通过从相邻的盒子中选择不同颜色的钻石来最大化钻石

给定两个整数N和M ，其中N是放置在一行中的盒子的数量， M是分布在这些盒子中的钻石颜色的数量，这样每个盒子中至少包含1 个钻石。每个钻石都有一个颜色和一个由M * N矩阵表示的值，其中mat[i][j]表示第j^个框中颜色为i的钻石的数量。任务是从每个盒子中取出一种颜色类型的钻石，使得所选钻石的总价值最大，而从相邻盒子中取出的钻石颜色不同。如果不可能满足条件，则打印-1 。

例子：

Input: mat[][] = { 
{10, 2, 20, 0}, 
{0, 0, 5, 0}, 
{0, 0, 0, 6}, 
{4, 0, 11, 5}, 
{0, 0, 0, 0}, 
{0, 0, 0, 0}, 
{0, 0, 0, 0}} 
Output: 23 
Explanation: From box 1, we can choose either 10 diamonds of Colour1 or 4 diamonds of Colour4. But since from box 2, we have to choose 2 diamonds of Colour1, therefore choose 4 diamonds of Colour4 from box 1. 
Similarly, to maximize the diamonds count, choose 11 diamonds of Colour4 from box 3 and 6 diamonds of Colour3 from box 4. 
The total diamonds chosen = 4 + 2 + 11 + 6 = 23, which is the maximum count possible.

Input: mat[][] = { 
{1, 0, 0}, 
{0, 0, 5}, 
{0, 11, 5}, 
{0, 0, 0}, 
{0, 0, 0}, 
{0, 0, 0}, 
{0, 0, 0}} 
Output: 16 
Explanation: We choose 1 from the 1^st box of Colour1, 11 from the 2^nd box of Colour3, and 5 from the 3^rd box of Colour2. 

方法：

• 动态规划可以用来解决这个问题。
• 制作一个大小为M x N的二维数组，定义将通过选择^第j列的^第i^个颜色是可以获得的最大值。
• 递推关系将是dp[i][j] = arr[i][j] + max of (dp[1…M][i-1]) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
 
#include 
using namespace std;
 
// Function to return the maximized value
int maxSum(vector > arr)
{
 
    // Number of rows and columns
    int m = (int)arr.size();
    int n = (int)arr[0].size() - 1;
 
    // Creating the Dp array
    int dp[m][n + 1];
 
    // memset(arr, 0, sizeof(arr));
    memset(dp, 0, sizeof(dp));
 
    // Populating the first column
    for (int i = 1; i < m; ++i)
        dp[i][1] = arr[i][1];
 
    for (int i = 1; i < n + 1; ++i) {
 
        // Iterating over all the rows
        for (int j = 1; j < m; ++j) {
 
            int mx = 0;
 
            // Getting the (i-1)th max value
            for (int k = 1; k < m; ++k) {
                if (k != j) {
                    if (dp[k][i - 1] > mx) {
                        mx = dp[k][i - 1];
                    }
                }
            }
 
            // Adding it to the current cell
            if (mx and arr[j][i]) {
                dp[j][i] = arr[j][i] + mx;
            }
        }
    }
 
    // Getting the max sum
    // from the last column
    int ans = -1;
    for (int i = 1; i <= m; ++i) {
        if (dp[i][n])
            ans = max(ans, dp[i][n]);
    }
 
    return ans;
}
 
// Driver code
int main()
{
 
    // Columns are indexed 1-based
    vector > arr = {
        { 0, 0, 0, 0, 0 },
        { 0, 10, 2, 20, 0 },
        { 0, 0, 0, 5, 0 },
        { 0, 0, 0, 0, 6 },
        { 0, 4, 0, 11, 5 },
        { 0, 0, 0, 0, 0 },
        { 0, 0, 0, 0, 0 },
        { 0, 0, 0, 0, 0 }
    };
 
    cout << maxSum(arr);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
import java.util.*;
 
class GFG{
   
// Function to return the maximized value
static int maxSum(int[][] arr)
{
     
    // Number of rows and columns
    int m = (int)arr.length;
    int n = (int)arr[0].length - 1;
 
    // Creating the Dp array
    int[][] dp = new int[m + 1][n + 2];
 
    // memset(arr, 0, sizeof(arr));
    for(int i = 0; i <= m; i++)
    {
        for(int j = 0; j <= n + 1; j++)
        {
            dp[i][j] = 0;
        }
    }
 
