Dada una array ‘arr’ que contiene el peso de ‘N’ artículos distintos y dos mochilas que pueden soportar pesos ‘W1’ y ‘W2’, la tarea es encontrar la suma del subconjunto más grande de la array ‘arr’, que cabe en las dos mochilas. No está permitido romper ningún artículo en dos, es decir, un artículo debe colocarse en una de las bolsas como un todo.
Ejemplos:
Entrada: arr[] = {8, 3, 2}
W1 = 10, W2 = 3
Salida: 13
El primer y tercer objeto van en la primera mochila. El segundo objeto va en la segunda mochila. Por lo tanto, el peso total se convierte en 13.
Entrada: arr[] = {8, 5, 3}
W1 = 10, W2 = 3
Salida: 11
Solución:
Una solución recursiva es probar todas las formas posibles de llenar las dos mochilas y elegir la que da el peso máximo.
Para optimizar la idea anterior, necesitamos determinar los estados de DP, sobre los que construiremos nuestra solución. Después de un poco de observación, podemos determinar que esto se puede representar en tres estados (i, w1_r, w2_r) . Aquí ‘i’ significa el índice del elemento que estamos tratando de almacenar, w1_r significa el espacio restante de la primera mochila y w2_r significa el espacio restante de la segunda mochila. Por lo tanto, el problema se puede resolver utilizando una programación dinámica tridimensional con una relación de recurrencia
DP[i][w1_r][w2_r] = max( DP[i + 1][w1_r][w2_r], arr[i] + DP[i + 1][w1_r - arr[i]][w2_r], arr[i] + DP[i + 1][w1_r][w2_r - arr[i]])
La explicación de la relación de recurrencia anterior es la siguiente:
Para cada ‘i’, podemos:
- No seleccione el elemento ‘i’.
- Llene el artículo ‘i’ en la primera mochila.
- Rellena el artículo ‘i’ en la segunda mochila.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> #define maxN 31 #define maxW 31 using namespace std; // 3D array to store // states of DP int dp[maxN][maxW][maxW]; // w1_r represents remaining capacity of 1st knapsack // w2_r represents remaining capacity of 2nd knapsack // i represents index of the array arr we are working on int maxWeight(int* arr, int n, int w1_r, int w2_r, int i) { // Base case if (i == n) return 0; if (dp[i][w1_r][w2_r] != -1) return dp[i][w1_r][w2_r]; // Variables to store the result of three // parts of recurrence relation int fill_w1 = 0, fill_w2 = 0, fill_none = 0; if (w1_r >= arr[i]) fill_w1 = arr[i] + maxWeight(arr, n, w1_r - arr[i], w2_r, i + 1); if (w2_r >= arr[i]) fill_w2 = arr[i] + maxWeight(arr, n, w1_r, w2_r - arr[i], i + 1); fill_none = maxWeight(arr, n, w1_r, w2_r, i + 1); // Store the state in the 3D array dp[i][w1_r][w2_r] = max(fill_none, max(fill_w1, fill_w2)); return dp[i][w1_r][w2_r]; } // Driver code int main() { // Input array int arr[] = { 8, 2, 3 }; // Initializing the array with -1 memset(dp, -1, sizeof(dp)); // Number of elements in the array int n = sizeof(arr) / sizeof(arr[0]); // Capacity of knapsacks int w1 = 10, w2 = 3; // Function to be called cout << maxWeight(arr, n, w1, w2, 0); return 0; }
Java
// Java implementation of the above approach class GFG { static int maxN = 31; static int maxW = 31; // 3D array to store // states of DP static int dp [][][] = new int[maxN][maxW][maxW]; // w1_r represents remaining capacity of 1st knapsack // w2_r represents remaining capacity of 2nd knapsack // i represents index of the array arr we are working on static int maxWeight(int arr [] , int n, int w1_r, int w2_r, int i) { // Base case if (i == n) return 0; if (dp[i][w1_r][w2_r] != -1) return dp[i][w1_r][w2_r]; // Variables to store the result of three // parts of recurrence relation int fill_w1 = 0, fill_w2 = 0, fill_none = 0; if (w1_r >= arr[i]) fill_w1 = arr[i] + maxWeight(arr, n, w1_r - arr[i], w2_r, i + 1); if (w2_r >= arr[i]) fill_w2 = arr[i] + maxWeight(arr, n, w1_r, w2_r - arr[i], i + 1); fill_none = maxWeight(arr, n, w1_r, w2_r, i + 1); // Store the state in the 3D array dp[i][w1_r][w2_r] = Math.max(fill_none, Math.max(fill_w1, fill_w2)); return dp[i][w1_r][w2_r]; } // Driver code public static void main (String[] args) { // Input array int arr[] = { 8, 2, 3 }; // Initializing the array with -1 for (int i = 0; i < maxN ; i++) for (int j = 0; j < maxW ; j++) for (int k = 0; k < maxW ; k++) dp[i][j][k] = -1; // Number of elements in the array int n = arr.length; // Capacity of knapsacks int w1 = 10, w2 = 3; // Function to be called System.out.println(maxWeight(arr, n, w1, w2, 0)); } } // This code is contributed by ihritik
