算法（排列彼此相邻的配对所需的最小交换数）

本文概述

C ++
Java
Python3
C#

有n对, 因此有2n个人。每个人都有一个唯一的数字, 范围从1到2n。所有这些2n个人以大小为2n的数组随机排列。我们也知道谁是谁的伙伴。找到安排这些对以使所有对彼此相邻所需的最小交换次数。
例子：

Input:n = 3pairs[] = {1-> 3, 2-> 6, 4-> 5}// 1 is partner of 3 and so onarr[] = {3, 5, 6, 4, 1, 2}Output: 2We can get {3, 1, 5, 4, 6, 2} by swapping 5 & 6, and 6 & 1

我们强烈建议你最小化浏览器, 然后自己尝试。
这个想法是从第一个和第二个元素开始, 然后重复其余元素。以下是详细步骤/

1) If first and second elements are pair, then simply recur for remaining n-1 pairs and return the value returned by recursive call.2) If first and second are NOT pair, then there are two ways to arrange. So try both of them return the minimum of two.a) Swap second with pair of first and recur for n-1 elements. Let the value returned by recursive call be 'a'.b) Revert the changes made by previous step.c) Swap first with pair of second and recur for n-1 elements. Let the value returned by recursive call be 'b'.d) Revert the changes made by previous step before returningcontrol to parent call.e) Return 1 + min(a, b)

【算法（排列彼此相邻的配对所需的最小交换数）】下面是上述算法的实现。
C ++

// C++ program to find minimum number of swaps required so that // all pairs become adjacent. #include< bits/stdc++.h> using namespace std; // This function updates indexes of elements 'a' and 'b' void updateindex( int index[], int a, int ai, int b, int bi) { index[a] = ai; index[b] = bi; }// This function returns minimum number of swaps required to arrange // all elements of arr[i..n] become aranged int minSwapsUtil( int arr[], int pairs[], int index[], int i, int n) { // If all pairs procesed so no swapping needed return 0 if (i > n) return 0; // If current pair is valid so DO NOT DISTURB this pair // and move ahead. if (pairs[arr[i]] == arr[i+1]) return minSwapsUtil(arr, pairs, index, i+2, n); // If we reach here, then arr[i] and arr[i+1] don't form a pair// Swap pair of arr[i] with arr[i+1] and recursively compute // minimum swap required if this move is made. int one = arr[i+1]; int indextwo = i+1; int indexone = index[pairs[arr[i]]]; int two = arr[index[pairs[arr[i]]]]; swap(arr[i+1], arr[indexone]); updateindex(index, one, indexone, two, indextwo); int a = minSwapsUtil(arr, pairs, index, i+2, n); // Backtrack to previous configuration. Also restore the // previous indices, of one and two swap(arr[i+1], arr[indexone]); updateindex(index, one, indextwo, two, indexone); one = arr[i], indexone = index[pairs[arr[i+1]]]; // Now swap arr[i] with pair of arr[i+1] and recursively // compute minimum swaps required for the subproblem // after this move two = arr[index[pairs[arr[i+1]]]], indextwo = i; swap(arr[i], arr[indexone]); updateindex(index, one, indexone, two, indextwo); int b = minSwapsUtil(arr, pairs, index, i+2, n); // Backtrack to previous configuration.Also restore // the previous indices, of one and two swap(arr[i], arr[indexone]); updateindex(index, one, indextwo, two, indexone); // Return minimum of two cases return 1 + min(a, b); }// Returns minimum swaps required int minSwaps( int n, int pairs[], int arr[]) { int index[2*n + 1]; // To store indices of array elements// Store index of each element in array index for ( int i = 1; i < = 2*n; i++) index[arr[i]] = i; // Call the recursive function return minSwapsUtil(arr, pairs, index, 1, 2*n); }// Driver program int main() { // For simplicity, it is assumed that arr[0] is // not used.The elements from index 1 to n are // only valid elements int arr[] = {0, 3, 5, 6, 4, 1, 2}; // if (a, b) is pair than we have assigned elements // in array such that pairs[a] = b and pairs[b] = a int pairs[] = {0, 3, 6, 1, 5, 4, 2}; int m = sizeof (arr)/ sizeof (arr[0]); int n = m/2; // Number of pairs n is half of total elements// If there are n elements in array, then // there are n pairs cout < < "Min swaps required is " < < minSwaps(n, pairs, arr); return 0; }