    // Populating the first column
    for(int i = 1; i < m; ++i)
        dp[i][1] = arr[i][1];
 
    for(int i = 1; i < n + 1; ++i)
    {
         
        // Iterating over all the rows
        for(int j = 1; j < m; ++j)
        {
            int mx = 0;
 
            // Getting the (i-1)th max value
            for(int k = 1; k < m; ++k)
            {
                if (k != j)
                {
                    if (dp[k][i - 1] > mx)
                    {
                        mx = dp[k][i - 1];
                    }
                }
            }
 
            // Adding it to the current cell
            if (mx != 0 && arr[j][i] != 0)
            {
                dp[j][i] = arr[j][i] + mx;
            }
        }
    }
 
    // Getting the max sum
    // from the last column
    int ans = -1;
    for(int i = 1; i <= m; ++i)
    {
        if ((dp[i][n]) != 0)
            ans = Math.max(ans, dp[i][n]);
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Columns are indexed 1-based
    int[][] arr = { { 0, 0, 0, 0, 0 },
                    { 0, 10, 2, 20, 0 },
                    { 0, 0, 0, 5, 0 },
                    { 0, 0, 0, 0, 6 },
                    { 0, 4, 0, 11, 5 },
                    { 0, 0, 0, 0, 0 },
                    { 0, 0, 0, 0, 0 },
                    { 0, 0, 0, 0, 0 } };
 
    System.out.println(maxSum(arr));
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python 3 implementation of the approach
 
# Function to return the maximized value
def maxSum(arr):
   
    # Number of rows and columns
    m = len(arr)
    n = len(arr[0]) - 1
 
    # Creating the Dp array
    dp = [[0 for i in range(n+1)] for j in range(m)]
 
    # Populating the first column
    for i in range(1,m,1):
        dp[i][1] = arr[i][1]
 
    for i in range(1,n + 1,1):
       
        # Iterating over all the rows
        for j in range(1,m,1):
            mx = 0
 
            # Getting the (i-1)th max value
            for k in range(1,m,1):
                if (k != j):
                    if (dp[k][i - 1] > mx):
                        mx = dp[k][i - 1]
 
            # Adding it to the current cell
            if (mx and arr[j][i]):
                dp[j][i] = arr[j][i] + mx
 
    # Getting the max sum
    # from the last column
    ans = -1
    for i in range(1,m,1):
        if (dp[i][n]):
            ans = max(ans, dp[i][n])
 
    return ans
 
# Driver code
if __name__ == '__main__':
   
    # Columns are indexed 1-based
    arr = [[0, 0, 0, 0, 0],
           [0, 10, 2, 20, 0],
           [0, 0, 0, 5, 0],
           [0, 0, 0, 0, 6],
           [0, 4, 0, 11, 5],
           [0, 0, 0, 0, 0],
           [0, 0, 0, 0, 0],
           [0, 0, 0, 0, 0]]
 
    print(maxSum(arr))
     
    # This coe is contributed by bgangwar59.

Javascript

// Javascript implementation of the approach
 
// Function to return the maximized value
function maxSum(arr)
{
 
  // Number of rows and columns
  let m = arr.length;
  let n = arr[0].length - 1;
 
  // Creating the Dp array
  let dp = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
 
  // Populating the first column
  for (let i = 1; i < m; ++i) dp[i][1] = arr[i][1];
 
  for (let i = 1; i < n + 1; ++i) {
    // Iterating over all the rows
    for (let j = 1; j < m; ++j) {
      let mx = 0;
 
      // Getting the (i-1)th max value
      for (let k = 1; k < m; ++k) {
        if (k != j) {
          if (dp[k][i - 1] > mx) {
            mx = dp[k][i - 1];
          }
        }
      }
 
      // Adding it to the current cell
      if (mx && arr[j][i]) {
        dp[j][i] = arr[j][i] + mx;
      }
    }
  }
 
  // Getting the max sum
  // from the last column
  let ans = -1;
  for (let i = 1; i <= m; ++i) {
    if (dp[i][n]) {ans = Math.max(ans, dp[i][n])};
  }
 
  return ans;
}
 
// Driver code
// Columns are indexed 1-based
let arr = [
  [0, 0, 0, 0, 0],
  [0, 10, 2, 20, 0],
  [0, 0, 0, 5, 0],
  [0, 0, 0, 0, 6],
  [0, 4, 0, 11, 5],
  [0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0],
];
 
document.write(maxSum(arr));
 
// This code is contributed by gfgking.

输出：

23

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