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Description: 给定n 个物品, 物品i重为wi 并且价值为 vi ，背包所能承载的最大容量为 W.
0-1 背包问题即是选择含有着最大总价值的物品的子集且它的容量 ≤W ．
用动态规划实现-given n goods, items i weight of wi and value of vi, the backpack can carry a maximum capacity of W. 0-1 knapsack problem that is a choice with a maximum total value of the goods but a subset of the W capacity. Dynamic Programming
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Author: 叶黎 |
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Description: 1.1 背包问题
有一个徒步旅行者，已知他能承受的旅行背包的重量不超过a（kg）。设有n种物品可供他选择装入背包，这n种物品分别编号为1，2，…，n。其中第i种物品每件的重量为ai（kg），其使用价值（指一件第i种物品对旅行者来说所带来的好处的一种数量指标）为ci（i＝1，2，…，n）。问这位旅行者应如何选择携带这n种物品的件数，使得总价值最大？
-1.1 knapsack problem is a hikers. he is known to travel to withstand the weight of the backpack not exceed a (kg). N Species with articles for his choice loaded backpack, n product were numbered 1, 2, ..., n. I first of which product each ai of the weight (kg), its value (i refers to a section of the types of items travelers the benefits brought by a number of indicators) for the ci (i = 1, 2, ..., n). Asked how travelers who choose to bring this product to the n pieces, making the total value of the largest?
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Size: 14008 |
Author: 孙竟飞 |
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Description: 用遗传算法求解背包问题是南京航空航天大学信息与计算科学专业编写的.本程序利用遗传算法来求解背包问题.采用二进制字符串编码,1表示选择物体,0则不选择.
背包问题描述:在M件物品取出若干件放在空间为W的背包里，每件物品的重量为W1，W·2……Wn，与之相对应的价值为P1,P2……Pn。求出获得最大价值的方案。注意：在本题中，所有的重量值均为整数。-genetic algorithm knapsack problem is the Nanjing University of Aeronautics and Astronautics, Information and Computing Science prepared. The procedure the use of genetic algorithms to solve knapsack problem. binary string encoding, an option objects, 0 no choice. Knapsack Problem Description : M items removed several pieces on the space-W backpack, Each of the weight W1, W 2 ... Wn, the corresponding value of P1, P2 ... Pn. Sought to gain maximum value of the program. NOTE : In this issue, all the weight values are integers.
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Size: 25921 |
Author: 王杰 |
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Description: 解决背包的取舍问题，问选择多少个不超过背包的重量的物品．能使价值最大．-solve knapsack choice question about the choice of the number does not exceed the weight of the backpack items. Make the greatest value.
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Size: 1014 |
Author: ＹＵＪＵＮ |
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Description: 回溯（b a c k t r a c k i n g）是一种系统地搜索问题解答的方法。为了实现回溯，首先需要为问题定义一个解空间（ solution space），这个空间必须至少包含问题的一个解（可能是最优的）。在迷宫老鼠问题中，我们可以定义一个包含从入口到出口的所有路径的解空间；在具有n 个对象的0 / 1背包问题中（见1 . 4节和2 . 2节），解空间的一个合理选择是2n 个长度为n 的0 / 1向量的集合，这个集合表示了将0或1分配给x的所有可能方法。当n= 3时，解空间为{ ( 0 , 0 , 0 )，( 0 , 1 , 0 )，( 0 , 0 , 1 )，( 1 , 0 , 0 )，( 0 , 1 , 1 )，( 1 , 0 , 1 )，( 1 , 1 , 0 )，( 1 , 1 , 1 ) }。-retrospective (b a c k t r a c k i n g) is a systematic search to answer the question. To achieve retrospective, the first issue of the need for a definition of the solution space (solution space), The space must contain at least one solution to the problem (which may be optimal). Rats in a maze problem, we can contain a definition from the entrance to the export of the solution space trails; n is the object of the 0 / 1 knapsack problem (see 1. 4 and 2. 2) The solution space is a reasonable choice of two n length of the 0 n / a vector set, this assembly that it will 0 or 1 x allocated to all possible ways. When n = 3, the solution space for the ((0, 0, 0), (0, 1, 0), (0, 0, 1), (1, 0, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 1, 1)).
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Size: 29600 |
Author: john |
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Description: 该程序用贪心算法来求解0-1背包问题 采用贪婪准则：每次选择p/w最大的物品放入背包。-procedures with the greedy algorithm to solve 0-1 knapsack problem using greed criteria : Each choice p / w largest goods Add backpack.
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Size: 2195 |
Author: 杨晨 |
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Description: 0 / 1背包问题是一个N P-复杂问题，为了解决该问题，，将用回溯算法解决该问题。既然想选择一个对象的子集，将它们装入背包，以便获得的收益最大，则解空间应组织成子集树的形状（如图1 6 - 2所示）。该回溯算法与4 . 2节的装载问题很类似。首先形成一个递归算法，去找到可获得的最大收益。然后，对该算法加以改进，形成代码。改进后的代码可找到获得最大收益时包含在背包中的对象的集合。-0 / 1 knapsack problem is a P-complex issues, in order to solve the problem, and will be used backtracking algorithm to solve the problem. As to the choice of a subset of object, they will load a backpack, in order to obtain the greatest profits, the solution should be organized into space subset of the shape of the tree (Figure 1 6-2 below). The backtracking algorithms and 4. Two loading problem is very similar. Forming a recursive algorithm to get the maximum benefit available. Then, this algorithm is modified to form code. Improved code can be found at the maximum benefits included in the backpack of a collection of objects.
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Size: 8915 |
Author: 周易 |
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Description: 该程序用贪心算法来求解0-1背包问题 采用贪婪准则：每次选择p/w最大的物品放入背包。-procedures with the greedy algorithm to solve 0-1 knapsack problem using greed criteria : Each choice p/w largest goods Add backpack.
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Size: 2048 |
Author: 杨晨 |
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Description: 0 / 1背包问题是一个N P-复杂问题，为了解决该问题，，将用回溯算法解决该问题。既然想选择一个对象的子集，将它们装入背包，以便获得的收益最大，则解空间应组织成子集树的形状（如图1 6 - 2所示）。该回溯算法与4 . 2节的装载问题很类似。首先形成一个递归算法，去找到可获得的最大收益。然后，对该算法加以改进，形成代码。改进后的代码可找到获得最大收益时包含在背包中的对象的集合。-0/1 knapsack problem is a P-complex issues, in order to solve the problem, and will be used backtracking algorithm to solve the problem. As to the choice of a subset of object, they will load a backpack, in order to obtain the greatest profits, the solution should be organized into space subset of the shape of the tree (Figure 1 6-2 below). The backtracking algorithms and 4. Two loading problem is very similar. Forming a recursive algorithm to get the maximum benefit available. Then, this algorithm is modified to form code. Improved code can be found at the maximum benefits included in the backpack of a collection of objects.
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Size: 8192 |
Author: 周易 |
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Description: 给定n 个物品, 物品i重为wi 并且价值为 vi ，背包所能承载的最大容量为 W.
0-1 背包问题即是选择含有着最大总价值的物品的子集且它的容量 ≤W ．
用动态规划实现-given n goods, items i weight of wi and value of vi, the backpack can carry a maximum capacity of W. 0-1 knapsack problem that is a choice with a maximum total value of the goods but a subset of the W capacity. Dynamic Programming
Platform: |
Size: 1024 |
Author: 叶黎 |
Hits:
Description: 1.1 背包问题
有一个徒步旅行者，已知他能承受的旅行背包的重量不超过a（kg）。设有n种物品可供他选择装入背包，这n种物品分别编号为1，2，…，n。其中第i种物品每件的重量为ai（kg），其使用价值（指一件第i种物品对旅行者来说所带来的好处的一种数量指标）为ci（i＝1，2，…，n）。问这位旅行者应如何选择携带这n种物品的件数，使得总价值最大？
-1.1 knapsack problem is a hikers. he is known to travel to withstand the weight of the backpack not exceed a (kg). N Species with articles for his choice loaded backpack, n product were numbered 1, 2, ..., n. I first of which product each ai of the weight (kg), its value (i refers to a section of the types of items travelers the benefits brought by a number of indicators) for the ci (i = 1, 2, ..., n). Asked how travelers who choose to bring this product to the n pieces, making the total value of the largest?
Platform: |
Size: 13312 |
Author: 孙竟飞 |
Hits:
Description: 用遗传算法求解背包问题是南京航空航天大学信息与计算科学专业编写的.本程序利用遗传算法来求解背包问题.采用二进制字符串编码,1表示选择物体,0则不选择.
背包问题描述:在M件物品取出若干件放在空间为W的背包里，每件物品的重量为W1，W·2……Wn，与之相对应的价值为P1,P2……Pn。求出获得最大价值的方案。注意：在本题中，所有的重量值均为整数。-genetic algorithm knapsack problem is the Nanjing University of Aeronautics and Astronautics, Information and Computing Science prepared. The procedure the use of genetic algorithms to solve knapsack problem. binary string encoding, an option objects, 0 no choice. Knapsack Problem Description : M items removed several pieces on the space-W backpack, Each of the weight W1, W 2 ... Wn, the corresponding value of P1, P2 ... Pn. Sought to gain maximum value of the program. NOTE : In this issue, all the weight values are integers.