Python3
# Python3 implementation of the above approach # w1_r represents remaining capacity of 1st knapsack # w2_r represents remaining capacity of 2nd knapsack # i represents index of the array arr we are working on def maxWeight(arr, n, w1_r, w2_r, i): # Base case if i == n: return 0 if dp[i][w1_r][w2_r] != -1: return dp[i][w1_r][w2_r] # Variables to store the result of three # parts of recurrence relation fill_w1, fill_w2, fill_none = 0, 0, 0 if w1_r >= arr[i]: fill_w1 = arr[i] + maxWeight(arr, n, w1_r - arr[i], w2_r, i + 1) if w2_r >= arr[i]: fill_w2 = arr[i] + maxWeight(arr, n, w1_r, w2_r - arr[i], i + 1) fill_none = maxWeight(arr, n, w1_r, w2_r, i + 1) # Store the state in the 3D array dp[i][w1_r][w2_r] = max(fill_none, max(fill_w1, fill_w2)) return dp[i][w1_r][w2_r] # Driver code if __name__ == "__main__": # Input array arr = [8, 2, 3] maxN, maxW = 31, 31 # 3D array to store # states of DP dp = [[[-1] * maxW] * maxW] * maxN # Number of elements in the array n = len(arr) # Capacity of knapsacks w1, w2 = 10, 3 # Function to be called print(maxWeight(arr, n, w1, w2, 0)) # This code is contributed by Rituraj Jain
C#
// C# implementation of the above approach using System; class GFG { static int maxN = 31; static int maxW = 31; // 3D array to store // states of DP static int [ , , ] dp = new int[maxN, maxW, maxW]; // w1_r represents remaining capacity of 1st knapsack // w2_r represents remaining capacity of 2nd knapsack // i represents index of the array arr we are working on static int maxWeight(int [] arr, int n, int w1_r, int w2_r, int i) { // Base case if (i == n) return 0; if (dp[i ,w1_r, w2_r] != -1) return dp[i, w1_r, w2_r]; // Variables to store the result of three // parts of recurrence relation int fill_w1 = 0, fill_w2 = 0, fill_none = 0; if (w1_r >= arr[i]) fill_w1 = arr[i] + maxWeight(arr, n, w1_r - arr[i], w2_r, i + 1); if (w2_r >= arr[i]) fill_w2 = arr[i] + maxWeight(arr, n, w1_r, w2_r - arr[i], i + 1); fill_none = maxWeight(arr, n, w1_r, w2_r, i + 1); // Store the state in the 3D array dp[i, w1_r, w2_r] = Math.Max(fill_none, Math.Max(fill_w1, fill_w2)); return dp[i, w1_r, w2_r]; } // Driver code public static void Main () { // Input array int [] arr = { 8, 2, 3 }; // Initializing the array with -1 for (int i = 0; i < maxN ; i++) for (int j = 0; j < maxW ; j++) for (int k = 0; k < maxW ; k++) dp[i, j, k] = -1; // Number of elements in the array int n = arr.Length; // Capacity of knapsacks int w1 = 10, w2 = 3; // Function to be called Console.WriteLine(maxWeight(arr, n, w1, w2, 0)); } } // This code is contributed by ihritik
Javascript
<script> // Javascript implementation of // the above approach var maxN = 31 var maxW = 31 // 3D array to store // states of DP var dp = Array(maxN); for(var i=0;i<maxN;i++) { dp[i] = Array(maxW); for(var j =0; j<maxW; j++) { dp[i][j] = Array(maxW).fill(-1); } } // w1_r represents remaining // capacity of 1st knapsack // w2_r represents remaining // capacity of 2nd knapsack // i represents index of the array arr // we are working on function maxWeight(arr, n, w1_r, w2_r, i) { // Base case if (i == n) return 0; if (dp[i][w1_r][w2_r] != -1) return dp[i][w1_r][w2_r]; // Variables to store the result of three // parts of recurrence relation var fill_w1 = 0, fill_w2 = 0, fill_none = 0; if (w1_r >= arr[i]) fill_w1 = arr[i] + maxWeight(arr, n, w1_r - arr[i], w2_r, i + 1); if (w2_r >= arr[i]) fill_w2 = arr[i] + maxWeight(arr, n, w1_r, w2_r - arr[i], i + 1); fill_none = maxWeight(arr, n, w1_r, w2_r, i + 1); // Store the state in the 3D array dp[i][w1_r][w2_r] = Math.max(fill_none, Math.max(fill_w1, fill_w2)); return dp[i][w1_r][w2_r]; } // Driver code // Input array var arr = [8, 2, 3 ]; // Number of elements in the array var n = arr.length; // Capacity of knapsacks var w1 = 10, w2 = 3; // Function to be called document.write( maxWeight(arr, n, w1, w2, 0)); </script>
13
Complejidad de tiempo: O(N*W1*W2)
Espacio auxiliar: O(N*W1*W2))
Publicación traducida automáticamente
Artículo escrito por DivyanshuShekhar1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA