Given a knapsack weight W and a set of n items with certain value vali and weight wti, we need to calculate the maximum amount that could make up this quantity exactly. This is different from classical Knapsack problem, here we are allowed to use unlimited number of instances of an item.
Examples:
Input : W = 100
val[] = {1, 30}
wt[] = {1, 50}
Output : 100
There are many ways to fill knapsack.
1) 2 instances of 50 unit weight item.
2) 100 instances of 1 unit weight item.
3) 1 instance of 50 unit weight item and 50
instances of 1 unit weight items.
We get maximum value with option 2.
Input : W = 8
val[] = {10, 40, 50, 70}
wt[] = {1, 3, 4, 5}
Output : 110
We get maximum value with one unit of
weight 5 and one unit of weight 3.Its an unbounded knapsack problem as we can use 1 or more instances of any resource. A simple 1D array, say dp[W+1] can be used such that dp[i] stores the maximum value which can achieved using all items and i capacity of knapsack. Note that we use 1D array here which is different from classical knapsack where we used 2D array. Here number of items never changes. We always have all items available.
We can recursively compute dp[] using below formula
dp[i] = 0
dp[i] = max(dp[i], dp[i-wt[j]] + val[j]
where j varies from 0
to n-1 such that:
wt[j] <= i
result = d[W]Below is the implementation of above idea.
C++
#include<bits/stdc++.h>
using namespace std;
int unboundedKnapsack(int W, int n,
int val[], int wt[])
{
int dp[W+1];
memset(dp, 0, sizeof dp);
for (int i=0; i<=W; i++)
for (int j=0; j<n; j++)
if (wt[j] <= i)
dp[i] = max(dp[i], dp[i-wt[j]] + val[j]);
return dp[W];
}
int main()
{
int W = 100;
int val[] = {10, 30, 20};
int wt[] = {5, 10, 15};
int n = sizeof(val)/sizeof(val[0]);
cout << unboundedKnapsack(W, n, val, wt);
return 0;
}
|
Java
public class UboundedKnapsack
{
private static int max(int i, int j)
{
return (i > j) ? i : j;
}
private static int unboundedKnapsack(int W, int n,
int[] val, int[] wt)
{
int dp[] = new int[W + 1];
for(int i = 0; i <= W; i++){
for(int j = 0; j < n; j++){
if(wt[j] <= i){
dp[i] = max(dp[i], dp[i - wt[j]] +
val[j]);
}
}
}
return dp[W];
}
public static void main(String[] args)
{
int W = 100;
int val[] = {10, 30, 20};
int wt[] = {5, 10, 15};
int n = val.length;
System.out.println(unboundedKnapsack(W, n, val, wt));
}
}
|
Python3
def unboundedKnapsack(W, n, val, wt):
dp = [0 for i in range(W + 1)]
ans = 0
for i in range(W + 1):
for j in range(n):
if (wt[j] <= i):
dp[i] = max(dp[i], dp[i - wt[j]] + val[j])
return dp[W]
W = 100
val = [10, 30, 20]
wt = [5, 10, 15]
n = len(val)
print(unboundedKnapsack(W, n, val, wt))
|
C#
using System;
class UboundedKnapsack {
private static int max(int i, int j)
{
return (i > j) ? i : j;
}
private static int unboundedKnapsack(int W, int n,
int []val, int []wt)
{
int []dp = new int[W + 1];
for(int i = 0; i <= W; i++){
for(int j = 0; j < n; j++){
if(wt[j] <= i){
dp[i] = Math.Max(dp[i], dp[i -
wt[j]] + val[j]);
}
}
}
return dp[W];
}
public static void Main()
{
int W = 100;
int []val = {10, 30, 20};
int []wt = {5, 10, 15};
int n = val.Length;
Console.WriteLine(unboundedKnapsack(W, n, val, wt));
}
}
|
PHP
<?php
function unboundedKnapsack($W, $n,
$val, $wt)
{
for($i = 0; $i <= $W; $i++)
$dp[$i] = 0;
$ans = 0;
for ($i = 0; $i <= $W; $i++)
for ($j = 0; $j < $n; $j++)
if ($wt[$j] <= $i)
$dp[$i] = max($dp[$i],
$dp[$i - $wt[$j]] +
$val[$j]);
return $dp[$W];
}
$W = 100;
$val = array(10, 30, 20);
$wt = array(5, 10, 15);
$n = count($val);
echo unboundedKnapsack($W, $n,
$val, $wt);
?>
|
Javascript
<script>
function max(i, j)
{
return (i > j) ? i : j;
}
function unboundedKnapsack(W, n, val, wt)
{
let dp = new Array(W + 1);
dp.fill(0);
for(let i = 0; i <= W; i++){
for(let j = 0; j < n; j++){
if(wt[j] <= i){
dp[i] = Math.max(dp[i], dp[i - wt[j]] + val[j]);
}
}
}
return dp[W];
}
let W = 100;
let val = [10, 30, 20];
let wt = [5, 10, 15];
let n = val.length;
document.write(unboundedKnapsack(W, n, val, wt));
</script>
|
Output:
300
This article is compiled using inputs from Shubham Gupta, Shubham Joshi and Ashish kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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