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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,47 @@ | ||
| W = 8 | ||
| N = 4 | ||
| wi = [3, 3, 5, 6] | ||
| ci = [3, 5, 10, 14] | ||
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| d = [[0] * (W + 1) for i in range(N + 1)] | ||
| for i in range(1, N + 1): | ||
| for w in range(1, W + 1): | ||
| if wi[i-1] > w: | ||
| d[i][w] = d[i-1][w] | ||
| else: | ||
| d[i][w] = max(d[i-1][w], d[i-1][w-wi[i-1]]+ci[i-1]) | ||
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| print('\n'.join([' '.join(map(str, d[i])) for i in range(N + 1)])) | ||
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| # Python3 program to find maximum | ||
| # achievable value with a knapsack | ||
| # of weight W and multiple instances allowed. | ||
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| # Returns the maximum value | ||
| # with knapsack of W capacity | ||
| def unboundedKnapsack(W, n, val, wt): | ||
| # dp[i] is going to store maximum | ||
| # value with knapsack capacity i. | ||
| dp = [0 for i in range(W + 1)] | ||
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| ans = 0 | ||
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| # Fill dp[] using above recursive formula | ||
| 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]) | ||
| print(dp) | ||
| return dp[W] | ||
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| # Driver program | ||
| W = 5 | ||
| val = [10] | ||
| wt = [2] | ||
| n = len(val) | ||
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| print(unboundedKnapsack(W, n, val, wt)) | ||
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| # This code is contributed by Anant Agarwal. |