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| 1 | +// Solution to Rod cutting problem |
| 2 | +// https://en.wikipedia.org/wiki/Cutting_stock_problem |
| 3 | +// http://www.geeksforgeeks.org/dynamic-programming-set-13-cutting-a-rod/ |
| 4 | + |
| 5 | +package dpRodCutting |
| 6 | + |
| 7 | +// package main |
| 8 | + |
| 9 | +import "fmt" |
| 10 | + |
| 11 | +func max(a, b int) int { |
| 12 | + if a > b { |
| 13 | + return a |
| 14 | + } else { |
| 15 | + return b |
| 16 | + } |
| 17 | + |
| 18 | +} |
| 19 | + |
| 20 | +// solve the problem recursively: initial approach |
| 21 | +func cutRodRec(price []int, length int) int { |
| 22 | + if length == 0 { |
| 23 | + return 0 |
| 24 | + } |
| 25 | + |
| 26 | + q := -1 |
| 27 | + for i := 1; i <= length; i++ { |
| 28 | + q = max(q, price[i]+cutRodRec(price, length-i)) |
| 29 | + } |
| 30 | + return q |
| 31 | +} |
| 32 | + |
| 33 | +// solve the same problem using dynamic programming |
| 34 | +func cutRodDp(price []int, length int) int { |
| 35 | + r := make([]int, length+1) // a.k.a the memoization array |
| 36 | + r[0] = 0 // cost of 0 length rod is 0 |
| 37 | + |
| 38 | + for j := 1; j <= length; j++ { // for each length (subproblem) |
| 39 | + q := -1 |
| 40 | + for i := 1; i <= j; i++ { |
| 41 | + q = max(q, price[i]+r[j-i]) // avoiding recursive call |
| 42 | + } |
| 43 | + r[j] = q |
| 44 | + } |
| 45 | + |
| 46 | + return r[length] |
| 47 | +} |
| 48 | + |
| 49 | +/* |
| 50 | +func main() { |
| 51 | + length := 10 |
| 52 | + price := []int{0, 1, 5, 8, 9, 17, 17, 17, 20, 24, 30} |
| 53 | + // price := []int{0, 10, 5, 8, 9, 17, 17, 17, 20, 24, 30} |
| 54 | +
|
| 55 | + // fmt.Print(price[5]+price[length-5], "\n") |
| 56 | +
|
| 57 | + fmt.Print(cutRodRec(price, length), "\n") |
| 58 | + fmt.Print(cutRodDp(price, length), "\n") |
| 59 | +} |
| 60 | +*/ |
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