本文概述
- 建议:在继续解决方案之前, 请先在{IDE}上尝试使用你的方法。
- C ++
- Java
- Python3
- C#
在从arr [0] [0]toarr [N – 1] [M – 1], 你可以在上下方向以及从arr [N – 1] [M – 1]toarr [0] [0], 你可以上下移动。
注意:两条路径都不能相等, 即必须至少有一个像元arr [i] [j]这在两条路径中都不常见。
例子:
Input: mat[][]= {{1, 0, 3, -1}, {3, 5, 1, -2}, {-2, 0, 1, 1}, {2, 1, -1, 1}}Output: 16Maximum sum on path from arr[0][0] to arr[3][3] = 1 + 3 + 5 + 1 + 1 + 1 + 1 = 13Maximum sum on path from arr[3][3] to arr[0][0] = 3Total path sum = 13 + 3 = 16Input: mat[][]= {{1, 0}, {1, 1}}Output: 3
推荐:请尝试使用{IDE}首先, 在继续解决方案之前。方法:
这个问题有点类似于
最低成本路径
问题是, 除了在本问题中, 将找到两个具有最大和的路径。另外, 我们需要注意两条路径上的单元仅对总和贡献一次。
首先要注意的是
arr [N – 1] [M – 1]
to
arr [0] [0]
只是别的途径而已
arr [0] [0]
to
arr [N – 1] [M – 1]
。因此, 我们必须找到两条路径
arr [0] [0]
to
arr [N – 1] [M – 1]
最大金额。
与最小成本路径问题类似, 我们从
arr [0] [0]
【矩阵从上到下以及到后的最大求和路径】在一起并递归到矩阵的相邻单元, 直到我们到达
arr [N – 1] [M – 1]
。为了确保一个单元格的贡献不止一次, 我们检查两条路径上的当前单元格是否相同。如果它们相同, 则仅将其添加到答案一次。
下面是上述方法的实现:
C ++
// C++ implementation of the approach
#include <
bits/stdc++.h>
using namespace std;
// Input matrix
int n = 4, m = 4;
int arr[4][4] = { { 1, 0, 3, -1 }, { 3, 5, 1, -2 }, { -2, 0, 1, 1 }, { 2, 1, -1, 1 } };
// DP matrix
int cache[5][5][5];
// Function to return the sum of the cells
// arr[i1][j1] and arr[i2][j2]
int sum( int i1, int j1, int i2, int j2)
{
if (i1 == i2 &
&
j1 == j2) {
return arr[i1][j1];
}
return arr[i1][j1] + arr[i2][j2];
}// Recursive function to return the
// required maximum cost path
int maxSumPath( int i1, int j1, int i2)
{// Column number of second path
int j2 = i1 + j1 - i2;
// Base Case
if (i1 >
= n || i2 >
= n || j1 >
= m || j2 >
= m) {
return 0;
}// If already calculated, return from DP matrix
if (cache[i1][j1][i2] != -1) {
return cache[i1][j1][i2];
}
int ans = INT_MIN;
// Recurring for neighbouring cells of both paths together
ans = max(ans, maxSumPath(i1 + 1, j1, i2 + 1) + sum(i1, j1, i2, j2));
ans = max(ans, maxSumPath(i1, j1 + 1, i2) + sum(i1, j1, i2, j2));
ans = max(ans, maxSumPath(i1, j1 + 1, i2 + 1) + sum(i1, j1, i2, j2));
ans = max(ans, maxSumPath(i1 + 1, j1, i2) + sum(i1, j1, i2, j2));
// Saving result to the DP matrix for current state
cache[i1][j1][i2] = ans;
return ans;
}// Driver code
int main()
{
memset (cache, -1, sizeof (cache));
cout <
<
maxSumPath(0, 0, 0);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{ // Input matrix
static int n = 4 , m = 4 ;
static int arr[][] = { { 1 , 0 , 3 , - 1 }, { 3 , 5 , 1 , - 2 }, { - 2 , 0 , 1 , 1 }, { 2 , 1 , - 1 , 1 } };
// DP matrix
static int cache[][][] = new int [ 5 ][ 5 ][ 5 ];
// Function to return the sum of the cells
// arr[i1][j1] and arr[i2][j2]
static int sum( int i1, int j1, int i2, int j2)
{
if (i1 == i2 &
&
j1 == j2)
{
return arr[i1][j1];
}
return arr[i1][j1] + arr[i2][j2];
}// Recursive function to return the
// required maximum cost path
static int maxSumPath( int i1, int j1, int i2)
{// Column number of second path
int j2 = i1 + j1 - i2;
// Base Case
if (i1 >
= n || i2 >
= n || j1 >
= m || j2 >
= m)
{
return 0 ;
}// If already calculated, return from DP matrix
if (cache[i1][j1][i2] != - 1 )
{
return cache[i1][j1][i2];
}
int ans = Integer.MIN_VALUE;
// Recurring for neighbouring cells of both paths together
ans = Math.max(ans, maxSumPath(i1 + 1 , j1, i2 + 1 ) + sum(i1, j1, i2, j2));
ans = Math.max(ans, maxSumPath(i1, j1 + 1 , i2) + sum(i1, j1, i2, j2));
ans = Math.max(ans, maxSumPath(i1, j1 + 1 , i2 + 1 ) + sum(i1, j1, i2, j2));