Java

// Java program to find minimum number // of swaps required so that // all pairs become adjacent.class GFG {// This function updates indexes // of elements 'a' and 'b' static void updateindex( int index[], int a, int ai, int b, int bi) { index[a] = ai; index[b] = bi; }// This function returns minimum number // of swaps required to arrange // all elements of arr[i..n] become aranged static int minSwapsUtil( int arr[], int pairs[], int index[], int i, int n) { // If all pairs procesed so // no swapping needed return 0 if (i > n) return 0 ; // If current pair is valid so // DO NOT DISTURB this pair // and move ahead. if (pairs[arr[i]] == arr[i + 1 ]) return minSwapsUtil(arr, pairs, index, i + 2 , n); // If we reach here, then arr[i] and // arr[i+1] don't form a pair// Swap pair of arr[i] with arr[i+1] // and recursively compute minimum swap // required if this move is made. int one = arr[i + 1 ]; int indextwo = i + 1 ; int indexone = index[pairs[arr[i]]]; int two = arr[index[pairs[arr[i]]]]; arr[i + 1 ] = arr[i + 1 ] ^ arr[indexone] ^ (arr[indexone] = arr[i + 1 ]); updateindex(index, one, indexone, two, indextwo); int a = minSwapsUtil(arr, pairs, index, i + 2 , n); // Backtrack to previous configuration. // Also restore the previous // indices, of one and two arr[i + 1 ] = arr[i + 1 ] ^ arr[indexone] ^ (arr[indexone] = arr[i + 1 ]); updateindex(index, one, indextwo, two, indexone); one = arr[i]; indexone = index[pairs[arr[i + 1 ]]]; // Now swap arr[i] with pair of arr[i+1] // and recursively compute minimum swaps // required for the subproblem // after this move two = arr[index[pairs[arr[i + 1 ]]]]; indextwo = i; arr[i] = arr[i] ^ arr[indexone] ^ (arr[indexone] = arr[i]); updateindex(index, one, indexone, two, indextwo); int b = minSwapsUtil(arr, pairs, index, i + 2 , n); // Backtrack to previous configuration. Also restore // the previous indices, of one and two arr[i] = arr[i] ^ arr[indexone] ^ (arr[indexone] = arr[i]); updateindex(index, one, indextwo, two, indexone); // Return minimum of two cases return 1 + Math.min(a, b); }// Returns minimum swaps required static int minSwaps( int n, int pairs[], int arr[]) { // To store indices of array elements int index[] = new int [ 2 * n + 1 ]; // Store index of each element in array index for ( int i = 1 ; i < = 2 * n; i++) index[arr[i]] = i; // Call the recursive function return minSwapsUtil(arr, pairs, index, 1 , 2 * n); }// Driver code public static void main(String[] args) {// For simplicity, it is assumed that arr[0] is // not used. The elements from index 1 to n are // only valid elements int arr[] = { 0 , 3 , 5 , 6 , 4 , 1 , 2 }; // if (a, b) is pair than we have assigned elements // in array such that pairs[a] = b and pairs[b] = a int pairs[] = { 0 , 3 , 6 , 1 , 5 , 4 , 2 }; int m = pairs.length; // Number of pairs n is half of total elements int n = m / 2 ; // If there are n elements in array, then // there are n pairs System.out.print( "Min swaps required is " + minSwaps(n, pairs, arr)); } }// This code is contributed by Anant Agarwal.

Python3

# Python program to find # minimum number of swaps # required so that # all pairs become adjacent.# This function updates # indexes of elements 'a' and 'b' def updateindex(index, a, ai, b, bi):index[a] = ai index[b] = bi# This function returns minimum # number of swaps required to arrange # all elements of arr[i..n] # become aranged def minSwapsUtil(arr, pairs, index, i, n):# If all pairs procesed so # no swapping needed return 0 if (i > n): return 0# If current pair is valid so # DO NOT DISTURB this pair # and move ahead. if (pairs[arr[i]] = = arr[i + 1 ]): return minSwapsUtil(arr, pairs, index, i + 2 , n)# If we reach here, then arr[i] # and arr[i+1] don't form a pair# Swap pair of arr[i] with # arr[i+1] and recursively compute # minimum swap required # if this move is made. one = arr[i + 1 ] indextwo = i + 1 indexone = index[pairs[arr[i]]] two = arr[index[pairs[arr[i]]]] arr[i + 1 ], arr[indexone] = arr[indexone], arr[i + 1 ]updateindex(index, one, indexone, two, indextwo)a = minSwapsUtil(arr, pairs, index, i + 2 , n)# Backtrack to previous configuration. # Also restore the # previous indices, # of one and two arr[i + 1 ], arr[indexone] = arr[indexone], arr[i + 1 ] updateindex(index, one, indextwo, two, indexone) one = arr[i] indexone = index[pairs[arr[i + 1 ]]]# Now swap arr[i] with pair # of arr[i+1] and recursively # compute minimum swaps # required for the subproblem # after this move two = arr[index[pairs[arr[i + 1 ]]]] indextwo = i arr[i], arr[indexone] = arr[indexone], arr[i] updateindex(index, one, indexone, two, indextwo) b = minSwapsUtil(arr, pairs, index, i + 2 , n)# Backtrack to previous # configuration.Also restore # 3 the previous indices, # of one and two arr[i], arr[indexone] = arr[indexone], arr[i] updateindex(index, one, indextwo, two, indexone)# Return minimum of two cases return 1 + min (a, b)# Returns minimum swaps required def minSwaps(n, pairs, arr):index = [] # To store indices of array elements for i in range ( 2 * n + 1 + 1 ): index.append( 0 )# Store index of each # element in array index for i in range ( 1 , 2 * n + 1 ): index[arr[i]] = i# Call the recursive function return minSwapsUtil(arr, pairs, index, 1 , 2 * n)# Driver code# For simplicity, it is # assumed that arr[0] is # not used.The elements # from index 1 to n are # only valid elements arr = [ 0 , 3 , 5 , 6 , 4 , 1 , 2 ]# if (a, b) is pair than # we have assigned elements # in array such that # pairs[a] = b and pairs[b] = a pairs = [ 0 , 3 , 6 , 1 , 5 , 4 , 2 ] m = len (pairs)n = m / / 2# Number of pairs n # is half of total elements# If there are n # elements in array, then # there are n pairs print ( "Min swaps required is " , minSwaps(n, pairs, arr))# This code is contributed # by Anant Agarwal.

