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Siirrä painoasteikkoa vuorotellen annettujen rajoitusten mukaisesti

Annettu painotusasteikko ja joukko erilaisia ​​positiivisia painoja, joissa jokaista painoa on ääretön määrä. Tehtävämme on laittaa painot vaa'an vasempaan ja oikeaan astiaan yksitellen siten, että kattilat siirtyvät sille puolelle, jolle paino asetetaan, eli joka kerta, kun vaaka-astiat siirtyvät toiselle puolelle.

  • Meille annetaan toinen kokonaisluku "askelaikoja", jotka meidän on suoritettava tämä toiminto.
  • Toinen rajoitus on se, että emme voi laittaa samaa painoa peräkkäin, eli jos paino w otetaan, emme voi ottaa w uudelleen seuraavassa vaiheessa, kun laitamme painoa vastakkaiselle astialle.

Esimerkkejä:

Let weight array is [7 11] and steps = 3 then 7 11 7 is the sequence in which weights should be kept in order to move scale alternatively. Let another weight array is [2 3 5 6] and steps = 10 then 3 2 3 5 6 5 3 2 3 is the sequence in which weights should be kept in order to move scale alternatively.

Tämä ongelma voidaan ratkaista tekemällä DFS mittakaavatilojen joukossa.



  1. Kuljemme eri DFS-tilojen välillä ratkaisua varten, jossa jokainen DFS-tila vastaa todellista erotusarvoa vasemman ja oikean puolen välissä ja nykyistä askelmäärää.
  2. Sen sijaan, että tallentaisimme molempien kattiloiden painot, tallennamme vain erotusjäännösarvon ja joka kerta valitun painoarvon tulee olla suurempi kuin tämä ero, eikä se saa olla sama kuin aiemmin valittu painoarvo.
  3. Jos on, kutsumme DFS-menetelmää rekursiivisesti uudella erotusarvolla ja vielä yhdellä askeleella.