ans = Math.max(ans, maxSumPath(i1 + 1 , j1, i2) + sum(i1, j1, i2, j2));
// Saving result to the DP matrix for current state
cache[i1][j1][i2] = ans;
return ans;
}// Driver code
public static void main(String args[])
{
//set initial value
for ( int i= 0 ;
i<
5 ;
i++)
for ( int i1= 0 ;
i1<
5 ;
i1++)
for ( int i2= 0 ;
i2<
5 ;
i2++)
cache[i][i1][i2]=- 1 ;
System.out.println( maxSumPath( 0 , 0 , 0 ));
}
}// This code is contributed by Arnab Kundu
Python3
# Python 3 implementation of the approach
import sys# Input matrix
n = 4
m = 4
arr = [[ 1 , 0 , 3 , - 1 ], [ 3 , 5 , 1 , - 2 ], [ - 2 , 0 , 1 , 1 ], [ 2 , 1 , - 1 , 1 ]]# DP matrix
cache = [[[ - 1 for i in range ( 5 )] for j in range ( 5 )] for k in range ( 5 )]# Function to return the sum of the cells
# arr[i1][j1] and arr[i2][j2]
def sum (i1, j1, i2, j2):
if (i1 = = i2 and j1 = = j2):
return arr[i1][j1]
return arr[i1][j1] + arr[i2][j2]# Recursive function to return the
# required maximum cost path
def maxSumPath(i1, j1, i2):# Column number of second path
j2 = i1 + j1 - i2# Base Case
if (i1 >
= n or i2 >
= n or j1 >
= m or j2 >
= m):
return 0# If already calculated, return from DP matrix
if (cache[i1][j1][i2] ! = - 1 ):
return cache[i1][j1][i2]
ans = - sys.maxsize - 1# Recurring for neighbouring cells of both paths together
ans = max (ans, maxSumPath(i1 + 1 , j1, i2 + 1 ) + sum (i1, j1, i2, j2))
ans = max (ans, maxSumPath(i1, j1 + 1 , i2) + sum (i1, j1, i2, j2))
ans = max (ans, maxSumPath(i1, j1 + 1 , i2 + 1 ) + sum (i1, j1, i2, j2))
ans = max (ans, maxSumPath(i1 + 1 , j1, i2) + sum (i1, j1, i2, j2))# Saving result to the DP matrix for current state
cache[i1][j1][i2] = ansreturn ans# Driver code
if __name__ = = '__main__' :
print (maxSumPath( 0 , 0 , 0 ))# This code is contributed by
# Surendra_Gangwar
C#
// C# implementation of the approach
using System;
class GFG
{ // Input matrix
static int n = 4, m = 4;
static int [, ]arr = { { 1, 0, 3, -1 }, { 3, 5, 1, -2 }, { -2, 0, 1, 1 }, { 2, 1, -1, 1 } };
// DP matrix
static int [, , ]cache = new int [5, 5, 5];
// Function to return the sum of the cells
// arr[i1][j1] and arr[i2][j2]
static int sum( int i1, int j1, int i2, int j2)
{
if (i1 == i2 &
&
j1 == j2)
{
return arr[i1, j1];
}
return arr[i1, j1] + arr[i2, j2];
}// Recursive function to return the
// required maximum cost path
static int maxSumPath( int i1, int j1, int i2)
{// Column number of second path
int j2 = i1 + j1 - i2;
// Base Case
if (i1 >
= n || i2 >
= n || j1 >
= m || j2 >
= m)
{
return 0;
}// If already calculated, return from DP matrix
if (cache[i1, j1, i2] != -1)
{
return cache[i1, j1, i2];
}
int ans = int .MinValue;
// Recurring for neighbouring cells of both paths together
ans = Math.Max(ans, maxSumPath(i1 + 1, j1, i2 + 1) + sum(i1, j1, i2, j2));
ans = Math.Max(ans, maxSumPath(i1, j1 + 1, i2) + sum(i1, j1, i2, j2));
ans = Math.Max(ans, maxSumPath(i1, j1 + 1, i2 + 1) + sum(i1, j1, i2, j2));
ans = Math.Max(ans, maxSumPath(i1 + 1, j1, i2) + sum(i1, j1, i2, j2));
// Saving result to the DP matrix for current state
cache[i1, j1, i2] = ans;
return ans;
}// Driver code
public static void Main(String []args)
{
//set initial value
for ( int i = 0;
i <
5;
i++)
for ( int i1 = 0;
i1 <
5;
i1++)
for ( int i2 = 0;
i2 <
5;
i2++)
cache[i, i1, i2]=-1;
Console.WriteLine( maxSumPath(0, 0, 0));
}
}// This code contributed by Rajput-Ji
输出如下:
16
时间复杂度:上2)* M)
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