为偏斜树着色的方法，以使父级和子级具有不同的颜色 _偏斜树着色

本文概述

C ++
Java
Python3
C#
的PHP

给定一个带有N个节点和K种颜色的偏斜树(每个节点最多有一个孩子)。你必须为每个节点分配从1到K的颜色, 以便父级和子级具有不同的颜色。找出最大数量方法为节点着色。
例子 -

Input : N = 2, K = 2. Output : Let A1 and A2 be the two nodes. Let A1 is parent of A2. Colors are Red and Blue. Case 1: A1 is colored Red and A2 is colored Blue. Case 2: A1 is colored Blue and A2 is colored Red. No. of ways : 2Input : N = 3, K = 3. Output : A1, A2, A3 are the nodes. A1 is parent of A2 and A2 is parent of A3. Let colors be R, B, G. A1 can choose any three colors and A2 can choose any other two colors and A3 can choose any other two colors than its parents. No. of ways: 12

注意，只有根节点和子节点(子节点、孙节点、孙节点....)和所有)应该有不同的颜色。树的根可以选择K种颜色中的任意一种K种方法。每个其他节点都可以选择除其父节点以外的K-1种颜色。每个节点有K-1个选项。
在这里, 我们选择树作为每个节点, 只有一个孩子。我们可以选择K种颜色中的任何一种作为树的根, 因此采用K种方式。我们为它的孩子留下了K-1颜色。因此, 对于每个孩子, 我们都可以为其父母分配其他颜色。因此, 对于N-1个节点中的每一个, 我们剩下K-1种颜色。因此答案是K *(K-1)^(N-1).
我们可以通过使用耗费O(N)时间复杂度的普通幂函数来找到答案。但是为了获得更好的时间复杂度, 我们使用了快速指数函数, 该函数需要O(log N)时间复杂度。
C ++

// C++ program to count number of ways to color // a N node skewed tree with k colors such that // parent and children have different colors. #include < bits/stdc++.h> using namespace std; // fast_way is recursive // method to calculate power int fastPow( int N, int K) { if (K == 0) return 1; int temp = fastPow(N, K / 2); if (K % 2 == 0) return temp * temp; else return N * temp * temp; }int countWays( int N, int K) { return K * fastPow(K - 1, N - 1); }// driver program int main() { int N = 3, K = 3; cout < < countWays(N, K); return 0; }

Java

// Java program to count number of ways to color // a N node skewed tree with k colors such that // parent and children have different colors. import java.io.*; class GFG { // fast_way is recursive // method to calculate power static int fastPow( int N, int K) { if (K == 0 ) return 1 ; int temp = fastPow(N, K / 2 ); if (K % 2 == 0 ) return temp * temp; else return N * temp * temp; }static int countWays( int N, int K) { return K * fastPow(K - 1 , N - 1 ); }// Driver program public static void main(String[] args) { int N = 3 , K = 3 ; System.out.println(countWays(N, K)); } }// This code is contributed by vt_m.

Python3

# Python3 program to count # number of ways to color # a N node skewed tree with # k colors such that parent # and children have different # colors.# fast_way is recursive # method to calculate power def fastPow(N, K): if (K = = 0 ): return 1 ; temp = fastPow(N, int (K / 2 )); if (K % 2 = = 0 ): return temp * temp; else : return N * temp * temp; def countWays(N, K): return K * fastPow(K - 1 , N - 1 ); # Driver Code N = 3 ; K = 3 ; print (countWays(N, K)); # This code is contributed by mits

// C# program to count number of ways // to color a N node skewed tree with // k colors such that parent and // childrenhave different colors using System; class GFG {// fast_way is recursive // method to calculate power static int fastPow( int N, int K) { if (K == 0) return 1; int temp = fastPow(N, K / 2); if (K % 2 == 0) return temp * temp; else return N * temp * temp; }static int countWays( int N, int K) { return K * fastPow(K - 1, N - 1); }// Driver code public static void Main() { int N = 3, K = 3; Console.WriteLine(countWays(N, K)); } }// This code is contributed by vt_m.

的PHP

< ?php // PHP program to count number // of ways to color a N node // skewed tree with k colors // such that parent and children // have different colors.// fast_way is recursive // method to calculate power function fastPow( $N , $K ) { if ( $K == 0) return 1; $temp = fastPow( $N , $K / 2); if ( $K % 2 == 0) return $temp * $temp ; else return $N * $temp * $temp ; }function countWays( $N , $K ) { return $K * fastPow( $K - 1, $N - 1); }// Driver Code $N = 3; $K = 3; echo countWays( $N , $K ); // This code is contributed by ajit ?>

输出：