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Description: Knapsack 0-1背包问题实现源码
Knapsack 0-1背包问题实现源码-Knapsack 0-1 knapsack problem to achieve source Knapsack 0-1 knapsack problem to achieve source
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Size: 1024 |
Author: chen |
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Description: 很方便解决背包问题的好源码相信会给大家带来方便
-Very convenient to solve knapsack problem a good source I believe will bring convenience to everyone
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Size: 20480 |
Author: 付梅彦 |
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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: 01背包问题的C++源程序,写的比较简单-01 knapsack problem C++ Source code, written in relatively simple
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Size: 49152 |
Author: sandy chen |
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Description: 背包问题是一个经典的动态规划模型。它既简单形象容易理解,又在某种程度上能够揭示动态规划的本质-Knapsack problem is a classic dynamic programming model. It is simple and easy to understand images, but also to some extent, able to reveal the essence of dynamic programming
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Size: 330752 |
Author: 于维川 |
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Description: 解0/1背包问题-Solutions of 0/1 knapsack problem
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Size: 1024 |
Author: |
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Description: 背包算法中用遗传算法解决VB代码
希望能帮助有用得着的朋友-Knapsack algorithm using genetic algorithm to solve VB code to help a friend has useful
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Size: 1024 |
Author: adsd |
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Description: 基于遗传算法的背包问题求解,有基本的说明和代码
其他人不需帐号就可自由下载此源码 -Based on genetic algorithms for solving knapsack problem, some basic instructions and code other people without accounts can download this free source
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Size: 1024 |
Author: ZHANGWEN |
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Description: 背包问题非常有用的程序代码打包,包括贪心,回溯,动态规划-Knapsack problem is very useful package of program code, including greedy, backtracking, dynamic programming
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Size: 6128640 |
Author: adsd |
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Description: 用蚁群算法解决0-1背包问题,matlab实现,自己亲自写的,没问题-Using ant colony algorithm to solve 0-1 knapsack problem, matlab realize that he personally wrote, no problem
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Size: 3072 |
Author: 有偿编程,挣点生活费 |
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Description: knap2
背包问题非递归,运用栈及回溯算法优化了一下,可以找到所有满足条件的解。
如果遇到什么问题,请发邮件至horsewhite32@hotmail.com,笔者将不胜感激
uuhorse-non-recursive knap2 knapsack problem, the use of stack and the backtracking algorithm to optimize the look, you can find all the solutions to meet the conditions. If any problems, please e-mail horsewhite32@hotmail.com, I would be grateful uuhorse
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Size: 6144 |
Author: uuhorse |
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Description: 01背包问题的优先队列式分枝限界算法程序LCKNAP-01 knapsack problem the priority queue algorithm Branch and Bound-style program LCKNAP
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Size: 8192 |
Author: lv |
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Description: 动态规划求01背包问题. 动态规划求01背包问题.-Dynamic programming for knapsack problem 01. Dynamic programming for knapsack problem 01.
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Size: 1024 |
Author: aa |
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Description: 0-1背包问题(0-1 Knapsack Problem)的定义为:设集合 代表m件物品,正整数 分别表示第 件物品的价值与重量,那么0-1背包问题KNAP(A,c)定义为,求A的子集,使得重量之和小于背包的容量c,并使得价值和最大。-0-1 knapsack problem (0-1 Knapsack Problem) is defined as: set up a collection on behalf of m items, respectively, the first positive integer value of items and weight, then the 0-1 knapsack problem KNAP (A, c) is defined as, A subset of demand, making less than the weight of the backpack and the capacity c, and make the value and the greatest.
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Size: 2048 |
Author: chj |
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Description: 此程序是使用Java编写的,实现了可分割的0/1背包问题。程序主要是是使用了经典的贪心算法,能够很好的解决此问题。-This procedure is to use Java prepared to achieve that can be separated from 0/1 knapsack problem. Procedure is mainly the use of a classic greedy algorithm that can solve this problem.