Katso alla oleva koodi ymmärtääksesi paremmin 

C++
// C++ program to print weights for alternating // the weighting scale #include    using namespace std; // DFS method to traverse among states of weighting scales bool dfs(int residue int curStep int wt[] int arr[]  int N int steps) {  // If we reach to more than required steps  // return true  if (curStep > steps)  return true;  // Try all possible weights and choose one which  // returns 1 afterwards  for (int i = 0; i < N; i++)  {  /* Try this weight only if it is greater than  current residueand not same as previous chosen  weight */  if (arr[i] > residue && arr[i] != wt[curStep - 1])  {  // assign this weight to array and recur for  // next state  wt[curStep] = arr[i];  if (dfs(arr[i] - residue curStep + 1 wt  arr N steps))  return true;  }  }  // if any weight is not possible return false  return false; } // method prints weights for alternating scale and if // not possible prints 'not possible' void printWeightsOnScale(int arr[] int N int steps) {  int wt[steps];  // call dfs with current residue as 0 and current  // steps as 0  if (dfs(0 0 wt arr N steps))  {  for (int i = 0; i < steps; i++)  cout << wt[i] << ' ';  cout << endl;  }  else  cout << 'Not possiblen'; } // Driver code to test above methods int main() {  int arr[] = {2 3 5 6};  int N = sizeof(arr) / sizeof(int);  int steps = 10;  printWeightsOnScale(arr N steps);  return 0; } 
Java
// Java program to print weights for alternating  // the weighting scale class GFG  {  // DFS method to traverse among   // states of weighting scales  static boolean dfs(int residue int curStep   int[] wt int[] arr  int N int steps)   {  // If we reach to more than required steps  // return true  if (curStep >= steps)  return true;  // Try all possible weights and   // choose one which returns 1 afterwards  for (int i = 0; i < N; i++)   {  /*  * Try this weight only if it is   * greater than current residue   * and not same as previous chosen weight  */  if (curStep - 1 < 0 ||   (arr[i] > residue &&  arr[i] != wt[curStep - 1]))  {  // assign this weight to array and   // recur for next state  wt[curStep] = arr[i];  if (dfs(arr[i] - residue curStep + 1  wt arr N steps))  return true;  }  }  // if any weight is not possible  // return false  return false;  }  // method prints weights for alternating scale   // and if not possible prints 'not possible'  static void printWeightOnScale(int[] arr   int N int steps)   {  int[] wt = new int[steps];  // call dfs with current residue as 0   // and current steps as 0  if (dfs(0 0 wt arr N steps))   {  for (int i = 0; i < steps; i++)  System.out.print(wt[i] + ' ');  System.out.println();  }   else  System.out.println('Not Possible');  }  // Driver Code  public static void main(String[] args)  {  int[] arr = { 2 3 5 6 };  int N = arr.length;  int steps = 10;  printWeightOnScale(arr N steps);  } } // This code is contributed by // sanjeev2552 
Python3
# Python3 program to print weights for  # alternating the weighting scale  # DFS method to traverse among states  # of weighting scales  def dfs(residue curStep wt arr N steps): # If we reach to more than required  # steps return true  if (curStep >= steps): return True # Try all possible weights and choose  # one which returns 1 afterwards for i in range(N): # Try this weight only if it is greater  # than current residueand not same as  # previous chosen weight  if (arr[i] > residue and arr[i] != wt[curStep - 1]): # assign this weight to array and  # recur for next state  wt[curStep] = arr[i] if (dfs(arr[i] - residue curStep + 1 wt arr N steps)): return True # if any weight is not possible # return false  return False # method prints weights for alternating scale  # and if not possible prints 'not possible'  def printWeightsOnScale(arr N steps): wt = [0] * (steps) # call dfs with current residue as 0  # and current steps as 0  if (dfs(0 0 wt arr N steps)): for i in range(steps): print(wt[i] end = ' ') else: print('Not possible') # Driver Code if __name__ == '__main__': arr = [2 3 5 6] N = len(arr) steps = 10 printWeightsOnScale(arr N steps) # This code is contributed by PranchalK 
C#
// C# program to print weights for alternating  // the weighting scale using System; namespace GFG {  class Program  {  // DFS method to traverse among states of weighting scales  static bool dfs(int residue int curStep   int[] wt int[] arr  int N int steps)   {  // If we reach to more than required steps return true  if (curStep >= steps)  return true;  // Try all possible weights and choose one which returns 1 afterwards  for (int i = 0; i < N; i++)   {  /*  * Try this weight only if it is greater than current residue   * and not same as previous chosen weight  */  if (curStep - 1 < 0 || (arr[i] > residue && arr[i] != wt[curStep - 1]))  {  // assign this weight to array and recur for next state  wt[curStep] = arr[i];  if (dfs(arr[i] - residue curStep + 1 wt arr N steps))  return true;  }  }  // if any weight is not possible return false  return false;  }  // method prints weights for alternating scale and   // if not possible prints 'not possible'  static void printWeightOnScale(int[] arr int N int steps)   {  int[] wt = new int[steps];  // call dfs with current residue as 0 and current steps as 0  if (dfs(0 0 wt arr N steps))   {  for (int i = 0; i < steps; i++)  Console.Write(wt[i] + ' ');  Console.WriteLine();  }   else  Console.WriteLine('Not Possible');  }  static void Main(string[] args)  {  int[] arr = { 2 3 5 6 };  int N = arr.Length;  int steps = 10;  printWeightOnScale(arr N steps);  }  } } 
JavaScript
function dfs(residue curStep wt arr N steps) {  // If we reach to more than required steps  // return true  if (curStep > steps) {  return true;  }  // Try all possible weights and choose one which  // returns 1 afterwards  for (let i = 0; i < N; i++)   {    /* Try this weight only if it is greater than  current residue and not same as previous chosen  weight */  if (arr[i] > residue && arr[i] !== wt[curStep - 1])  {    // assign this weight to array and recur for  // next state  wt[curStep] = arr[i];  if (dfs(arr[i] - residue curStep + 1 wt arr N steps)) {  return true;  }  }  }  // if any weight is not possible return false  return false; } function printWeightsOnScale(arr N steps) {  const wt = new Array(steps);  // call dfs with current residue as 0 and current  // steps as 0  if (dfs(0 1 wt arr N steps)) {  for (let i = 1; i <= steps; i++) {  process.stdout.write(`${wt[i]} `);  }  console.log();  } else {  console.log('Not possible');  } } const arr = [2 3 5 6]; const N = arr.length; const steps = 10; printWeightsOnScale(arr N steps); // This code is contributed by divyansh2212 

Lähtö:

2 3 2 3 5 6 5 3 2 3

 

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