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Impressió de la subseqüència bitònica més llarga

El problema de la subseqüència bitònica més llarga és trobar la subseqüència més llarga d'una seqüència donada de manera que primer creixi i després decreixi. Una seqüència ordenada en ordre creixent es considera Bitònica amb la part decreixent com a buida. De la mateixa manera, la seqüència d'ordre decreixent es considera bitònica amb la part creixent com a buida. Exemples:

vaja
  Input:    [1 11 2 10 4 5 2 1]   Output:   [1 2 10 4 2 1] OR [1 11 10 5 2 1] OR [1 2 4 5 2 1]   Input:    [12 11 40 5 3 1]   Output:   [12 11 5 3 1] OR [12 40 5 3 1]   Input:    [80 60 30 40 20 10]   Output:   [80 60 30 20 10] OR [80 60 40 20 10]

En anterior publicació que hem comentat sobre el problema de la subseqüència bitonica més llarga. Tanmateix, la publicació només cobria codi relacionat amb la cerca de la suma màxima de la subseqüència creixent, però no amb la construcció de la subseqüència. En aquesta publicació parlarem de com construir la subseqüència bitonica més llarga. Sigui arr[0..n-1] la matriu d'entrada. Definim el vector LIS de manera que LIS[i] és en si mateix un vector que emmagatzema la subseqüència creixent més llarga de arr[0..i] que acaba amb arr[i]. Per tant, per a un índex i LIS[i] es pot escriure de manera recursiva com -



LIS[0] = {arr[O]} LIS[i] = {Max(LIS[j])} + arr[i] where   j < i   and arr[j] < arr[i] = arr[i] if there is no such j

També definim un vector LDS de manera que LDS[i] és en si mateix un vector que emmagatzema la subseqüència decreixent més llarga d'arr[i..n] que comença amb arr[i]. Per tant, per a un índex i, LDS[i] es pot escriure de manera recursiva com -

LDS[n] = {arr[n]} LDS[i] = arr[i] + {Max(LDS[j])} where   j > i   and arr[j] < arr[i] = arr[i] if there is no such j

Per exemple, per a la matriu [1 11 2 10 4 5 2 1]

LIS[0]: 1 LIS[1]: 1 11 LIS[2]: 1 2 LIS[3]: 1 2 10 LIS[4]: 1 2 4 LIS[5]: 1 2 4 5 LIS[6]: 1 2 LIS[7]: 1
LDS[0]: 1 LDS[1]: 11 10 5 2 1 LDS[2]: 2 1 LDS[3]: 10 5 2 1 LDS[4]: 4 2 1 LDS[5]: 5 2 1 LDS[6]: 2 1 LDS[7]: 1

Per tant, la subseqüència bitònica més llarga pot ser



LIS[1] + LDS[1] = [1 11 10 5 2 1] OR LIS[3] + LDS[3] = [1 2 10 5 2 1] OR LIS[5] + LDS[5] = [1 2 4 5 2 1]

A continuació es mostra la implementació de la idea anterior: 

C++
/* Dynamic Programming solution to print Longest  Bitonic Subsequence */ #include    using namespace std; // Utility function to print Longest Bitonic // Subsequence void print(vector<int>& arr int size) {  for(int i = 0; i < size; i++)  cout << arr[i] << ' '; } // Function to construct and print Longest // Bitonic Subsequence void printLBS(int arr[] int n) {  // LIS[i] stores the length of the longest  // increasing subsequence ending with arr[i]  vector<vector<int>> LIS(n);  // initialize LIS[0] to arr[0]  LIS[0].push_back(arr[0]);  // Compute LIS values from left to right  for (int i = 1; i < n; i++)  {  // for every j less than i  for (int j = 0; j < i; j++)  {  if ((arr[j] < arr[i]) &&  (LIS[j].size() > LIS[i].size()))  LIS[i] = LIS[j];  }  LIS[i].push_back(arr[i]);  }  /* LIS[i] now stores Maximum Increasing  Subsequence of arr[0..i] that ends with  arr[i] */  // LDS[i] stores the length of the longest  // decreasing subsequence starting with arr[i]  vector<vector<int>> LDS(n);  // initialize LDS[n-1] to arr[n-1]  LDS[n - 1].push_back(arr[n - 1]);  // Compute LDS values from right to left  for (int i = n - 2; i >= 0; i--)  {  // for every j greater than i  for (int j = n - 1; j > i; j--)  {  if ((arr[j] < arr[i]) &&  (LDS[j].size() > LDS[i].size()))  LDS[i] = LDS[j];  }  LDS[i].push_back(arr[i]);  }  // reverse as vector as we're inserting at end  for (int i = 0; i < n; i++)  reverse(LDS[i].begin() LDS[i].end());  /* LDS[i] now stores Maximum Decreasing Subsequence  of arr[i..n] that starts with arr[i] */  int max = 0;  int maxIndex = -1;  for (int i = 0; i < n; i++)  {  // Find maximum value of size of LIS[i] + size  // of LDS[i] - 1  if (LIS[i].size() + LDS[i].size() - 1 > max)  {  max = LIS[i].size() + LDS[i].size() - 1;  maxIndex = i;  }  }  // print all but last element of LIS[maxIndex] vector  print(LIS[maxIndex] LIS[maxIndex].size() - 1);  // print all elements of LDS[maxIndex] vector  print(LDS[maxIndex] LDS[maxIndex].size()); } // Driver program int main() {  int arr[] = { 1 11 2 10 4 5 2 1 };  int n = sizeof(arr) / sizeof(arr[0]);  printLBS(arr n);  return 0; } 
Java
/* Dynamic Programming solution to print Longest  Bitonic Subsequence */ import java.util.*; class GFG  {  // Utility function to print Longest Bitonic  // Subsequence  static void print(Vector<Integer> arr int size)   {  for (int i = 0; i < size; i++)  System.out.print(arr.elementAt(i) + ' ');  }  // Function to construct and print Longest  // Bitonic Subsequence  static void printLBS(int[] arr int n)   {  // LIS[i] stores the length of the longest  // increasing subsequence ending with arr[i]  @SuppressWarnings('unchecked')  Vector<Integer>[] LIS = new Vector[n];  for (int i = 0; i < n; i++)  LIS[i] = new Vector<>();  // initialize LIS[0] to arr[0]  LIS[0].add(arr[0]);  // Compute LIS values from left to right  for (int i = 1; i < n; i++)   {  // for every j less than i  for (int j = 0; j < i; j++)   {  if ((arr[i] > arr[j]) &&   LIS[j].size() > LIS[i].size())   {  for (int k : LIS[j])  if (!LIS[i].contains(k))  LIS[i].add(k);  }  }  LIS[i].add(arr[i]);  }  /*  * LIS[i] now stores Maximum Increasing Subsequence   * of arr[0..i] that ends with arr[i]  */  // LDS[i] stores the length of the longest  // decreasing subsequence starting with arr[i]  @SuppressWarnings('unchecked')  Vector<Integer>[] LDS = new Vector[n];  for (int i = 0; i < n; i++)  LDS[i] = new Vector<>();  // initialize LDS[n-1] to arr[n-1]  LDS[n - 1].add(arr[n - 1]);  // Compute LDS values from right to left  for (int i = n - 2; i >= 0; i--)   {  // for every j greater than i  for (int j = n - 1; j > i; j--)   {  if (arr[j] < arr[i] &&   LDS[j].size() > LDS[i].size())  for (int k : LDS[j])  