对网格进行着色，以使所有相同颜色的像元水平或垂直连接

给定三个整数R，C，N和大小为N的数组arr [] 。任务是为R行和C列的网格中的所有单元着色，以使所有相同的颜色单元水平或垂直连接。 N表示从1到N的颜色，arr []表示每种颜色的数量。颜色的总数恰好等于网格的像元总数。
方法：

Input: R = 3, C = 5, N = 5, arr[] = {1, 2, 3, 4, 5} 
Output: 
1 4 4 4 3 
2 5 4 5 3 
2 5 5 5 3 
Explanation: Available colors are 1(count = 1), 2(count = 2), 3(count = 3) etc. 
For color 5: we can reach all color 5s by going horizontally or vertically through the same color 5. 
Similarly for color 3, the rightmost row contains all 3 etc. 
Similarly, for the rest of the colors 1, 2, 4. 
Below is an invalid grid: 
1 4 3 4 4 
2 5 4 5 3 
2 5 5 5 3 
This is because the connection for the colors 3 and 4 has been broken by the invalid position of 3 
in the position(0, 2).We can no longer traverse through all the 4s or all the 3s, horizontally or vertically, by passing through the respective 3s and 4s only.
Input: R = 2, C = 2, N = 3, arr[] = {2, 1, 1} 
Output: 
1 1 
2 3 
 

方法：

乍一看，似乎需要图形算法。但是，我们将遵循优化的贪婪算法。

1. 创建一个新的2D数组，这将是我们的最终网格。让我们将其称为dp [] []。
2. 遍历颜色数组A []
3. 对于每种颜色，我有A [i]个量
   • 如果该行是奇数行，请从左到右填充dp数组
   • 否则，如果是偶数行，则从右向左填充
4. 如果颜色数量用完，请贪婪地移至下一个颜色

下面是上述方法的实现：

C++

// C++ Program to Color a grid
// such that all same color cells
// are connected either
// horizontally or vertically
 
#include 
using namespace std;
 
void solve(vector& arr,
        int r, int c)
{
    // Current color
    int idx = 1;
 
    // final grid
    int dp[r];
 
    for (int i = 0; i < r; i++) {
 
        // if even row
        if (i % 2 == 0) {
 
            // traverse from left to
            // right
            for (int j = 0; j < c; j++) {
 
                // if color has been exhausted
                //, move to the next color
                if (arr[idx - 1] == 0)
                    idx++;
 
                // color the grid at
                // this position
                dp[i][j] = idx;
 
                // reduce the color count
                arr[idx - 1]--;
            }
        }
        else {
 
            // traverse from right to
            // left for odd rows
            for (int j = c - 1; j >= 0; j--) {
                if (arr[idx - 1] == 0)
                    idx++;
                dp[i][j] = idx;
                arr[idx - 1]--;
            }
        }
    }
 
    // print the grid
    for (int i = 0; i < r; ++i) {
        for (int j = 0; j < c; ++j) {
            cout << dp[i][j] << " ";
        }
        cout << endl;
    }
}
 
// Driver code
int main()
{
    int r = 3, c = 5;
    int n = 5;
    vector arr
        = { 1, 2, 3, 4, 5 };
    solve(arr, r, c);
    return 0;
}

Java

// Java program to color a grid       
// such that all same color cells       
// are connected either       
// horizontally or vertically       
import java.util.*;   
 
class GFG{       
             
static void solve(List arr,
                int r, int c)       
{
     
    // Current color       
    int idx = 1;       
             
    // Final grid       
    int[][] dp = new int[r];       
             
    for(int i = 0; i < r; i++)
    {       
         
        // If even row       
        if (i % 2 == 0)
        {
             
            // Traverse from left to       
            // right       
            for(int j = 0; j < c; j++)
            {
                 
                // If color has been exhausted
                //, move to the next color
                if (arr.get(idx - 1) == 0)   
                    idx++;
                 
                // Color the grid at
                // this position
                dp[i][j] = idx;
                 
                // Reduce the color count
                arr.set(idx - 1,
                arr.get(idx - 1) - 1);
            }
        }
        else
        {
             
            // Traverse from right to
            // left for odd rows
            for(int j = c - 1; j >= 0; j--)
            {
                if (arr.get(idx - 1) == 0)   
                    idx++;
             
                dp[i][j] = idx;
             
                arr.set(idx - 1,
                arr.get(idx - 1) - 1);
            }       
        }       
    }       
             
    // Print the grid       
    for(int i = 0; i < r; ++i)
    {       
        for(int j = 0; j < c; ++j)
        {       
            System.out.print(dp[i][j] + " ");       
        }       
        System.out.println();       
    }       
}       
         