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Size: 4096 |
Author: gengxin |
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Description: c++经典小程序。包括河内塔
费式数列
巴斯卡三角形
三色棋
老鼠走迷官(一)
老鼠走迷官(二)
骑士走棋盘
八个皇后
八枚银币
生命游戏
字串核对
双色、三色河内塔
背包问题(Knapsack Problem)
数、运算
蒙地卡罗法求 PI
Eratosthenes筛选求质数
超长整数运算(大数运算)
长 PI
最大公因数、最小公倍数、因式分解
完美数
阿姆斯壮数
最大访客数
中序式转后序式(前序式)
后序式的运算
关于赌博
洗扑克牌(乱数排列)
Craps赌博游戏
约瑟夫问题(Josephus Problem)
集合问题
排列组合
格雷码(Gray Code)
产生可能的集合
m元素集合的n个元素子集
数字拆解
排序
得分排行
选择、插入、气泡排序
Shell 排序法 - 改良的插入排序
Shaker 排序法 - 改良的气泡排序
Heap 排序法 - 改良的选择排序
快速排序法等
-err
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Size: 450560 |
Author: sunny |
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Description: 用回溯解背包问题 假设有n件物品,定义一个结构体a[]来存储,结构体有两个成员weight和value(weight表示重量,value表示价值)先定义一个数组col[]表示每个物品当前状态(为1表示被选,为0表示未被选),其初值全为1,从下标为0开始遍历,当前所选物品总重和总价值分别设为tw和tv(初值均为0),背包的限重设为limit,若第i个物品满足tw+a[i].weight<=limit且col[i]==1 就将a[i].weight和value加入tw和tv,否则col[i]设为0。-Knapsack problem using backtracking solution assumption has n items, the definition of a structure a [] to store, structure, body weight and has two members value (weight, said weight, value, said value) before the definition of an array col [] mean that each and every item current status (as one said to be elected, not selected to express to 0), all of its initial value 1, from the beginning subscript 0 ergodicity, the currently selected items and the total value of gross weight, respectively, as tw and tv (early values are 0), backpack weight limit is set to limit, if paragraph i of goods to meet tw+ a [i]. weight
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Size: 180224 |
Author: STartBoy |
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Description: 背包问题
采用分枝限界法解决0/1背包问题! 本人上机实习作业,通过老师验收,合格! 针对部分上机实习的同学可以来下~
-Knapsack problem using Branch and Bound method to solve 0/1 Knapsack Problem! My internship on the machine operation, through the teacher acceptance, qualified! The machine for some internship students can come to the next ~
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Size: 1024 |
Author: ohyoung |
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Description: 背包问题运用贪婪算法的matlab 程序实现-Use of greedy algorithm knapsack problem of matlab program
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Size: 1024 |
Author: shangsheng |
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Description: 程序设计思路
在动态规划中,可将一个问题的解决方案视为一系列决策的结果,要考察每个最优决策序列中是否包含一个最优子序列。所以在最短路径问题中,假如在的第一次决策时到达了某个节点v,那么不管v 是怎样确定的,此后选择从v 到d 的路径时,都必须采用最优策略。利用最优序列由最优子序列构成的结论,可得到f 的递归式。f ( 1 ,c) 是初始时背包问题的最优解。可使用(1)中所示公式通过递归或迭代来求解f ( 1 ,c)。从f (n, * )开始迭式, f (n, * )由第一个式子得出,然后由第二式递归计算f (i,*) ( i=n- 1,n- 2,⋯ , 2 ),最后得出f ( 1 ,c)。动态规划方法采用最优原则( principle of optimality)来建立用于计算最优解的递归式。所谓最优原则即不管前面的策略如何,此后的决策必须是基于当前状态(由上一次决策产生)的最优决策。由于对于有些问题的某些递归式来说并不一定能保证最优原则,因此在求解问题时有必要对它进行验证。若不能保持最优原则,则不可应用动态规划方法。
-err
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Size: 22528 |
Author: 王新峰 |
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