使用给定操作为 NK 块着色的方法数

给定N个块，其中K是彩色的。这些 K 色块由数组arr[] 表示。任务是计算为剩余的未着色块着色的方法的数量，以便在一个步骤中只有一个着色块的相邻块中的任何一个可以着色。用模 10 ⁹ +7 打印答案。

例子：

Input: N = 6, K = 3, arr[] = {1, 2, 6} 
Output: 4 
Explanation: 
The following are the 4 ways to color the blocks(each set reblockquotesents the order in which blocks are colored): 
1. {3, 4, 5} 
2. {3, 5, 4} 
3. {5, 3, 4} 
4. {5, 4, 3}

Input: N = 9, K = 3, A = [3, 6, 7] 
Output: 180 

朴素的方法：这个想法是使用递归。以下是步骤：

1. 从 1 到 N 遍历每个块。
2. 如果当前块(例如b )未着色，则检查相邻块之一是否着色。
3. 如果相邻块被着色，则为当前块着色并递归迭代以找到下一个未着色的块。
4. 在上述递归调用结束后，对blockquotevious递归调用的block进行uncolored，对下一个uncolored block重复上述步骤。
5. 在所有上述递归调用中为块着色的计数给出了为未着色块着色的方法数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
const int mod = 1000000007;
 
// Recursive function to count the ways
int countWays(int colored[], int count,
              int n)
{
 
    // Base case
    if (count == n) {
        return 1;
    }
 
    // Initialise answer to 0
    int answer = 0;
 
    // Color each uncolored block according
    // to the given condition
    for (int i = 1; i < n + 1; i++) {
 
        // If any block is uncolored
        if (colored[i] == 0) {
 
            // Check if adjacent blocks
            // are colored or not
            if (colored[i - 1] == 1
                || colored[i + 1] == 1) {
 
                // Color the block
                colored[i] = 1;
 
                // recursively iterate for
                // next uncolored block
                answer = (answer
                          + countWays(colored,
                                      count + 1,
                                      n))
                         % mod;
 
                // Uncolored for the next
                // recursive call
                colored[i] = 0;
            }
        }
    }
 
    // Return the final count
    return answer;
}
 
// Function to count the ways to color
// block
int waysToColor(int arr[], int n, int k)
{
 
    // Mark which blocks are colored in
    // each recursive step
    int colored[n + 2] = { 0 };
 
    for (int i = 0; i < k; i++) {
        colored[arr[i]] = 1;
    }
 
    // Function call to count the ways
    return countWays(colored, k, n);
}
 
// Driver Code
int main()
{
    // Number of blocks
    int N = 6;
 
    // Number of colored blocks
    int K = 3;
    int arr[K] = { 1, 2, 6 };
 
    // Function call
    cout << waysToColor(arr, N, K);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
static int mod = 1000000007;
 
// Recursive function to count the ways
static int countWays(int colored[],
                     int count, int n)
{
 
    // Base case
    if (count == n)
    {
        return 1;
    }
 
    // Initialise answer to 0
    int answer = 0;
 
    // Color each uncolored block according
    // to the given condition
    for (int i = 1; i < n + 1; i++)
    {
 
        // If any block is uncolored
        if (colored[i] == 0)
        {
 
            // Check if adjacent blocks
            // are colored or not
            if (colored[i - 1] == 1 ||
                colored[i + 1] == 1)
            {
 
                // Color the block
                colored[i] = 1;
 
                // recursively iterate for
                // next uncolored block
                answer = (answer +
                          countWays(colored,
                                    count + 1,
                                    n)) % mod;
 
                // Uncolored for the next
                // recursive call
                colored[i] = 0;
            }
        }
    }
 
    // Return the final count
    return answer;
}
 
// Function to count the ways to color
// block
static int waysToColor(int arr[],
                       int n, int k)
{
 
    // Mark which blocks are colored in
    // each recursive step
    int colored[] = new int[n + 2];
 
    for (int i = 0; i < k; i++)
    {
        colored[arr[i]] = 1;
    }
 
    // Function call to count the ways
    return countWays(colored, k, n);
}
 
// Driver Code
public static void main(String[] args)
{
    // Number of blocks
    int N = 6;
 
    // Number of colored blocks
    int K = 3;
    int arr[] = { 1, 2, 6 };
 
    // Function call
    System.out.print(waysToColor(arr, N, K));
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program for the above approach
mod = 1000000007
 
# Recursive function to count the ways
def countWays(colored, count, n):
 
    # Base case
    if (count == n):
        return 1
 
    # Initialise answer to 0
    answer = 0
 
    # Color each uncolored block according
    # to the given condition
    for i in range(1, n + 1):
 
        # If any block is uncolored
        if (colored[i] == 0):
 
            # Check if adjacent blocks
            # are colored or not
            if (colored[i - 1] == 1 or
                colored[i + 1] == 1):
 