// C# program to find minimum number // of swaps required so that // all pairs become adjacent. using System; class GFG {// This function updates indexes // of elements 'a' and 'b' public static void updateindex( int [] index, int a, int ai, int b, int bi) { index[a] = ai; index[b] = bi; }// This function returns minimum number // of swaps required to arrange // all elements of arr[i..n] become aranged public static int minSwapsUtil( int [] arr, int [] pairs, int [] index, int i, int n) {// If all pairs procesed so // no swapping needed return 0 if (i > n) { return 0; }// If current pair is valid so // DO NOT DISTURB this pair // and move ahead. if (pairs[arr[i]] == arr[i + 1]) { return minSwapsUtil(arr, pairs, index, i + 2, n); }// If we reach here, then arr[i] and // arr[i+1] don't form a pair // Swap pair of arr[i] with arr[i+1] // and recursively compute minimum swap // required if this move is made. int one = arr[i + 1]; int indextwo = i + 1; int indexone = index[pairs[arr[i]]]; int two = arr[index[pairs[arr[i]]]]; arr[i + 1] = arr[i + 1] ^ arr[indexone] ^ (arr[indexone] = arr[i + 1]); updateindex(index, one, indexone, two, indextwo); int a = minSwapsUtil(arr, pairs, index, i + 2, n); // Backtrack to previous configuration. // Also restore the previous // indices, of one and two arr[i + 1] = arr[i + 1] ^ arr[indexone] ^ (arr[indexone] = arr[i + 1]); updateindex(index, one, indextwo, two, indexone); one = arr[i]; indexone = index[pairs[arr[i + 1]]]; // Now swap arr[i] with pair of arr[i+1] // and recursively compute minimum swaps // required for the subproblem // after this move two = arr[index[pairs[arr[i + 1]]]]; indextwo = i; arr[i] = arr[i] ^ arr[indexone] ^ (arr[indexone] = arr[i]); updateindex(index, one, indexone, two, indextwo); int b = minSwapsUtil(arr, pairs, index, i + 2, n); // Backtrack to previous configuration. // Also restore the previous indices, // of one and two arr[i] = arr[i] ^ arr[indexone] ^ (arr[indexone] = arr[i]); updateindex(index, one, indextwo, two, indexone); // Return minimum of two cases return 1 + Math.Min(a, b); }// Returns minimum swaps required public static int minSwaps( int n, int [] pairs, int [] arr) { // To store indices of array elements int [] index = new int [2 * n + 1]; // Store index of each element in array index for ( int i = 1; i < = 2 * n; i++) { index[arr[i]] = i; }// Call the recursive function return minSwapsUtil(arr, pairs, index, 1, 2 * n); }// Driver code public static void Main( string [] args) {// For simplicity, it is assumed that arr[0] is // not used. The elements from index 1 to n are // only valid elements int [] arr = new int [] { 0, 3, 5, 6, 4, 1, 2 }; // if (a, b) is pair than we have assigned elements // in array such that pairs[a] = b and pairs[b] = a int [] pairs = new int [] { 0, 3, 6, 1, 5, 4, 2 }; int m = pairs.Length; // Number of pairs n is half of total elements int n = m / 2; // If there are n elements in array, then // there are n pairs Console.Write( "Min swaps required is " + minSwaps(n, pairs, arr)); } }// This code is contributed by Shrikant13

输出如下：