计数方法来选择 N 对不同颜色的糖果（动态编程 + 位掩码）

给定一个表示红色和蓝色糖果数量的整数N和一个大小为N * N的矩阵mat[][] ，其中mat[i][j] = 1表示第i^个红色糖果和^第j^个之间存在一对蓝色糖果，任务是找到选择N对糖果的方法的数量，使得每对包含不同颜色的不同糖果。

例子：

Input: N = 2, mat[][] = { { 1, 1 }, { 1, 1 } }
Output: 2
Explanation:
Possible ways to select N (= 2) pairs of candies are { { (1, 1), (2, 2) }, { (1, 2), (2, 1) } }.
Therefore, the required output is 2.

Input: N = 3, mat[][] = { { 0, 1, 1 }, { 1, 0, 1 }, { 1, 1, 1 } }
Output: 3
Explanation:
Possible ways to select N (= 3) pairs of candies are: { { (1, 2), (2, 1), (3, 3) }, { (1, 2), (2, 3), (3, 1) }, { (1, 3), (2, 1), (3, 2) } }
Therefore, the required output is 2.

编程需要懂一点英语

朴素方法：解决此问题的最简单方法是生成包含不同颜色的不同糖果的N对的所有可能排列。最后，打印获得的计数。

下面是上述方法的实现：

C++14

// C++14 program to implement
// the above approach
#include 
using namespace std;
 
// Function to count ways to select N distinct
// pairs of candies with different colours
int numOfWays(vector> a, int n,
                 int i, set &blue)
{
     
    // If n pairs are selected
    if (i == n)
        return 1;
 
    // Stores count of ways
    // to select the i-th pair
    int count = 0;
 
    // Iterate over the range [0, n]
    for(int j = 0; j < n; j++)
    {
         
        // If pair (i, j) is not included
        if (a[i][j] == 1 && blue.find(j) == blue.end())
        {
            blue.insert(j);
            count += numOfWays(a, n, i + 1, blue);
            blue.erase(j);
        }
    }
    return count;
}
 
// Driver Code
int main()
{
    int n = 3;
    vector> mat = { { 0, 1, 1 },
                                { 1, 0, 1 },
                                { 1, 1, 1 } };
    set mpp;
     
    cout << (numOfWays(mat, n, 0, mpp));
}
 
// This code is contributed by mohit kumar 29

Java

// Java program to implement
// the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to count ways to select N distinct
// pairs of candies with different colours
static int numOfWays(int a[][], int n, int i,
                     HashSet blue)
{
     
    // If n pairs are selected
    if (i == n)
        return 1;
 
    // Stores count of ways
    // to select the i-th pair
    int count = 0;
 
    // Iterate over the range [0, n]
    for(int j = 0; j < n; j++)
    {
         
        // If pair (i, j) is not included
        if (a[i][j] == 1 && !blue.contains(j))
        {
            blue.add(j);
            count += numOfWays(a, n, i + 1, blue);
            blue.remove(j);
        }
    }
    return count;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 3;
    int mat[][] = { { 0, 1, 1 },
                    { 1, 0, 1 },
                    { 1, 1, 1 } };
    HashSet mpp = new HashSet<>();
 
    System.out.println((numOfWays(mat, n, 0, mpp)));
}
}
 
// This code is contributed by Kingash

Python3

# Python3 program to implement
# the above approach
 
 
# Function to count ways to select N distinct
# pairs of candies with different colours
def numOfWays(a, n, i = 0, blue = []):
 
    # If n pairs are selected
    if i == n:
        return 1
 
    # Stores count of ways
    # to select the i-th pair
    count = 0
 
    # Iterate over the range [0, n]
    for j in range(n):
 
        # If pair (i, j) is not included
        if mat[i][j] == 1 and j not in blue:
            count += numOfWays(mat, n, i + 1,
                                blue + [j])
                                 
    return count
 
# Driver Code
if __name__ == "__main__":
    n = 3
    mat = [ [0, 1, 1],
            [1, 0, 1],
            [1, 1, 1] ]
    print(numOfWays(mat, n))

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to count ways to select N distinct
// pairs of candies with different colours
static int numOfWays(int[,] a, int n, int i,
                     HashSet blue)
{
     
    // If n pairs are selected
    if (i == n)
        return 1;
 
    // Stores count of ways
    // to select the i-th pair
    int count = 0;
 
    // Iterate over the range [0, n]
    for(int j = 0; j < n; j++)
    {
         
        // If pair (i, j) is not included
        if (a[i, j] == 1 && !blue.Contains(j))
        {
            blue.Add(j);
            count += numOfWays(a, n, i + 1, blue);
            blue.Remove(j);
        }
    }
    return count;
}
 