                # Color the block
                colored[i] = 1
 
                # recursively iterate for
                # next uncolored block
                answer = ((answer +
                           countWays(colored,
                                     count + 1,
                                       n)) % mod)
 
                # Uncolored for the next
                # recursive call
                colored[i] = 0
 
    # Return the final count
    return answer
 
# Function to count the ways to color
# block
def waysToColor( arr, n, k):
 
    # Mark which blocks are colored in
    # each recursive step
    colored = [0] * (n + 2)
     
    for i in range(k):
        colored[arr[i]] = 1
 
    # Function call to count the ways
    return countWays(colored, k, n)
 
# Driver Code
if __name__ == "__main__":
     
    # Number of blocks
    N = 6
 
    # Number of colored blocks
    K = 3
    arr = [ 1, 2, 6 ]
 
    # Function call
    print(waysToColor(arr, N, K))
 
# This code is contributed by chitranayal

C#

// C# program for the above approach
using System;
class GFG{
 
static int mod = 1000000007;
 
// Recursive function to count the ways
static int countWays(int []colored,
                     int count, int n)
{
 
    // Base case
    if (count == n)
    {
        return 1;
    }
 
    // Initialise answer to 0
    int answer = 0;
 
    // Color each uncolored block according
    // to the given condition
    for (int i = 1; i < n + 1; i++)
    {
 
        // If any block is uncolored
        if (colored[i] == 0)
        {
 
            // Check if adjacent blocks
            // are colored or not
            if (colored[i - 1] == 1 ||
                colored[i + 1] == 1)
            {
 
                // Color the block
                colored[i] = 1;
 
                // recursively iterate for
                // next uncolored block
                answer = (answer +
                          countWays(colored,
                                    count + 1,
                                    n)) % mod;
 
                // Uncolored for the next
                // recursive call
                colored[i] = 0;
            }
        }
    }
 
    // Return the final count
    return answer;
}
 
// Function to count the ways to color
// block
static int waysToColor(int []arr,
                    int n, int k)
{
 
    // Mark which blocks are colored in
    // each recursive step
    int []colored = new int[n + 2];
 
    for (int i = 0; i < k; i++)
    {
        colored[arr[i]] = 1;
    }
 
    // Function call to count the ways
    return countWays(colored, k, n);
}
 
// Driver Code
public static void Main()
{
    // Number of blocks
    int N = 6;
 
    // Number of colored blocks
    int K = 3;
    int []arr = { 1, 2, 6 };
 
    // Function call
    Console.Write(waysToColor(arr, N, K));
}
}
 
// This code is contributed by Code_Mech

Javascript

 
// Javascript program for the above approach
 
let mod = 1000000007;
  
// Recursive function to count the ways
function countWays(colored,
                     count, n)
{
  
    // Base case
    if (count == n)
    {
        return 1;
    }
  
    // Let initialise answer to 0
    let answer = 0;
  
    // Color each uncolored block according
    // to the given condition
    for (let i = 1; i < n + 1; i++)
    {
  
        // If any block is uncolored
        if (colored[i] == 0)
        {
  
            // Check if adjacent blocks
            // are colored or not
            if (colored[i - 1] == 1 ||
                colored[i + 1] == 1)
            {
  
                // Color the block
                colored[i] = 1;
  
                // recursively iterate for
                // next uncolored block
                answer = (answer +
                          countWays(colored,
                                    count + 1,
                                    n)) % mod;
  
                // Uncolored for the next
                // recursive call
                colored[i] = 0;
            }
        }
    }
  
    // Return the final count
    return answer;
}
  
// Function to count the ways to color
// block
function waysToColor(arr, n, k)
{
  
    // Mark which blocks are colored in
    // each recursive step
    let colored = Array.from({length: n+2}, (_, i) => 0);
  
    for (let i = 0; i < k; i++)
    {
        colored[arr[i]] = 1;
    }
  
    // Function call to count the ways
    return countWays(colored, k, n);
}
 
// Driver Code
     
    // Number of blocks
    let N = 6;
  
    // Number of colored blocks
    let K = 3;
    let arr = [ 1, 2, 6 ];
  
    // Function call
    document.write(waysToColor(arr, N, K));
         

输出：

4

时间复杂度： O(N ^NK )

