排列球以使相邻球为不同类型的方式

有P型的“ p”个球，Q型的“ q”个球和R型的“ r”个球。使用这些球，我们要创建一条直线，以使没有两个相同类型的球相邻。
例子 ：

Input  : p = 1, q = 1, r = 0
Output : 2
There are only two arrangements PQ and QP

Input  : p = 1, q = 1, r = 1
Output : 6
There are only six arrangements PQR, QPR,
QRP, RQP, PRQ and RPQ

Input  : p = 2, q = 1, r = 1
Output : 6
There are only six arrangements PQRP, QPRP,
PRQP, RPQP, PRPQ and PQPR

我们强烈建议您单击此处并进行实践，然后再继续解决方案。

天真的解决方案：
这个问题的天真的解决方案是递归解决方案。我们递归地呼吁三种情况
1)最后要放置的球是P型
2)最后要放置的球是Q型
3)最后要放置的球是R型
以下是上述想法的实现。

C++

// C++ program to count number of ways to arrange three
// types of balls such that no two balls of same color
// are adjacent to each other
#include
using namespace std;
 
// Returns count of arrangements where last placed ball is
// 'last'.  'last' is 0 for 'p', 1 for 'q' and 2 for 'r'
int countWays(int p, int q, int r, int last)
{
    // if number of balls of any color becomes less
    // than 0 the number of ways arrangements is 0.
    if (p<0 || q<0 || r<0)
        return 0;
 
    // If last ball required is of type P and the number
    // of balls of P type is 1 while number of balls of
    // other color is 0 the number of ways is 1.
    if (p==1 && q==0 && r==0 && last==0)
        return 1;
 
    // Same case as above for 'q' and 'r'
    if (p==0 && q==1 && r==0 && last==1)
        return 1;
    if (p==0 && q==0 && r==1 && last==2)
        return 1;
 
    // if last ball required is P and the number of ways is
    // the sum of number of ways to form sequence with 'p-1' P
    // balls, q Q Balls and r R balls ending with Q and R.
    if (last==0)
        return countWays(p-1,q,r,1) + countWays(p-1,q,r,2);
 
    // Same as above case for 'q' and 'r'
    if (last==1)
        return countWays(p,q-1,r,0) + countWays(p,q-1,r,2);
    if (last==2)
        return countWays(p,q,r-1,0) + countWays(p,q,r-1,1);
}
 
// Returns count of required arrangements
int countUtil(int p, int q, int r)
{
   // Three cases arise:
   return countWays(p, q, r, 0) +  // Last required balls is type P
          countWays(p, q, r, 1) +  // Last required balls is type Q
          countWays(p, q, r, 2); // Last required balls is type R
}
 
// Driver code to test above
int main()
{
    int p = 1, q = 1, r = 1;
    printf("%d", countUtil(p, q, r));
    return 0;
}

Java

// Java program to count number
// of ways to arrange three types of
// balls such that no two balls of
// same color are adjacent to each other
class GFG {
     
    // Returns count of arrangements
    // where last placed ball is
    // 'last'. 'last' is 0 for 'p',
    // 1 for 'q' and 2 for 'r'
    static int countWays(int p, int q, int r, int last)
    {
        // if number of balls of any
        // color becomes less than 0
        // the number of ways arrangements is 0.
        if (p < 0 || q < 0 || r < 0)
        return 0;
     
        // If last ball required is
        // of type P and the number
        // of balls of P type is 1
        // while number of balls of
        // other color is 0 the number
        // of ways is 1.
        if (p == 1 && q == 0 && r == 0 && last == 0)
            return 1;
     
        // Same case as above for 'q' and 'r'
        if (p == 0 && q == 1 && r == 0 && last == 1)
            return 1;
        if (p == 0 && q == 0 && r == 1 && last == 2)
            return 1;
     
        // if last ball required is P
        // and the number of ways is
        // the sum of number of ways
        // to form sequence with 'p-1' P
        // balls, q Q Balls and r R balls
        // ending with Q and R.
        if (last == 0)
        return countWays(p - 1, q, r, 1) +
               countWays(p - 1, q, r, 2);
     