// Driver Code
public static void Main()
{
    int n = 3;
    int[,] mat = { { 0, 1, 1 },
                   { 1, 0, 1 },
                   { 1, 1, 1 } };
    HashSet mpp = new HashSet();
 
    Console.WriteLine((numOfWays(mat, n, 0, mpp)));
}
}
 
// This code is contributed by ukasp

Javascript

Python3

# Python program to implement
# the above approach
  
  
# Function to count ways to select N distinct
# pairs of candies with different colors
def numOfWays(a, n):
 
 
    # dp[i][j]: Stores count of ways to make
    # pairs between i red candies and N blue candies
    dp = [[0]*((1 << n)+1) for _ in range(n + 1)]
 
    # If there is no red and blue candy,
    # the count of ways to make n pairs is 1
    dp[0][0] = 1
 
    # i: Stores count of red candy
    for i in range(n):
 
        # j generates a permutation of blue
        # candy of selected / not selected
        # as a binary number with n bits
        for j in range(1 << n):
 
            if dp[i][j] == 0:
                continue
    
            # Iterate over the range [0, n]   
            for k in range(n):
 
                # Create a mask with only
                # the k-th bit as set
                mask = 1 << k
                 
                # If Kth bit of mask is
                # unset  and mat[i][k] = 1
                if not(mask & j) and a[i][k]:
                    dp[i + 1][j | mask] += dp[i][j]
                     
         
                     
    # Return the dp states, where n red
    # and n blue candies are selected
    return dp[n][(1 << n)-1]
 
 
 
# Driver Code
if __name__ == "__main__":
    n = 3
    mat = [[0, 1, 1],
           [1, 0, 1],
           [1, 1, 1]]
    print(numOfWays(mat, n))

输出：

时间复杂度： O(N!)
辅助空间： O(1)

高效方法：上述方法可以针对使用位屏蔽的动态编程进行优化。对于每个红色糖果，不是生成N 个蓝色糖果的所有排列，而是使用mask ，其中^第j位掩码检查^第j^个蓝色糖果是否可用于选择当前对。
求解该问题的递推关系如下：

If K^th bit of mask is unset and mat[i][k] = 1:
dp[i + 1][j | (1 << k)] += dp[i][j]

where, (j | (1 << k)) marks the kth blue candy as selected.
dp[i][j] = Count of ways to make pairs between i red candy and N blue candies, where j is a permutation of N bit number ranging from 0 to 2^N – 1).

编程需要懂一点英语

请按照以下步骤解决问题：

初始化一个二维数组，比如dp[][] ，其中dp[i][j]存储在i 个红色糖果和N 个蓝色糖果之间配对的方法计数。 j表示从0到2 ^N -1的N位数的排列。
使用上面的递推关系并填充递推关系的所有可能的dp状态。
最后，打印选择了N 个红色糖果和N 个蓝色糖果的 dp 状态，即dp[i][2 ⁿ -1] 。

下面是上述方法的实现：

蟒蛇3

# Python program to implement
# the above approach
  
  
# Function to count ways to select N distinct
# pairs of candies with different colors
def numOfWays(a, n):
 
 
    # dp[i][j]: Stores count of ways to make
    # pairs between i red candies and N blue candies
    dp = [[0]*((1 << n)+1) for _ in range(n + 1)]
 
    # If there is no red and blue candy,
    # the count of ways to make n pairs is 1
    dp[0][0] = 1
 
    # i: Stores count of red candy
    for i in range(n):
 
        # j generates a permutation of blue
        # candy of selected / not selected
        # as a binary number with n bits
        for j in range(1 << n):
 
            if dp[i][j] == 0:
                continue
    
            # Iterate over the range [0, n]   
            for k in range(n):
 
                # Create a mask with only
                # the k-th bit as set
                mask = 1 << k
                 
                # If Kth bit of mask is
                # unset  and mat[i][k] = 1
                if not(mask & j) and a[i][k]:
                    dp[i + 1][j | mask] += dp[i][j]
                     
         
                     
    # Return the dp states, where n red
    # and n blue candies are selected
    return dp[n][(1 << n)-1]
 
 
 
# Driver Code
if __name__ == "__main__":
    n = 3
    mat = [[0, 1, 1],
           [1, 0, 1],
           [1, 1, 1]]
    print(numOfWays(mat, n))

输出：

时间复杂度： O(N ² * 2 ^N )
辅助空间： O(N * 2 ^N )