有效的方法：为了有效地解决这个问题，我们将使用排列和组合的概念。以下是步骤：

1. 如果两个连续着色块之间的块数为 x，则为这组块着色的方法数由下式给出：

ways = 2^x-1  

2. 给每组未着色的块着色是相互独立的。假设一个部分有x 个块，另一部分有y 个块。要找到合并两个部分时的总组合，由下式给出：

total combinations = ${n \choose x}*2^{x-1}*2^{y-1}$ 

3. 对有色块索引进行排序，找出每个无色块部分的长度，并使用上述公式迭代并找到每两部分的组合。

4. 使用本文中讨论的方法找到二项式系数。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
const int mod = 1000000007;
 
// Function to count the ways to color
// block
int waysToColor(int arr[], int n, int k)
{
    // For storing powers of 2
    int powOf2[500] = { 0 };
 
    // For storing binomial coefficient
    // values
    int c[500][500];
 
    // Calculating binomial coefficient
    // using DP
    for (int i = 0; i <= n; i++) {
 
        c[i][0] = 1;
        for (int j = 1; j <= i; j++) {
            c[i][j] = (c[i - 1][j]
                       + c[i - 1][j - 1])
                      % mod;
        }
    }
 
    powOf2[0] = powOf2[1] = 1;
 
    // Calculating powers of 2
    for (int i = 2; i <= n; i++) {
 
        powOf2[i] = powOf2[i - 1] * 2 % mod;
    }
 
    int rem = n - k;
    arr[k++] = n + 1;
 
    // Sort the indices to calculate
    // length of each section
    sort(arr, arr + k);
 
    // Initialise answer to 1
    int answer = 1;
 
    for (int i = 0; i < k; i++) {
 
        // Find the length of each section
        int x = arr[i] - (i - 1 >= 0
                              ? arr[i - 1]
                              : 0)
                - 1;
 
        // Merge this section
        answer *= c[rem][x] % mod * (i != 0
                                             && i != k - 1
                                         ? powOf2[x]
                                         : 1)
                  % mod;
        rem -= x;
    }
 
    // Return the final count
    return answer;
}
 
// Driver Code
int main()
{
    // Number of blocks
    int N = 6;
 
    // Number of colored blocks
    int K = 3;
    int arr[K] = { 1, 2, 6 };
 
    // Function call
    cout << waysToColor(arr, N, K);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
static int mod = 1000000007;
 
// Function to count the ways to color
// block
static int waysToColor(int arr[], int n, int k)
{
     
    // For storing powers of 2
    int powOf2[] = new int[500];
 
    // For storing binomial coefficient
    // values
    int [][]c = new int[500][500];
 
    // Calculating binomial coefficient
    // using DP
    for(int i = 0; i <= n; i++)
    {
       c[i][0] = 1;
       for(int j = 1; j <= i; j++)
       {
          c[i][j] = (c[i - 1][j] +
                     c[i - 1][j - 1]) % mod;
       }
    }
 
    powOf2[0] = powOf2[1] = 1;
 
    // Calculating powers of 2
    for(int i = 2; i <= n; i++)
    {
       powOf2[i] = powOf2[i - 1] * 2 % mod;
    }
 
    int rem = n - k;
    arr[k++] = n + 1;
     
    // Sort the indices to calculate
    // length of each section
    Arrays.sort(arr);
 
    // Initialise answer to 1
    int answer = 1;
 
    for(int i = 0; i < k; i++)
    {
         
       // Find the length of each section
       int x = arr[i] - (i - 1 >= 0 ?
                     arr[i - 1] : 0) - 1;
        
       // Merge this section
       answer *= c[rem][x] % mod * (i != 0 &&
                                    i != k - 1 ?
                                    powOf2[x] : 1) %
                                    mod;
       rem -= x;
    }
     
    // Return the final count
    return answer;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Number of blocks
    int N = 6;
 
    // Number of colored blocks
    int K = 3;
    int arr[] = { 1, 2, 6 ,0 };
 
    // Function call
    System.out.print(waysToColor(arr, N, K));
}
}
 
// This code is contributed by 29AjayKumar

蟒蛇3

# Python3 program for the above approach
mod = 1000000007
  
# Function to count the ways to color
# block
def waysToColor(arr, n, k):
     
    global mod
 
    # For storing powers of 2
    powOf2 = [0 for i in range(500)]
  
    # For storing binomial coefficient
    # values
    c = [[0 for i in range(500)] for j in range(500)]
  
    # Calculating binomial coefficient
    # using DP
    for i in range(n + 1):
  
        c[i][0] = 1;
         
        for j in range(1, i + 1):
         
            c[i][j] = (c[i - 1][j]+ c[i - 1][j - 1])% mod;
  
    powOf2[0] = 1
    powOf2[1] = 1;
  