        // Same as above case for 'q' and 'r'
        if (last == 1)
            return countWays(p, q - 1, r, 0) +
                   countWays(p, q - 1, r, 2);
         
        if (last == 2)
        return countWays(p, q, r - 1, 0) +
               countWays(p, q, r - 1, 1);
     
        return 0;
    }
     
    // Returns count of required arrangements
    static int countUtil(int p, int q, int r) {
        // Three cases arise:
        return countWays(p, q, r, 0) + // Last required balls is type P
               countWays(p, q, r, 1) + // Last required balls is type Q
               countWays(p, q, r, 2);  // Last required balls is type R
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int p = 1, q = 1, r = 1;
        System.out.print(countUtil(p, q, r));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to count
# number of ways to arrange
# three types of balls such 
# that no two balls of same
# color are adjacent to each
# other
 
# Returns count of arrangements
# where last placed ball is
# 'last'. 'last' is 0 for 'p',
# 1 for 'q' and 2 for 'r'
def countWays(p, q, r, last):
     
    # if number of balls of
    # any color becomes less
    # than 0 the number of
    # ways arrangements is 0.
    if (p < 0 or q < 0 or r < 0):
        return 0;
 
    # If last ball required is
    # of type P and the number
    # of balls of P type is 1
    # while number of balls of
    # other color is 0 the number
    # of ways is 1.
    if (p == 1 and q == 0 and
        r == 0 and last == 0):
        return 1;
 
    # Same case as above
    # for 'q' and 'r'
    if (p == 0 and q == 1 and
        r == 0 and last == 1):
        return 1;
         
    if (p == 0 and q == 0 and
        r == 1 and last == 2):
        return 1;
 
    # if last ball required is P
    # and the number of ways is
    # the sum of number of ways
    # to form sequence with 'p-1' P
    # balls, q Q Balls and r R
    # balls ending with Q and R.
    if (last == 0):
        return (countWays(p - 1, q, r, 1) +
                countWays(p - 1, q, r, 2));
 
    # Same as above case
    # for 'q' and 'r'
    if (last == 1):
        return (countWays(p, q - 1, r, 0) +
                countWays(p, q - 1, r, 2));
    if (last == 2):
        return (countWays(p, q, r - 1, 0) +
                countWays(p, q, r - 1, 1));
 
# Returns count of
# required arrangements
def countUtil(p, q, r):
     
    # Three cases arise:
    # Last required balls is type P
    # Last required balls is type Q
    # Last required balls is type R
    return (countWays(p, q, r, 0) +
            countWays(p, q, r, 1) +
            countWays(p, q, r, 2));
 
# Driver Code
p = 1;
q = 1;
r = 1;
print(countUtil(p, q, r));
     
# This code is contributed by mits

C#

// C# program to count number
// of ways to arrange three types of
// balls such that no two balls of
// same color are adjacent to each other
using System;
 
class GFG {
     
    // Returns count of arrangements
    // where last placed ball is
    // 'last'. 'last' is 0 for 'p',
    // 1 for 'q' and 2 for 'r'
    static int countWays(int p, int q,
                            int r, int last)
    {
         
        // if number of balls of any
        // color becomes less than 0
        // the number of ways
        // arrangements is 0.
        if (p < 0 || q < 0 || r < 0)
            return 0;
     
        // If last ball required is
        // of type P and the number
        // of balls of P type is 1
        // while number of balls of
        // other color is 0 the number
        // of ways is 1.
        if (p == 1 && q == 0 && r == 0
                              && last == 0)
            return 1;
     
        // Same case as above for 'q' and 'r'
        if (p == 0 && q == 1 && r == 0
                               && last == 1)
            return 1;
        if (p == 0 && q == 0 && r == 1
                               && last == 2)
            return 1;
     
        // if last ball required is P
        // and the number of ways is
        // the sum of number of ways
        // to form sequence with 'p-1' P
        // balls, q Q Balls and r R balls
        // ending with Q and R.
        if (last == 0)
            return countWays(p - 1, q, r, 1) +
                    countWays(p - 1, q, r, 2);
     
        // Same as above case for 'q' and 'r'
        if (last == 1)
            return countWays(p, q - 1, r, 0) +
                countWays(p, q - 1, r, 2);
         
        if (last == 2)
            return countWays(p, q, r - 1, 0) +
                    countWays(p, q, r - 1, 1);
     
        return 0;
    }
     
    // Returns count of required arrangements
    static int countUtil(int p, int q, int r)
    {
         
        // Three cases arise:
        // 1. Last required balls is type P
        // 2. Last required balls is type Q
        // 3. Last required balls is type R
        return countWays(p, q, r, 0) +
               countWays(p, q, r, 1) +
               countWays(p, q, r, 2);
    }
     
    // Driver code
    public static void Main()
    {
        int p = 1, q = 1, r = 1;
         
        Console.Write(countUtil(p, q, r));
    }
}
 
// This code is contributed by nitin mittal.

Javascript

 
// JavaScript program to count number
// of ways to arrange three
// types of balls such that no
// two balls of same color
// are adjacent to each other
  
    // Returns count of arrangements
    // where last placed ball is
    // 'last'. 'last' is 0 for 'p',
    // 1 for 'q' and 2 for 'r'
    function countWays(p, q, r, last)
    {
     
        // if number of balls of any
        // color becomes less than 0
        // the number of ways arrangements is 0.
        if (p < 0 || q < 0 || r < 0)
        return 0;
       
        // If last ball required is
        // of type P and the number
        // of balls of P type is 1
        // while number of balls of
        // other color is 0 the number
        // of ways is 1.
        if (p == 1 && q == 0 && r == 0 && last == 0)
            return 1;
       
        // Same case as above for 'q' and 'r'
        if (p == 0 && q == 1 && r == 0 && last == 1)
            return 1;
        if (p == 0 && q == 0 && r == 1 && last == 2)
            return 1;
       
        // if last ball required is P
        // and the number of ways is
        // the sum of number of ways
        // to form sequence with 'p-1' P
        // balls, q Q Balls and r R balls
        // ending with Q and R.
        if (last == 0)
        return countWays(p - 1, q, r, 1) +
               countWays(p - 1, q, r, 2);
       
        // Same as above case for 'q' and 'r'
        if (last == 1)
            return countWays(p, q - 1, r, 0) +
                   countWays(p, q - 1, r, 2);
           
        if (last == 2)
        return countWays(p, q, r - 1, 0) +
               countWays(p, q, r - 1, 1);
       
        return 0;
    }
       
    // Returns count of required arrangements
    function countUtil(p, q, r)
    {
     
        // Three cases arise:
        return countWays(p, q, r, 0) + // Last required balls is type P
               countWays(p, q, r, 1) + // Last required balls is type Q
               countWays(p, q, r, 2);  // Last required balls is type R
    }
 
// Driver Code
 
        let p = 1, q = 1, r = 1;
        document.write(countUtil(p, q, r));
     
    // This code is contributed by target_2.

输出：

6

该解决方案的时间复杂度是指数级的。
我们可以观察到有很多子问题一次又一次被解决，因此可以使用动态编程(DP)解决该问题。我们可以轻松地为该问题提供记忆解决方案。

C++

// C++ program to count number of ways to arrange three
// types of balls such that no two balls of same color
// are adjacent to each other
#include
using namespace std;
#define MAX 100
 
// table to store to store results of subproblems
int dp[MAX][MAX][MAX][3];
 
// Returns count of arrangements where last placed ball is
// 'last'.  'last' is 0 for 'p', 1 for 'q' and 2 for 'r'
int countWays(int p, int q, int r, int last)
{
    // if number of balls of any color becomes less
    // than 0 the number of ways arrangements is 0.
    if (p<0 || q<0 || r<0)
        return 0;
 
    // If last ball required is of type P and the number
    // of balls of P type is 1 while number of balls of
    // other color is 0 the number of ways is 1.
    if (p==1 && q==0 && r==0 && last==0)
        return 1;
 
    // Same case as above for 'q' and 'r'
    if (p==0 && q==1 && r==0 && last==1)
        return 1;
    if (p==0 && q==0 && r==1 && last==2)
        return 1;
 
    // If this subproblem is already evaluated
    if (dp[p][q][r][last] != -1)
        return dp[p][q][r][last];
 
    // if last ball required is P and the number of ways is
    // the sum of number of ways to form sequence with 'p-1' P
    // balls, q Q Balls and r R balls ending with Q and R.
    if (last==0)
       dp[p][q][r][last] = countWays(p-1,q,r,1) + countWays(p-1,q,r,2);
 
