递归树，则可以观察到重叠的子问题。由于该问题具有重叠的子问题，因此我们可以使用动态编程有效地解决它。以下是基于动态编程的解决方案。
C++
// C++ program to find the path having the
// maximum weight in matrix
#include
using namespace std;
const int MAX = 1000;
/* Function which return the maximum weight
path sum */
int maxCost(int mat[][MAX], int N)
{
// creat 2D matrix to store the sum of the path
int dp[N][N];
memset(dp, 0, sizeof(dp));
dp[0][0] = mat[0][0];
// Initialize first column of total weight
// array (dp[i to N][0])
for (int i=1; i Java
// Java Code for Maximum weight path ending at
// any element of last row in a matrix
import java.util.*;
class GFG {
/* Function which return the maximum weight
path sum */
public static int maxCost(int mat[][], int N)
{
// create 2D matrix to store the sum of
// the path
int dp[][]=new int[N][N];
dp[0][0] = mat[0][0];
// Initialize first column of total
// weight array (dp[i to N][0])
for (int i = 1; i < N; i++)
dp[i][0] = mat[i][0] + dp[i-1][0];
// Calculate rest path sum of weight matrix
for (int i = 1; i < N; i++)
for (int j = 1; j < i + 1 && j < N; j++)
dp[i][j] = mat[i][j] +
Math.max(dp[i-1][j-1],
dp[i-1][j]);
// find the max weight path sum to reach
// the last row
int result = 0;
for (int i = 0; i < N; i++)
if (result < dp[N-1][i])
result = dp[N-1][i];
// return maximum weight path sum
return result;
}
/* Driver program to test above function */
public static void main(String[] args)
{
int mat[][] = { { 4, 1 ,5 ,6 , 1 },
{ 2 ,9 ,2 ,11 ,10 },
{ 15,1 ,3 ,15, 2 },
{ 16, 92, 41,4,3},
{ 8, 142, 6, 4, 8 }
};
int N = 5;
System.out.println("Maximum Path Sum : "+
maxCost(mat, N));
}
}
// This code is contributed by Arnav Kr. Mandal.Python3
# Python3 program to find the path
# having the maximum weight in matrix
MAX = 1000
# Function which return the
# maximum weight path sum
def maxCost(mat, N):
# creat 2D matrix to store the sum of the path
dp = [[0 for i in range(N)] for j in range(N)]
dp[0][0] = mat[0][0]
# Initialize first column of total weight
# array (dp[i to N][0])
for i in range(1, N):
dp[i][0] = mat[i][0] + dp[i - 1][0]
# Calculate rest path sum of weight matrix
for i in range(1, N):
for j in range(1, min(i + 1, N)):
dp[i][j] = mat[i][j] + \
max(dp[i - 1][j - 1],
dp[i - 1][j])
# find the max weight path sum to reach
# the last row
result = 0
for i in range(N):
if (result < dp[N - 1][i]):
result = dp[N - 1][i]
# return maximum weight path sum
return result
# Driver Program
mat = [ [4, 1 ,5 ,6 , 1],
[2 ,9 ,2 ,11 ,10],
[15,1 ,3 ,15, 2],
[16, 92, 41,4,3],
[8, 142, 6, 4, 8]]
N = 5
print('Maximum Path Sum :', maxCost(mat, N))
# This code is contributed by Soumen Ghosh.C#
// C# Code for Maximum weight path
// ending at any element of last
// row in a matrix
using System;
class GFG {
/* Function which return the
maximum weight path sum */
public static int maxCost(int [,] mat, int N)
{
// create 2D matrix to store the
// sum of the path
int [,] dp = new int[N,N];
dp[0,0] = mat[0,0];
// Initialize first column of total
// weight array (dp[i to N][0])
for (int i = 1; i < N; i++)
dp[i,0] = mat[i,0] + dp[i-1,0];
// Calculate rest path sum of weight matrix
for (int i = 1; i < N; i++)
for (int j = 1; j < i + 1 && j < N; j++)
dp[i,j] = mat[i,j] +
Math.Max(dp[i-1,j-1], dp[i-1,j]);
// find the max weight path sum to reach
// the last row
int result = 0;
for (int i = 0; i < N; i++)
if (result < dp[N-1,i])
result = dp[N-1,i];
// return maximum weight path sum
return result;
}
/* Driver program to test above function */
public static void Main()
{
int [,] mat = { { 4, 1 ,5 ,6 , 1 },
{ 2 ,9 ,2 ,11 ,10 },
{ 15,1 ,3 ,15, 2 },
{ 16, 92, 41,4,3},
{ 8, 142, 6, 4, 8 }
};
int N = 5;
Console.Write("Maximum Path Sum : "
+ maxCost(mat, N));
}
}
// This code is contributed by KRV.PHP
输出：
Maximum Path Sum : 255
时间复杂度： O(N * N)