    # Calculating powers of 2
    for i in range(2, n + 1):
  
        powOf2[i] = (powOf2[i - 1] * 2) % mod;
     
    rem = n - k;
    arr[k] = n + 1;
    k += 1
  
    # Sort the indices to calculate
    # length of each section
    arr.sort()
  
    # Initialise answer to 1
    answer = 1;
     
    for i in range(k):
         
        x = 0
         
        # Find the length of each section
        if i - 1 >= 0:
            x = arr[i] - arr[i - 1] -1
        else:
            x = arr[i] - 1
  
        # Merge this section
        answer = answer * (c[rem][x] % mod) * ((powOf2[x] if (i != 0 and i != k - 1) else 1))% mod
        rem -= x;
  
    # Return the final count
    return answer;
 
# Driver Code
if __name__=='__main__':
  
    # Number of blocks
    N = 6;
  
    # Number of colored blocks
    K = 3;
    arr = [ 1, 2, 6, 0]
  
    # Function call
    print(waysToColor(arr, N, K))
 
# This code is contributed by rutvik_56

C＃

// C# program for the above approach
using System;
class GFG{
 
static int mod = 1000000007;
 
// Function to count the ways to color
// block
static int waysToColor(int []arr, int n, int k)
{
     
    // For storing powers of 2
    int []powOf2 = new int[500];
 
    // For storing binomial coefficient
    // values
    int [,]c = new int[500, 500];
 
    // Calculating binomial coefficient
    // using DP
    for(int i = 0; i <= n; i++)
    {
        c[i, 0] = 1;
        for(int j = 1; j <= i; j++)
        {
            c[i, j] = (c[i - 1, j] +
                       c[i - 1, j - 1]) % mod;
        }
    }
 
    powOf2[0] = powOf2[1] = 1;
 
    // Calculating powers of 2
    for(int i = 2; i <= n; i++)
    {
        powOf2[i] = powOf2[i - 1] * 2 % mod;
    }
 
    int rem = n - k;
    arr[k++] = n + 1;
     
    // Sort the indices to calculate
    // length of each section
    Array.Sort(arr);
 
    // Initialise answer to 1
    int answer = 1;
 
    for(int i = 0; i < k; i++)
    {
         
        // Find the length of each section
        int x = arr[i] - (i - 1 >= 0 ?
                arr[i - 1] : 0) - 1;
             
        // Merge this section
        answer *= c[rem, x] % mod * (i != 0 &&
                                     i != k - 1 ?
                                     powOf2[x] : 1) %
                                     mod;
        rem -= x;
    }
     
    // Return the readonly count
    return answer;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Number of blocks
    int N = 6;
 
    // Number of colored blocks
    int K = 3;
    int []arr = { 1, 2, 6, 0 };
 
    // Function call
    Console.Write(waysToColor(arr, N, K));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// JavaScript program for the above approach
 
let mod = 1000000007;
 
// Function to count the ways to color
// block
function waysToColor(arr,n,k)
{
    // For storing powers of 2
    let powOf2 = new Array(500);
  
    // For storing binomial coefficient
    // values
    let c = new Array(500);
    for(let i=0;i<500;i++)
    {
        c[i]=new Array(500);
        for(let j=0;j<500;j++)
        {
            c[i][j]=0;
        }
    }
  
    // Calculating binomial coefficient
    // using DP
    for(let i = 0; i <= n; i++)
    {
       c[i][0] = 1;
       for(let j = 1; j <= i; j++)
       {
          c[i][j] = (c[i - 1][j] +
                     c[i - 1][j - 1]) % mod;
       }
    }
  
    powOf2[0] = powOf2[1] = 1;
  
    // Calculating powers of 2
    for(let i = 2; i <= n; i++)
    {
       powOf2[i] = powOf2[i - 1] * 2 % mod;
    }
  
    let rem = n - k;
    arr[k++] = n + 1;
      
    // Sort the indices to calculate
    // length of each section
    arr.sort(function(a,b){return a-b;});
  
    // Initialise answer to 1
    let answer = 1;
  
    for(let i = 0; i < k; i++)
    {
          
       // Find the length of each section
       let x = arr[i] - (i - 1 >= 0 ?
                     arr[i - 1] : 0) - 1;
         
       // Merge this section
       answer *= c[rem][x] % mod * (i != 0 &&
                                    i != k - 1 ?
                                    powOf2[x] : 1) %
                                    mod;
       rem -= x;
    }
      
    // Return the final count
    return answer;
}
 
// Driver Code
// Number of blocks
let N = 6;
 
// Number of colored blocks
let K = 3;
let arr=[ 1, 2, 6 ,0];
 
// Function call
document.write(waysToColor(arr, N, K));
 
// This code is contributed by avanitrachhadiya2155
 

输出：

4

时间复杂度： O(N ² )