if (!LDS[i].contains(k))  LDS[i].add(k);  }  LDS[i].add(arr[i]);  }  // reverse as vector as we're inserting at end  for (int i = 0; i < n; i++)  Collections.reverse(LDS[i]);  /*  * LDS[i] now stores Maximum Decreasing Subsequence   * of arr[i..n] that starts with arr[i]  */  int max = 0;  int maxIndex = -1;  for (int i = 0; i < n; i++)  {  // Find maximum value of size of   // LIS[i] + size of LDS[i] - 1  if (LIS[i].size() + LDS[i].size() - 1 > max)  {  max = LIS[i].size() + LDS[i].size() - 1;  maxIndex = i;  }  }  // print all but last element of LIS[maxIndex] vector  print(LIS[maxIndex] LIS[maxIndex].size() - 1);  // print all elements of LDS[maxIndex] vector  print(LDS[maxIndex] LDS[maxIndex].size());  }  // Driver Code  public static void main(String[] args)   {  int[] arr = { 1 11 2 10 4 5 2 1 };  int n = arr.length;  printLBS(arr n);  } } // This code is contributed by // sanjeev2552 
Python3
# Dynamic Programming solution to print Longest # Bitonic Subsequence def _print(arr: list size: int): for i in range(size): print(arr[i] end=' ') # Function to construct and print Longest # Bitonic Subsequence def printLBS(arr: list n: int): # LIS[i] stores the length of the longest # increasing subsequence ending with arr[i] LIS = [0] * n for i in range(n): LIS[i] = [] # initialize LIS[0] to arr[0] LIS[0].append(arr[0]) # Compute LIS values from left to right for i in range(1 n): # for every j less than i for j in range(i): if ((arr[j] < arr[i]) and (len(LIS[j]) > len(LIS[i]))): LIS[i] = LIS[j].copy() LIS[i].append(arr[i]) # LIS[i] now stores Maximum Increasing # Subsequence of arr[0..i] that ends with # arr[i] # LDS[i] stores the length of the longest # decreasing subsequence starting with arr[i] LDS = [0] * n for i in range(n): LDS[i] = [] # initialize LDS[n-1] to arr[n-1] LDS[n - 1].append(arr[n - 1]) # Compute LDS values from right to left for i in range(n - 2 -1 -1): # for every j greater than i for j in range(n - 1 i -1): if ((arr[j] < arr[i]) and (len(LDS[j]) > len(LDS[i]))): LDS[i] = LDS[j].copy() LDS[i].append(arr[i]) # reverse as vector as we're inserting at end for i in range(n): LDS[i] = list(reversed(LDS[i])) # LDS[i] now stores Maximum Decreasing Subsequence # of arr[i..n] that starts with arr[i] max = 0 maxIndex = -1 for i in range(n): # Find maximum value of size of LIS[i] + size # of LDS[i] - 1 if (len(LIS[i]) + len(LDS[i]) - 1 > max): max = len(LIS[i]) + len(LDS[i]) - 1 maxIndex = i # print all but last element of LIS[maxIndex] vector _print(LIS[maxIndex] len(LIS[maxIndex]) - 1) # print all elements of LDS[maxIndex] vector _print(LDS[maxIndex] len(LDS[maxIndex])) # Driver Code if __name__ == '__main__': arr = [1 11 2 10 4 5 2 1] n = len(arr) printLBS(arr n) # This code is contributed by # sanjeev2552 
C#
/* Dynamic Programming solution to print longest  Bitonic Subsequence */ using System; using System.Linq; using System.Collections.Generic; class GFG  {  // Utility function to print longest Bitonic  // Subsequence  static void print(List<int> arr int size)   {  for (int i = 0; i < size; i++)  Console.Write(arr[i] + ' ');  }  // Function to construct and print longest  // Bitonic Subsequence  static void printLBS(int[] arr int n)   {  // LIS[i] stores the length of the longest  // increasing subsequence ending with arr[i]  List<int>[] LIS = new List<int>[n];  for (int i = 0; i < n; i++)  LIS[i] = new List<int>();  // initialize LIS[0] to arr[0]  LIS[0].Add(arr[0]);  // Compute LIS values from left to right  for (int i = 1; i < n; i++)   {  // for every j less than i  for (int j = 0; j < i; j++)   {  if ((arr[i] > arr[j]) &&   LIS[j].Count > LIS[i].Count)   {  foreach (int k in LIS[j])  if (!LIS[i].Contains(k))  LIS[i].Add(k);  }  }  LIS[i].Add(arr[i]);  }  /*  * LIS[i] now stores Maximum Increasing Subsequence   * of arr[0..i] that ends with arr[i]  */  // LDS[i] stores the length of the longest  // decreasing subsequence starting with arr[i]  List<int>[] LDS = new List<int>[n];  for (int i = 0; i < n; i++)  LDS[i] = new List<int>();  // initialize LDS[n-1] to arr[n-1]  LDS[n - 1].Add(arr[n - 1]);  // Compute LDS values from right to left  for (int i = n - 2; i >= 0; i--)   {  // for every j greater than i  for (int j = n - 1; j > i; j--)   {  if (arr[j] < arr[i] &&   LDS[j].Count > LDS[i].Count)  foreach (int k in LDS[j])  if (!LDS[i].Contains(k))  LDS[i].Add(k);  }  LDS[i].Add(arr[i]);  }  // reverse as vector as we're inserting at end  for (int i = 0; i < n; i++)  LDS[i].Reverse();  /*  * LDS[i] now stores Maximum Decreasing Subsequence   * of arr[i..n] that starts with arr[i]  */  int max = 0;   int maxIndex = -1;  for (int i = 0; i < n; i++)  {  // Find maximum value of size of   // LIS[i] + size of LDS[i] - 1  if (LIS[i].Count + LDS[i].Count - 1 > max)  {  max = LIS[i].Count + LDS[i].Count - 1;  maxIndex = i;  }  }  // print all but last element of LIS[maxIndex] vector  print(LIS[maxIndex] LIS[maxIndex].Count - 1);  // print all elements of LDS[maxIndex] vector  print(LDS[maxIndex] LDS[maxIndex].Count);  }  // Driver Code  public static void Main(String[] args)   {  int[] arr = { 1 11 2 10 4 5 2 1 };  int n = arr.Length;  printLBS(arr n);  } } // This code is contributed by PrinciRaj1992 
JavaScript
// Function to print the longest bitonic subsequence function _print(arr size) {  for (let i = 0; i<size; i++) {  process.stdout.write(arr[i]+' ');  } } // Function to construct and print the longest bitonic subsequence function printLBS(arr n) {  // LIS[i] stores the length of the longest increasing subsequence ending with arr[i]  let LIS = new Array(n);  for (let i = 0; i < n; i++) {  LIS[i] = [];  }  // initialize LIS[0] to arr[0]  LIS[0].push(arr[0]);  // Compute LIS values from left to right  for (let i = 1; i < n; i++) {  // for every j less than i  for (let j = 0; j < i; j++) {  if (arr[j] < arr[i] && LIS[j].length > LIS[i].length) {  LIS[i] = LIS[j].slice();  }  }  LIS[i].push(arr[i]);  }  // LIS[i] now stores the Maximum Increasing Subsequence of arr[0..i] that ends with arr[i]  // LDS[i] stores the length of the longest decreasing subsequence starting with arr[i]  let LDS = new Array(n);  for (let i = 0; i < n; i++) {  LDS[i] = [];  }  // initialize LDS[n-1] to arr[n-1]  LDS[n - 1].push(arr[n - 1]);  // Compute LDS values from right to left  for (let i = n - 2; i >= 0; i--) {  // for every j greater than i  for (let j = n - 1; j > i; j--) {  if (arr[j] < arr[i] && LDS[j].length > LDS[i].length) {  LDS[i] = LDS[j].slice();  }  }  LDS[i].push(arr[i]);  }  // reverse the LDS vector as we're inserting at the end  for (let i = 0; i < n; i++) {  LDS[i].reverse();  }  // LDS[i] now stores the Maximum Decreasing Subsequence of arr[i..n] that starts with arr[i]  let max = 0;  let maxIndex = -1;  for (let i = 0; i < n; i++) {  // Find maximum value of size of LIS[i] + size of LDS[i] - 1  if (LIS[i].length + LDS[i].length - 1 > max) {  max = LIS[i].length + LDS[i].length - 1;  maxIndex = i;  }  }  // print all but  // print all but last element of LIS[maxIndex] array  _print(LIS[maxIndex].slice(0 -1) LIS[maxIndex].length - 1);  // print all elements of LDS[maxIndex] array  _print(LDS[maxIndex] LDS[maxIndex].length); } // Driver program const arr = [1 11 2 10 4 5 2 1]; const n = arr.length; printLBS(arr n); 

Sortida:

1 11 10 5 2 1

Complexitat temporal La solució de programació dinàmica anterior és O(n2). Espai auxiliar utilitzat pel programa és O(n2).