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[Data structs] 01package DL : 0: 背包问题优先队列分枝限界算法所谓的背包问题，可以描述如下：一个小偷打劫一个保险箱，发现柜子里有N类不同大小与价值的物品，但小偷只有一个容积为M的背包来装东西，背包问题就是要找出一个小偷选择所偷物品的组合，以使偷走的物品总价值最大。这个问题的求解有很多种方法，本程序使用分枝限界法求解。-knapsack problem Branch and Bound priority queue algorithm called knapsack problem can be described as follows : a thief robbed a safe, found N closets, a different type size and value of the items, but the thieves only one volume of M backpack to hold things. knapsack problem is to identify a thief stole items chosen by the portfolio, so that stolen goods worth largest. Solving this problem is many ways, the procedures used Branch and Bound method.
Date : 2025-12-17 Size : 233kb User : 王涛
[Data structs] fenzhixianjie DL : 0: 0/1背包问题的优先队列式分支限界算法程序-0/1 knapsack problem the priority queue-type branch and bound algorithm procedure
Date : 2025-12-17 Size : 2kb User : shanyan
[Data structs] knapbranch DL : 0: 0-1背包问题的分支定界算法实现。包含优先队列的实现细节。-0-1 knapsack problem branch and bound algorithm. Priority queue that contains the implementation details.
Date : 2025-12-17 Size : 228kb User : 黄诚
[Data structs] 01-knapsack-problem-- DL : 0: 01背包问题的优先队列式分支限界算法程序实现-01 knapsack problem with priority queues to achieve branch and bound algorithm program
Date : 2025-12-17 Size : 760kb User : hudongfang
[Data structs] knapsack(five) DL : 0: 实现0-1背包问题的优先队列分支限界算法 FIFO 分支限界算法递归法回溯法动态规划算法-0-1 knapsack problem to achieve the priority queue FIFO branch and bound algorithms branch and bound backtracking algorithm recursion dynamic programming algorithm
Date : 2025-12-17 Size : 14kb User : iwillgoon
[Data structs] BOX DL : 0: 最少背包问题：假设有许多盒子，每个盒子能保存的总重量为1.0。有N个项i1,i2,…,iN，它们的重量分别是w1,w2,…,wN。目的是用尽可能少的盒子放入所有的项，任何盒子的重量不能超过他的容量。例如，如果想的重量为0.4, 0.4, 0.6和0.6，用两个盒子就能解决。按如下策略解决此问题：按给定的次序扫描每一个项，把每一个项放入能够容纳他而不至于溢出的最满的盒子。用优先级队列选择要装入的盒子。-Minimum knapsack problem: Suppose there are many boxes, each box can hold a total weight of 1.0. There are N items i1, i2, ..., iN, their weight are w1, w2, ..., wN. Purpose is to use as little as possible of all the items into the box, any box weight can not exceed his capacity. For example, if you want the weight of 0.4, 0.4, 0.6 and 0.6, with the two boxes can be solved. Strategy to solve this problem as follows: given the order by scanning each item, put each item into him and will not be able to accommodate the overflow of the most full of boxes. With a priority queue to select into the box.
Date : 2025-12-17 Size : 2kb User : lijinping
[Data structs] leastpackage DL : 0: 实现了最小背包问题的解决，利用优先级队列来解决-Achieve a minimum knapsack problem, the use of priority queues to resolve
Date : 2025-12-17 Size : 2kb User : stendardo
[Data structs] src DL : 0: ACM题目用优先队列法解决0-1背包问题c++源代码-ACM topic with priority queuing method to solve 0-1 knapsack problem c++ source code
Date : 2025-12-17 Size : 1kb User : Denise