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[Data structs] PackageProblem DL : 0: 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
Date : 2026-01-09 Size : 9kb User : 李昭明
[Data structs] ACS.pdf DL : 0: 连续蚁群算法，对大家可能有帮助，希望大家互相学习。-Continuous ant colony algorithm, the U.S. may have to help, I hope everyone learn from each other.
Date : 2026-01-09 Size : 114kb User : 秦军立
[Data structs] Towers DL : 0: 学习数据结构的好例子,提供给大家参考学习.-Learning a good example of data structure to provide a reference to the U.S. study.
Date : 2026-01-09 Size : 9kb User : nn
[Data structs] IntroductionToAlgorithm DL : 0: 算法导论经典中的经典找了很久才找到内容我就不深入介绍了大家自己看吧不需要账号直接下载-Introduction to the classic algorithm to find the classic for a long time to find the contents of my depth not introduce himself and see the U.S. do not need to account direct download
Date : 2026-01-09 Size : 11.44mb User : 帽子
[Data structs] FourArithmetic DL : 0: 精心编写的带有4个基本算法的程序，能加深大家对希尔排序、冒泡排序、快速排序、选择排序的理解，并清楚其中的区别，以及算法优劣性-Carefully prepared with a basic algorithm for four procedures, to strengthen the U.S. on the Hill sort, bubble sort, quick sort, selection sort of understanding, and clearly one of the distinction, as well as the advantages and disadvantages of algorithm
Date : 2026-01-09 Size : 879kb User : 胡庆恩
[Data structs] Haffmancode DL : 0: 课程设计： 1.求出在一个n×n的棋盘上，放置n个不能互相捕捉的国际象棋“皇后”的所有布局。 2.设计一个利用哈夫曼算法的编码和译码系统，重复地显示并处理以下项目，直到选择退出为止。【基本要求】 1) 将权值数据存放在数据文件(文件名为data.txt，位于执行程序的当前目录中) 2) 分别采用动态和静态存储结构 3) 初始化：键盘输入字符集大小n、n个字符和n个权值，建立哈夫曼树； 4) 编码：利用建好的哈夫曼树生成哈夫曼编码； 5) 输出编码； 6) 设字符集及频度如下表：字符空格 A B C D E F G H I J K L M 频度 186 64 13 22 32 103 21 15 47 57 1 5 32 20 字符 N O P Q R S T U V W X Y Z 频度 57 63 15 1 48 51 80 23 8 18 1 16 1 -Curriculum design: 1. Obtained in an n × n chessboard, the place to catch each other should not n个chess "Queen" of all the layout. 2. The design of a use of Huffman coding and decoding algorithms systems, and deal with duplicate to show the following items until the exit date selection. The basic requirements 【】 1) will be the right value data stored in data files (file named data.txt, located in the implementation of procedures in the current directory) 2), respectively, dynamic and static storage structure 3) Initialization: keyboard input character set size of n, n and n characters of the right value, set up Huffman tree 4) Coding: Using the built Huffman tree generated Huffman coding 5) output coding 6) The character set and the frequency of the following table: Space characters A B C D E F G H I J K L M Frequency of 186 64 13 22 32 103 21 15 47 57 1 5 32 20 Character N O P Q R S T U V W X Y Z Frequency 57 63 15 1 48 51 80 23 8 18 1 16 1
Date : 2026-01-09 Size : 538kb User : 赵刚