// Driver Code       
public static void main (String[] args)
{       
    int r = 3, c = 5;       
    int n = 5;       
    List arr = Arrays.asList(1, 2, 3, 4, 5);
     
    solve(arr, r, c);       
}       
}
 
// This code is contributed by offbeat

Python3

# Python3 program to color a grid
# such that all same color cells
# are connected either
# horizontally or vertically
def solve(arr, r, c):
     
    # Current color
    idx = 1
 
    # Final grid
    dp = [[0 for i in range(c)]
            for i in range(r)]
 
    for i in range(r):
 
        # If even row
        if (i % 2 == 0):
 
            # Traverse from left to
            # right
            for j in range(c):
 
                # If color has been exhausted,
                # move to the next color
                if (arr[idx - 1] == 0):
                    idx += 1
 
                # Color the grid at
                # this position
                # print(i,j)
                dp[i][j] = idx
 
                # Reduce the color count
                arr[idx - 1] -= 1
        else:
 
            # Traverse from right to
            # left for odd rows
            for j in range(c - 1, -1, -1):
                if (arr[idx - 1] == 0):
                    idx += 1
                     
                dp[i][j] = idx
                arr[idx - 1] -= 1
 
    # Print the grid
    for i in range(r):
        for j in range(c):
            print(dp[i][j], end = " ")
 
        print()
 
# Driver code
if __name__ == '__main__':
 
    r = 3
    c = 5
    n = 5
    arr = [ 1, 2, 3, 4, 5 ]
     
    solve(arr, r, c)
 
# This code is contributed by mohit kumar 29

C#

// C# program to color a grid       
// such that all same color cells       
// are connected either       
// horizontally or vertically       
using System;
using System.Collections.Generic;
 
class GFG{       
             
static void solve(List arr,
                int r, int c)       
{
     
    // Current color       
    int idx = 1;       
             
    // Final grid       
    int[,] dp = new int[r, c];       
             
    for(int i = 0; i < r; i++)
    {       
         
        // If even row       
        if (i % 2 == 0)
        {
             
            // Traverse from left to       
            // right       
            for(int j = 0; j < c; j++)
            {
                 
                // If color has been exhausted,
                // move to the next color
                if (arr[idx - 1] == 0)   
                    idx++;
                 
                // Color the grid at
                // this position
                dp[i, j] = idx;
                 
                // Reduce the color count
                arr[idx - 1] = arr[idx - 1] - 1;
            }
        }
        else
        {
             
            // Traverse from right to
            // left for odd rows
            for(int j = c - 1; j >= 0; j--)
            {
                if (arr[idx - 1] == 0)   
                    idx++;
             
                dp[i, j] = idx;
                arr[idx - 1] = arr[idx - 1] - 1;
            }       
        }       
    }       
             
    // Print the grid       
    for(int i = 0; i < r; ++i)
    {       
        for(int j = 0; j < c; ++j)
        {       
            Console.Write(dp[i, j] + " ");       
        }       
        Console.Write('\n');       
    }       
}       
         
// Driver Code       
public static void Main (string[] args)
{       
    int r = 3, c = 5;       
    //int n = 5;   
     
    List arr = new List();
    arr.Add(1);
    arr.Add(2);
    arr.Add(3);
    arr.Add(4);
    arr.Add(5);
     
    solve(arr, r, c);       
}       
}
 
// This code is contributed by rutvik_56

Javascript

 
//  Javascript Program to Color a grid
//  such that all same color cells
//  are connected either
//  horizontally or vertically
 
function solve(arr,r,c)
{
    // Current color
    var idx = 1;
 
    // final grid
    var dp = new Array(r);
     
    var i,j;
    for(i=0;i<r;i++)
      dp[i] = new Array(c);
    for(i = 0; i < r; i++) {
 
        // if even row
        if (i % 2 == 0) {
 
            // traverse from left to
            // right
            for(j = 0; j < c; j++) {
 
                // if color has been exhausted
                //, move to the next color
                if (arr[idx - 1] == 0)
                    idx++;
 
                // color the grid at
                // this position
                dp[i][j] = idx;
 
                // reduce the color count
                arr[idx - 1]--;
            }
        }
        else {
 
            // traverse from right to
            // left for odd rows
            for (j = c - 1; j >= 0; j--) {
                if (arr[idx - 1] == 0)
                    idx++;
                dp[i][j] = idx;
                arr[idx - 1]--;
            }
        }
    }
 
    // print the grid
    for(i = 0; i < r; ++i) {
        for(j = 0; j < c; ++j) {
            document.write(dp[i][j] + " ");
        }
        document.write("<br>");
    }
}
 
// Driver code
      
    var r = 3, c = 5;
    var n = 5;
    var arr = [1, 2, 3, 4, 5];
    solve(arr, r, c);
 

输出：

1 2 2 3 3
4 4 4 4 3
5 5 5 5 5

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