// Old code for simplified knapsack problemelse if (j >= weights[i]) {if (elementMatrix[i - 1][j - weights[i]].isExists) {elementMatrix[i][j].setExists(true);elementMatrix[i][j].setIncludes(true);}}// New code, note that we're searching for a solution in the same// row ...
public static int knapsack(int[] weights, int[] values, int capacity) { int n = weights.length; int[][] dp = new int[n + 1][capacity + 1]; for (int i = 1; i <= n; i++) { for (int j = 1; j <= capacity; j++) { if (j >= weights[i - 1]) { dp...
// W 为背包总体积 // N 为物品数量 // weights 数组存储 N 个物品的重量 // values 数组存储 N 个物品的价值 public int knapsack(int W, int N, int[] weights, int[] values) { int[][] dp=new int[N+1][W+1]; for(int i=1;i<=N;i++){ int weight=weights[i-1]; int value=...
packagecom.liuzhen.chapterThree;publicclassKnapsack {publicintmaxSumValue = 0;//定义满足背包问题子集的最大承重所得的总价值,初始化为0/** 数组A的行数为2^n,代表n个物品共有2^n个子集,列数为n。即每一行的排列为一个背包实例 * 数组weight存放每个物品的具体重量 * 数组value存放每个物品的具体价值 *...
pid=2602 */ public class M4_Knapsack { static int n, C; static int[] w, v; static int[][] dp; //记忆化 static int rec(int p, int curW) { if (p == n) return 0; if (dp[p][curW] != -1) return dp[p][curW]; if (curW + w[p] > C) return dp[p][curW] =...
for(int j=n;j>0;j--){ bestX[j]=(enode.leftChild)?1:0; enode=enode.parent; } return cp; } public static double knapsack(double pp[],double ww[],double cc,int xx[]){ //返回最大值,bestX返回最优解 c=cc; n=pp.length-1; ...
public class Knapsack { public static int knapsack(int[] weights, int[] values, int capacity) { int n = weights.length; int[][] dp = new int[n + 1][capacity + 1]; for (int i = 1; i <= n; i++) { for (int j = 0; j <= capacity; j++) { if (weights[i - 1] <...
4. 0/1 Knapsack Problem (Dynamic Programming) 5. Backtracking 6. Trie 7. Topological Sort 8. Union Find (Disjoint-Set Data Structure) 9. Ordered Set 3. Resources: I acknowledge the use of OpenAI's ChatGPT for topic explanation 🤖, LeetCode and HackerRank for coding interview qu...
avDyanamicProg AV DP {knapsack, MCM} Sep 4, 2024 basics whether string of a and b has first and last a or not Dec 11, 2024 condition_loops Cartision Product of String array Oct 14, 2024 countPairsForTargetInArray count Pairs For a Target In Array questions Jan 16, 2025 method < so...
Knapsack Problem 背包问题 Discrete Fourier Transform 离散Fourier变换 Combinatorial Problems 组合问题 Sorting 排序 Searching 查找 Median and Selection 中位数 Generating Permutations 排列生成 Generating Subsets 子集生成 Generating Partitions 划分生成 Generating Graphs 图的生成 ...