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Description: 实现背包问题
package problem
1． 问题描述
假设有一个能装入总体积为T的背包和n件体积分别为w1 , w2 , … , wn 的物品，能否从n件物品中挑选若干件恰好装满背包，即使w1 +w2 + … + wn=T，要求找出所有满足上述条件的解。例如：当T=10，各件物品的体积{1，8，4，3，5，2}时，可找到下列4组解： （1，4，3，2）、（1，4，5）、（8，2）、（3，5，2）。
2． 基本要求
读入T、n、w1 , w2 , … , wn
3．提示：
可利用递归方法：若选中w1 则问题变成在w2 , … , wn 中挑选若干件使得其重量之和为T- w1 ，若不选中w1，则问题变成在w2 , … , wn 中挑选若干件使得其重量之和为T 。依次类推。
也可利用回溯法的设计思想来解决背包问题。首先将物品排成一列，然后顺序选取物品装入背包，假设已选取了前i 件物品之后背包还没有装满，则继续选取第i+1件物品，若该件物品“太大”不能装入，则弃之而继续选取下一件，直至背包装满为止。但如果在剩余的物品中找不到合适的物品以填满背包，则说明“刚刚”装入背包的那件物品“不合适”，应将它取出“弃之一边”，继续再从“它之后”的物品中选取，如此重复，，直至求得满足条件的解，或者无解。
注：没压缩密码-knapsack problem achieve a package problem. Problem description one would assume that the total volume of loaded backpack and T n size pieces of w1, w2, ..., wn items Can n items from the selected pieces were just filled with backpacks, even w1 w2 ... wn = T, asked to identify all of the above conditions are met solution. For example : when T = 10, the volume of items (1,8,4,3,5,2), may find the following four solutions : (1,4,3,2), (1,4,5), (8,2), (3,5,2). 2. Basic requirements read T, n, w1, w2, ..., wn 3. Tip : Recursive methods can be used : If the problem were selected w1 into the w2, ..., wn selected several pieces make its weight and as a T-w1, if selected w1, w2 issue into the .... wn selected several pieces make its weight and as T. Renumbered accordingly. Retrospective can also u
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Size: 9296 |
Author: 李昭明 |
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Description: 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.
Platform: |
Size: 9084 |
Author: 王涛 |
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Description: 实现背包问题
package problem
1． 问题描述
假设有一个能装入总体积为T的背包和n件体积分别为w1 , w2 , … , wn 的物品，能否从n件物品中挑选若干件恰好装满背包，即使w1 +w2 + … + wn=T，要求找出所有满足上述条件的解。例如：当T=10，各件物品的体积{1，8，4，3，5，2}时，可找到下列4组解： （1，4，3，2）、（1，4，5）、（8，2）、（3，5，2）。
2． 基本要求
读入T、n、w1 , w2 , … , wn
3．提示：
可利用递归方法：若选中w1 则问题变成在w2 , … , wn 中挑选若干件使得其重量之和为T- w1 ，若不选中w1，则问题变成在w2 , … , wn 中挑选若干件使得其重量之和为T 。依次类推。
也可利用回溯法的设计思想来解决背包问题。首先将物品排成一列，然后顺序选取物品装入背包，假设已选取了前i 件物品之后背包还没有装满，则继续选取第i+1件物品，若该件物品“太大”不能装入，则弃之而继续选取下一件，直至背包装满为止。但如果在剩余的物品中找不到合适的物品以填满背包，则说明“刚刚”装入背包的那件物品“不合适”，应将它取出“弃之一边”，继续再从“它之后”的物品中选取，如此重复，，直至求得满足条件的解，或者无解。
注：没压缩密码-knapsack problem achieve a package problem. Problem description one would assume that the total volume of loaded backpack and T n size pieces of w1, w2, ..., wn items Can n items from the selected pieces were just filled with backpacks, even w1 w2 ... wn = T, asked to identify all of the above conditions are met solution. For example : when T = 10, the volume of items (1,8,4,3,5,2), may find the following four solutions : (1,4,3,2), (1,4,5), (8,2), (3,5,2). 2. Basic requirements read T, n, w1, w2, ..., wn 3. Tip : Recursive methods can be used : If the problem were selected w1 into the w2, ..., wn selected several pieces make its weight and as a T-w1, if selected w1, w2 issue into the .... wn selected several pieces make its weight and as T. Renumbered accordingly. Retrospective can also u
Platform: |
Size: 9216 |
Author: 李昭明 |
Hits:
Description: 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.
Platform: |
Size: 238592 |
Author: 王涛 |
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Description: 使用蒙特卡罗方法计算积分和体积的Matlab代码,适合初学者。-calculated using the Monte Carlo method and size of the integration of Matlab code for beginners.
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Size: 1024 |
Author: 宋仁栋 |
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Description: 货物配装问题实质上是背包问题与装箱问题的混合，有着广泛的应用。配装,是指充分利用运输工具(如火车、货车、轮船等,本文主要是指货车) 的载重量和容积,采用先进的装载方法,合理安排货物的装载计划，最大限度地发挥货车容积和承载能力，从而降低配送成本。-Cargo loading problem and knapsack problem is essentially a mixture of bin-packing problem, with a wide range of applications. Loading refers to the full use of means of transport (such as trains, trucks, ships, etc., this paper mainly refers to goods) and the load volume, using advanced loading methods, rational arrangements for cargo loading plan maximizing the volume of goods and carrying capacity, thereby reducing distribution costs.
