logo

Izmenično premikajte lestvico tehtanja pod danimi omejitvami

Podana je utežna lestvica in niz različnih pozitivnih uteži, kjer imamo neskončno zalogo vsake uteži. Naša naloga je, da eno za drugo položimo uteži na levo in desno ploščo tehtnice tako, da se ponvi premakneta na tisto stran, kjer je postavljena teža, tj. vsakič, ko se plošča tehtnice premakne na drugo stran.

  • Dobimo še eno celo število 'korakov', ki jih potrebujemo za izvedbo te operacije.
  • Druga omejitev je, da ne moremo dati iste teže zaporedoma, tj. če vzamemo utež w, potem v naslednjem koraku, medtem ko postavimo utež na nasprotno ploščo, ne moremo ponovno vzeti w.

Primeri:

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.

To težavo je mogoče rešiti z delom DFS med državami lestvice.



  1. Prehajamo med različnimi stanji DFS za rešitev, kjer bo vsako stanje DFS ustrezalo dejanski vrednosti razlike med levim in desnim pomikom ter trenutnim številom korakov.
  2. Namesto shranjevanja teže obeh ponv samo shranimo razliko vrednosti ostanka in vsakič izbrana vrednost teže mora biti večja od te razlike in ne sme biti enaka predhodno izbrani vrednosti teže.
  3. Če je, potem rekurzivno pokličemo metodo DFS z novo vrednostjo razlike in še enim korakom.

Za boljše razumevanje si oglejte spodnjo kodo 

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 

Izhod:

2 3 2 3 5 6 5 3 2 3

 

Ustvari kviz