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Size: 25600 |
Author: 王杰 |
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Description: 解决背包的取舍问题，问选择多少个不超过背包的重量的物品．能使价值最大．-solve knapsack choice question about the choice of the number does not exceed the weight of the backpack items. Make the greatest value.
Platform: |
Size: 1024 |
Author: solid |
Hits:
Description: 回溯（b a c k t r a c k i n g）是一种系统地搜索问题解答的方法。为了实现回溯，首先需要为问题定义一个解空间（ solution space），这个空间必须至少包含问题的一个解（可能是最优的）。在迷宫老鼠问题中，我们可以定义一个包含从入口到出口的所有路径的解空间；在具有n 个对象的0 / 1背包问题中（见1 . 4节和2 . 2节），解空间的一个合理选择是2n 个长度为n 的0 / 1向量的集合，这个集合表示了将0或1分配给x的所有可能方法。当n= 3时，解空间为{ ( 0 , 0 , 0 )，( 0 , 1 , 0 )，( 0 , 0 , 1 )，( 1 , 0 , 0 )，( 0 , 1 , 1 )，( 1 , 0 , 1 )，( 1 , 1 , 0 )，( 1 , 1 , 1 ) }。-retrospective (b a c k t r a c k i n g) is a systematic search to answer the question. To achieve retrospective, the first issue of the need for a definition of the solution space (solution space), The space must contain at least one solution to the problem (which may be optimal). Rats in a maze problem, we can contain a definition from the entrance to the export of the solution space trails; n is the object of the 0/1 knapsack problem (see 1. 4 and 2. 2) The solution space is a reasonable choice of two n length of the 0 n/a vector set, this assembly that it will 0 or 1 x allocated to all possible ways. When n = 3, the solution space for the ((0, 0, 0), (0, 1, 0), (0, 0, 1), (1, 0, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 1, 1)).
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Size: 29696 |
Author: john |
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Description: 设有一个背包可以放入的物品重量最重为s，现有n件物品，它们的重量分别为w[0]、 w[1]、w[2]、…、w[n-1]。问能否从这n件物品中选择若干件放入此背包中，使得放入的重量之和正好为s。如果存在一种符合上述要求的选择，则称此背包问题有解（或称其解为真）；否则称此背包问题无解（或称其解为假）。试用递归方法设计求解背包问题的算法。-There is a backpack weight items can be placed the most weight of s, the existing n items, and their respective weights for w [0], w [1], w [2], ..., w [n-1]. Asked whether the items from the n number of pieces of select Add this backpack makes Add the weight of and an opportunity for s. If there is a choice of these requirements, then the knapsack problem solvability (or its solution is true) otherwise said no solution to this knapsack problem (or its solution to be false). Trial designed recursive algorithm for solving knapsack problem.
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Size: 249856 |
Author: 张巨松 |
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Description: 遗传算法用于求解多目标背包问题，学包括基本的选择、杂交、变异等遗传算子.-Genetic algorithm for solving multi-objective knapsack problem, learning the basic choice, hybridization, mutation and other genetic operators.
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Size: 44032 |
Author: daniel |
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Description: 基本遗传算法带最优保存思想的背包问题，其中，目标值那段代码使用的是惩罚函数法，选择是概率选择，交叉是双点随机交叉，变异是概率变异-The basic genetic algorithm with elitist thinking knapsack problem, which is a target that part of the code using penalty function method, choice is the probability of selection, crossover is a two-point random crossover and mutation is the probability of mutation
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Size: 2048 |
Author: 田文杰 |
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Description: 背包问题的贪婪算法，主要是通过性价比来决定物品的选择。-The greedy algorithm knapsack problem, mainly through cost-effective to determine the choice of items.
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Size: 1024 |
Author: 覃亮 |
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Description: 关于多选择背包的很好的文章，提供了各种设计思路-Multiple Choice Knapsack Problem Example of planning choice in transportation
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Size: 390144 |
Author: amy |
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Description: 背包问题即给定一组物品，每种物品都有自己的重量和价格，在限定的总重量内，我们如何选择，才能使得物品的总价格最高。-A hybrid algorithm combining ant colony system with multi-choice Knapsack problem was proposed.
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Size: 1024 |
Author: 刘芳 |
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