    // Same as above case for 'q' and 'r'
    else if (last==1)
       dp[p][q][r][last] = countWays(p,q-1,r,0) + countWays(p,q-1,r,2);
    else //(last==2)
       dp[p][q][r][last] =  countWays(p,q,r-1,0) + countWays(p,q,r-1,1);
 
    return dp[p][q][r][last];
}
 
// Returns count of required arrangements
int countUtil(int p, int q, int r)
{
   // Initialize 'dp' array
   memset(dp, -1, sizeof(dp));
 
   // Three cases arise:
   return countWays(p, q, r, 0) +  // Last required balls is type P
          countWays(p, q, r, 1) +  // Last required balls is type Q
          countWays(p, q, r, 2); // Last required balls is type R
}
 
// Driver code to test above
int main()
{
    int p = 1, q = 1, r = 1;
    printf("%d", countUtil(p, q, r));
    return 0;
}

Java

// Java program to count number
// of ways to arrange three
// types of balls such that no
// two balls of same color
// are adjacent to each other
import java.util.Arrays;
 
class GFG
{
 
    static final int MAX = 100;
     
    // table to store to store results of subproblems
    static int dp[][][][] = new int[MAX][MAX][MAX][3];
     
    // Returns count of arrangements
    // where last placed ball is
    // 'last'. 'last' is 0 for 'p',
    // 1 for 'q' and 2 for 'r'
    static int countWays(int p, int q, int r, int last)
    {
        // if number of balls of any
        // color becomes less than 0
        // the number of ways arrangements is 0.
        if (p < 0 || q < 0 || r < 0)
        return 0;
     
        // If last ball required is
        // of type P and the number
        // of balls of P type is 1
        // while number of balls of
        // other color is 0 the number
        // of ways is 1.
        if (p == 1 && q == 0 && r == 0 && last == 0)
            return 1;
     
        // Same case as above for 'q' and 'r'
        if (p == 0 && q == 1 && r == 0 && last == 1)
            return 1;
         
        if (p == 0 && q == 0 && r == 1 && last == 2)
            return 1;
     
        // If this subproblem is already evaluated
        if (dp[p][q][r][last] != -1)
            return dp[p][q][r][last];
     
        // if last ball required is P and
        // the number of ways is the sum
        // of number of ways to form sequence
        // with 'p-1' P balls, q Q balss and
        // r R balls ending with Q and R.
        if (last == 0)
        dp[p][q][r][last] = countWays(p - 1, q, r, 1) +
                            countWays(p - 1, q, r, 2);
     
        // Same as above case for 'q' and 'r'
        else if (last == 1)
        dp[p][q][r][last] = countWays(p, q - 1, r, 0) +
                            countWays(p, q - 1, r, 2);
        //(last==2)
        else
        dp[p][q][r][last] = countWays(p, q, r - 1, 0) +
                            countWays(p, q, r - 1, 1);
     
        return dp[p][q][r][last];
    }
     
    // Returns count of required arrangements
    static int countUtil(int p, int q, int r)
    {
        // Initialize 'dp' array
        for (int[][][] row : dp)
        {
            for (int[][] innerRow : row)
            {
                for (int[] innerInnerRow : innerRow)
                {
                    Arrays.fill(innerInnerRow, -1);
                }
            }
        };
     
        // Three cases arise:
        return countWays(p, q, r, 0) + // Last required balls is type P
            countWays(p, q, r, 1) +    // Last required balls is type Q
            countWays(p, q, r, 2);       // Last required balls is type R
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int p = 1, q = 1, r = 1;
        System.out.print(countUtil(p, q, r));
    }
}
 
// This code is contributed by Anant Agarwal.

C＃

// C# program to count number
// of ways to arrange three
// types of balls such that no
// two balls of same color
// are adjacent to each other
using System;
 
class GFG
{
 
    static int MAX = 101;
     
    // table to store to store results of subproblems
    static int[,,,] dp = new int[MAX, MAX, MAX, 4];
     
    // Returns count of arrangements
    // where last placed ball is
    // 'last'. 'last' is 0 for 'p',
    // 1 for 'q' and 2 for 'r'
    static int countWays(int p, int q, int r, int last)
    {
        // if number of balls of any
        // color becomes less than 0
        // the number of ways arrangements is 0.
        if (p < 0 || q < 0 || r < 0)
        return 0;
     
        // If last ball required is
        // of type P and the number
        // of balls of P type is 1
        // while number of balls of
        // other color is 0 the number
        // of ways is 1.
        if (p == 1 && q == 0 && r == 0 && last == 0)
            return 1;
     
