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TOT

  • 用分支限界法求解背包问题(0/1背包) 1.问题描述:已知有N个物品和一个可以容纳TOT重量的背包

    用分支限界法求解背包问题(0/1背包) 1.问题描述:已知有N个物品和一个可以容纳TOT重量的背包,每种物品I的重量为Weight,价值为Value。一个只能全放入或者不放入,求解如何放入物品,可以使背包里的物品的总价值最大。 2.设计思想与分析:对物品的选取与否构成一棵解树,左子树表示装入,右表示不装入,通过检索问题的解树得出最优解,并用结点上界杀死不符合要求的结点。

    标签: TOT 分支 背包问题

    上传时间: 2016-02-09

    上传用户:我们的船长

  • Chuong trinh dong bang may tinh TOT nhat hien nay

    Chuong trinh dong bang may tinh TOT nhat hien nay

    标签: Chuong trinh dong bang

    上传时间: 2017-07-12

    上传用户:youke111

  • We have a group of N items (represented by integers from 1 to N), and we know that there is some TOT

    We have a group of N items (represented by integers from 1 to N), and we know that there is some TOTal order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the TOTal cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the TOTal cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.

    标签: represented integers group items

    上传时间: 2016-01-17

    上传用户:jeffery