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Size: 6144 |
Author: 廖志 |
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Description: 对背包问题进行的求解，能够处理体积相同，但是不同物品的问题-The solution on the knapsack problem that can handle the same volume, but the problem of different items
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Size: 1024 |
Author: liufeng |
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Description: 假设有一个能装入总体积为T的背包和n件体积分别为w1 , w2 , … , wn 的物品，能否从n件物品中挑选若干件恰好装满背包，即使w1 +w2 + … + wn=T，要求找出所有满足上述条件的解。-Suppose a total volume of T can be packed into a backpack and n pieces of volume, respectively w1, w2, ..., wn items, items can be selected from n number of pieces is just filled with backpacks, even if w1+ w2+ ...+ wn = T, required to find all the solutions to meet the above conditions.
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Size: 14336 |
Author: 潇洒 |
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Description: 粒子群算法解决0-1背包问题，对于n个体积为aj、价值分别为cj的物品，如何将它们装入总体积为b的背包中，使得所选物品的总价值最大。-Particle swarm algorithm to solve the 0-1 knapsack problem, for n volume for aj, value for cj items, how they are loaded in the total volume of the backpack, b, making the total value of the selected items.
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Size: 1024 |
Author: whl |
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Description: 背包问题，用PSO解背包a=[95 4 60 32 23 72 80 62 65 46] 物品的体积
c=[55 10 47 5 4 50 8 61 85 87] 物品的价值
b=269 背包的重量限制-Knapsack problem, PSO solutions backpack a = [95 4 60 32 23 72 80 62 65 46] the volume of goods c = 55 10 47 5 4 50 8 61 85 87] b = 269 value of the items backpack weight limit
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Size: 1024 |
Author: 程序菜鸟 |
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Description: 01背包问题是在M件物品取出若干件放在空间为W的背包里，每件物品的体积为W1，W2……Wn，与之相对应的价值为P1,P2……Pn。-01 knapsack problem is the M items removed several pieces on the space W of a backpack, each item has a volume of W1, W2 ...... Wn, with the corresponding value of P1, P2 ...... Pn.
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Size: 24576 |
Author: 杨鹤凌 |
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Description: PSO背包問題
用PSO求解背包問題
對於n個體積為aj、價值分別為cj的物品，如何將它們裝入總體積為b的背包中，使得所選物品的總價值最大。-PSO PSO knapsack problem using knapsack problem for n volume aj, respectively cj value items, how they can be loaded into a total volume of b backpack, making the total value of the selected items maximum.
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Size: 1024 |
Author: 黃鼎翔 |
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Description: 用粒子群算法解决0/1背包问题，解决在背包体积有限的情况下，如何最大化的实现有效资源分配-Solve knapsack problem with particle swarm algorithm to solve the limited volume of the backpack, how to maximize the efficient allocation of resources to achieve
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Size: 1024 |
Author: liupan |
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Description: 假设有一个能装入总体体积为T的背包和n件体积分别为w1,w2,…,wn的物品，能否从n件物品中挑若干件恰好装满背包，即使w1+w2+…+wn=T，要求找出所有满足上述条件的解。-Suppose there is a fit into the overall volume of T knapsack and N volume were W1, w2,... , WN items, whether the n items in the pick some parts just loaded backpack, even if w1+w2+...+wn=T, asked to find all solutions to satisfy the above conditions.
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Size: 2048 |
Author: 程瑶 |
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Description: 编写背包问题的动态规划求解代码。
背包问题 ，背包容量为W，物品个数为n，随机产生n个物品的体积（物品的体积不可大于W）与价值，求解该实例的最优解。
-Write knapsack problem dynamic Solver code. Knapsack problem, knapsack capacity of W, the number of items is n, randomly generated volume of n items (bulky items can not be greater than W) and the value of the optimal solution to solve this instance.
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Size: 1024 |
Author: 圣经 |
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Description: 给定n种物品和一背包。物品i的重量是wi，体积是bi，其价值为vi，背包的容量为c，容积为d。问应如何选择装入背包中的物品，使得装入背包中物品的总价值最大？在选择装入背包的物品时，对每种物品i只有两种选择，即装入背包或者不装入背包。不能将物品i装入背包多次，也不能只装入部分的物品i。试设计一个解此问题的动态规划算法，并分析算法的计算复杂性。(Given n items and a knapsack. The weight of the item I is wi, the volume is Bi, its value is VI, the capacity of the knapsack is C, the volume is d. How do you choose to choose the items in the backpack to make the most of the total value of the items in the backpack? When choosing a backpack, there are only two options for each item I, that is, to be loaded into a backpack or not to be loaded into a backpack. You can't load an item I into a backpack many times, and you can't just load a part of the item I. A dynamic programming algorithm for solving this problem is designed and the computational complexity of the algorithm is analyzed.)
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Size: 740352 |
Author: 为了
|
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Description: 采用多目标粒子群算法求解多目标背包问题 问题：假设存在五类物品，每类物品又包含四种具体物品，要求从五类物品中分别选择一种放入背包，使得背包总价值最大，总体积最小，总质量不超过92kg(The problem is solved by multi objective particle swarm optimization algorithm, multi-objective knapsack problem: suppose there are five categories of goods, items in each category contains four specific items from five items were selected into a backpack, making the total value of the maximum minimum total volume, backpack, total quality does not exceed 92kg)
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Size: 2048 |
Author: wyj2277 |
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