        // Same case as above for 'q' and 'r'
        if (p == 0 && q == 1 && r == 0 && last == 1)
            return 1;
         
        if (p == 0 && q == 0 && r == 1 && last == 2)
            return 1;
     
        // If this subproblem is already evaluated
        if (dp[p, q, r, last] != -1)
            return dp[p, q, r, last];
     
        // if last ball required is P and
        // the number of ways is the sum
        // of number of ways to form sequence
        // with 'p-1' P balls, q Q balss and
        // r R balls ending with Q and R.
        if (last == 0)
        dp[p, q, r, last] = countWays(p - 1, q, r, 1) +
                            countWays(p - 1, q, r, 2);
     
        // Same as above case for 'q' and 'r'
        else if (last == 1)
        dp[p, q, r, last] = countWays(p, q - 1, r, 0) +
                            countWays(p, q - 1, r, 2);
        //(last==2)
        else
        dp[p, q, r, last] = countWays(p, q, r - 1, 0) +
                            countWays(p, q, r - 1, 1);
     
        return dp[p, q, r, last];
    }
     
    // Returns count of required arrangements
    static int countUtil(int p, int q, int r)
    {
        // Initialize 'dp' array
        for(int i = 0; i < MAX; i++)
        for(int j = 0; j < MAX; j++)
        for(int k = 0; k < MAX; k++)
        for(int l = 0; l < 4; l++)
        dp[i, j, k, l] = -1;
     
        // Three cases arise:
        return countWays(p, q, r, 0) + // Last required balls is type P
            countWays(p, q, r, 1) + // Last required balls is type Q
            countWays(p, q, r, 2); // Last required balls is type R
    }
 
    // Driver code
    static void Main()
    {
        int p = 1, q = 1, r = 1;
        Console.WriteLine(countUtil(p, q, r));
    }
}
 
// This code is contributed by mits.

Python3

# Python3 program to count number of ways to
# arrange three types of balls such that no
# two balls of same color are adjacent to each other
MAX = 100;
 
# table to store to store results of subproblems
dp = [[[[-1] * 4 for i in range(MAX)]
                 for j in range(MAX)]
                 for k in range(MAX)];
 
# Returns count of arrangements where last
# placed ball is 'last'. 'last' is 0 for 'p',
# 1 for 'q' and 2 for 'r'
def countWays(p, q, r, last):
 
    # if number of balls of any color becomes less
    # than 0 the number of ways arrangements is 0.
    if (p < 0 or q < 0 or r < 0):
        return 0;
 
    # If last ball required is of type P and the
    # number of balls of P type is 1 while number
    # of balls of other color is 0 the number of
    # ways is 1.
    if (p == 1 and q == 0 and
        r == 0 and last == 0):
        return 1;
 
    # Same case as above for 'q' and 'r'
    if (p == 0 and q == 1 and
        r == 0 and last == 1):
        return 1;
    if (p == 0 and q == 0 and
        r == 1 and last == 2):
        return 1;
 
    # If this subproblem is already evaluated
    if (dp[p][q][r][last] != -1):
        return dp[p][q][r][last];
 
    # if last ball required is P and the number
    # of ways is the sum of number of ways to 
    # form sequence with 'p-1' P balls, q Q Balls
    # and r R balls ending with Q and R.
    if (last == 0):
        dp[p][q][r][last] = (countWays(p - 1, q, r, 1) +
                             countWays(p - 1, q, r, 2));
 
    # Same as above case for 'q' and 'r'
    elif (last == 1):
        dp[p][q][r][last] = (countWays(p, q - 1, r, 0) +
                             countWays(p, q - 1, r, 2));
    else:
         
        #(last==2)
        dp[p][q][r][last] = (countWays(p, q, r - 1, 0) +
                             countWays(p, q, r - 1, 1));
 
    return dp[p][q][r][last];
 
# Returns count of required arrangements
def countUtil(p, q, r):
     
    # Three cases arise:
    # Last required balls is type P
    # Last required balls is type Q
    # Last required balls is type R
    return (countWays(p, q, r, 0) +
            countWays(p, q, r, 1) +
            countWays(p, q, r, 2));
 
# Driver Code
p, q, r = 1, 1, 1;
print(countUtil(p, q, r));
 
# This code is contributed by mits

输出：

6