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| 1 | +/** |
| 2 | + * There are N objects and a bag with capacity of V. Evey type of object can be |
| 3 | + * used with unlimited number of times. |
| 4 | + * The ith object is having costs[i] and weights[i]. |
| 5 | + * Find out which objects can be put into the bag in order to have maximum value. |
| 6 | + */ |
| 7 | + |
| 8 | + |
| 9 | +public class CompleteKnapsack { |
| 10 | + |
| 11 | + /** |
| 12 | + * dp[i][v] = max{dp[i-1][v], dp[i][v-c[i]] + w[i]} |
| 13 | + */ |
| 14 | + public int maxValue(int[] costs, int[] weights, int V) { |
| 15 | + if (costs == null || weights == null) return 0; |
| 16 | + int N = costs.length; |
| 17 | + int[][] dp = new int[N + 1][V + 1]; |
| 18 | + |
| 19 | + for (int i=1; i<=N; i++) { |
| 20 | + for (int v=1; v<=V; v++) { |
| 21 | + if (v < costs[i-1]) { |
| 22 | + dp[i][v] = dp[i-1][v]; |
| 23 | + } else { |
| 24 | + dp[i][v] = Math.max(dp[i-1][v], dp[i][v - costs[i-1]] + weights[i-1]); |
| 25 | + } |
| 26 | + } |
| 27 | + } |
| 28 | + return dp[N][V]; |
| 29 | + } |
| 30 | + |
| 31 | + |
| 32 | + /** |
| 33 | + * dp[v] = max{dp[v], dp[v-c[i]] + w[i]} |
| 34 | + */ |
| 35 | + public int maxValue2(int[] costs, int[] weights, int V) { |
| 36 | + if (costs == null || weights == null) return 0; |
| 37 | + int N = costs.length; |
| 38 | + int[] dp = new int[V + 1]; |
| 39 | + |
| 40 | + for (int i=0; i<N; i++) { |
| 41 | + for (int v=costs[i]; v<=V; v++) { |
| 42 | + dp[v] = Math.max(dp[v], dp[v - costs[i]] + weights[i]); |
| 43 | + } |
| 44 | + } |
| 45 | + return dp[V]; |
| 46 | + } |
| 47 | + |
| 48 | + |
| 49 | + public static void main(String[] args) { |
| 50 | + CompleteKnapsack slt = new CompleteKnapsack(); |
| 51 | + |
| 52 | + int[] costs = new int[]{}; |
| 53 | + int[] weights = new int[]{}; |
| 54 | + System.out.println(slt.maxValue(costs, weights, 10)); |
| 55 | + System.out.println(slt.maxValue2(costs, weights, 10)); |
| 56 | + |
| 57 | + costs = new int[]{2}; |
| 58 | + weights = new int[]{2}; |
| 59 | + System.out.println(slt.maxValue(costs, weights, 2)); |
| 60 | + System.out.println(slt.maxValue2(costs, weights, 2)); |
| 61 | + |
| 62 | + costs = new int[]{2}; |
| 63 | + weights = new int[]{2}; |
| 64 | + System.out.println(slt.maxValue(costs, weights, 3)); |
| 65 | + System.out.println(slt.maxValue2(costs, weights, 3)); |
| 66 | + |
| 67 | + costs = new int[]{2}; |
| 68 | + weights = new int[]{2}; |
| 69 | + System.out.println(slt.maxValue(costs, weights, 1)); |
| 70 | + System.out.println(slt.maxValue2(costs, weights, 1)); |
| 71 | + |
| 72 | + |
| 73 | + |
| 74 | + |
| 75 | + |
| 76 | + } |
| 77 | + |
| 78 | + |
| 79 | + // procedure CompletePack(cost,weight) |
| 80 | + // for v=cost..V |
| 81 | + // f[v]=max{f[v],f[v-c[i]]+w[i]} |
| 82 | + |
| 83 | +} |
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