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ZeroOnePack
- 分支界限法求解0-1背包问题。有N件物品和一个容量为V的背包，第i件物品的费用是c[i]，价值是w[i]，求解将哪些物品装入背包可使这些物品的费用总和不超过背包容量，且价值总和最大。-Branch and bound method for solving 0-1 knapsack problem. N items and a capacity V of the backpack, the cost of items i C [i], the value of W [i], the sum of
mochila_BB
- Branch and Bound program for the backpack problem
mochila_BK
- Backtracking program to solve backpack problem
01backpack
- 算法设计与分析01背包优化问题实验 0-1背包问题： : 有N个物品和一个背包，其中: 物品具有重量 (w1, w2, …, wn) 和价值 (v1, v2, …, vn) 背包的最大重量承受限制为c 如何放置物品可得最高价值? 思路：动态规划 -Algorithm Design and Analysis 01 knapsack optimization experiments 0-1 knapsack problem :: with N items and a back
01
- 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.
01packet
- 01背包是在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 backpack 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.
Scripts
- unity3D 角色背包 拾取 记录点 的源代码-unity3D role backpack pickup point source records
beibaoPSO
- 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
fzxj
- 分支限界法是一种优化方法，背包问题是NP完全问题，本程序通过分支限界实现了一种优化的背包。-Branch and bound method is an optimization method, knapsack problem is NP-complete problems, this program through the branch and bound to achieve an optimized backpack.
hs
- 背包问题是NP完全问题，本程序通过回溯法实现了一种优化的背包。-Knapsack problem is NP-complete problems, this program through backtracking realize an optimized backpack.
Deformation-knapsack-problem
- 背包问题的变形，可以更好的理解，01背包，完全背包，多重背包问题等问题-Deformation knapsack problem can be better understood, 01 backpacks, completely backpack, multiple knapsack problem and other issues
Knapsack
- 算法设计与分析中背包问题。 首先将最优解矩阵的第一行和第一列均赋值为0；然后从上至下、从左至右依次计算m[i][j]的值，如果第i个物品 的重量大于j的值，即第i个物品不放入背包中，则将m[i][j]赋值为m[i-1][j]，否则取第i个物品放入背包和不放入背 包所得价值的最大值；最后从后往前计算最优解向量的值，如果m[i][j]=m[i-1][j]，则将x[i]赋值为0，否则将x[i] 赋值为1，并将j减去第i个物品的重量。-The knapsack problem al
0-1bag
- 0-1背包问题：有N件物品和一个容量为V的背包。第i件物品的花费是c[i]，价值是w[i]。求解将哪些物品装入背包可使价值总和最大。-The 0-1 knapsack problem: N items and a capacity of V backpack. Take the first I items is c[i], the value is w[i]. For which goods loaded backpack can make the value of the sum of the
beibao
- c++实现的背包问题代码，可能不全，但是是自己写的，有需要的可以参考一下-c++ backpack problem
tanlanbeibao
- 进的背包问题：给定一个超递增序列和一个背包的容量，然后在超递增序列中选（只能选一次）或不选每一个数值，使得选中的数值的和正好等于背包的容量。 代码思路：从最大的元素开始遍历超递增序列中的每个元素，若背包还有大于或等于当前元素值的空间，则放入，然后继续判断下一个元素；若背包剩余空间小于当前元素值，则判断下一个元素-Into the knapsack problem: Given a super-increasing sequence and a capacity backpack, an
0-1bag
- 假设有n件物品，每件物品有各自的重量W1,W2,……,Wn和 与之对应的价值V1,V2,……,Vn。设背包的容量为c，在不超过背包容量的前提下，求出获得最大价值总和的方案。（0-1背包的情况下物品不可分割，只能选择放入，或者不放入背包中）。 -Suppose there are n items, each item has its own weight W1, W2, ......, Wn and the corresponding value of V1, V2, ......, Vn
greedy
- 贪心法解决背包问题 可自己设置物品数量和背包总容量-Greedy method to solve the knapsack problem and the number of items can set up their own backpack with a total capacity
DP
- 动态分配法解决背包问题，自己设置物品总数,背包总容量-Dynamic allocation method to solve the knapsack problem, the total number of items set up their own backpack with a total capacity
backpack-problem
- 描述《算法引论》上最基本的背包问题，PPT以及算法源码的实现-The most basic descr iption of the knapsack problem, " Introduction to Algorithms" on, PPT and algorithm realization source
01beibao
- 01背包的完整版，采用动态规划和贪婪算法解决-01 backpack full version, using dynamic programming and greedy algorithm to solve