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Description: Computer Networks 4th Edition Problem Solutions.pdf计算机网络第四版的习题解答。是原版，但是有些章节的有几个问题的解答存在错误。本身书上的问题也有错误。-Computer Networks 4th Edition of Computer Network Problem Solutions.pdf fourth edition of the questions to answer. Yes original, but some sections of a number of problems exist in the wrong answer. The book itself is the wrong question.
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Size: 258906 |
Author: 王斐 |
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Description: 这是数据结构与问题求解的Java版的源码-This is the data structure and problem solving version of the Java source code
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Size: 423884 |
Author: 舒海林 |
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Description: 吃苹果eating apple problem问题代码-eat apples problem was eating apple code
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Size: 4715 |
Author: ws |
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Description: dining philosophers problem
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Size: 6144 |
Author: 邓飞 |
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Description: Computer Networks 4th Edition Problem Solutions.pdf计算机网络第四版的习题解答。是原版，但是有些章节的有几个问题的解答存在错误。本身书上的问题也有错误。-Computer Networks 4th Edition of Computer Network Problem Solutions.pdf fourth edition of the questions to answer. Yes original, but some sections of a number of problems exist in the wrong answer. The book itself is the wrong question.
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Size: 259072 |
Author: 王斐 |
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Description: reader-writer-problem读者写者问题代码-reader-writer-problem readers to write code problem
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Size: 8192 |
Author: ws |
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Description: 车辆行驶路线优化（Vehicle Routing Problem）-traffic route optimization (Vehicle Routing Problem)
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Size: 13312 |
Author: 王则 |
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Description: 遗传算法求解01背包问题＋实验报告＋参考文献。如果你看了这个程序还是不能明白遗传算法的巧妙，那么还是不要再看这个问题了。-Genetic Algorithm for Knapsack Problem 01 experimental report references. If you read this procedure should not understand the genetic algorithm or clever, then it is better not to look at this issue.
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Size: 201728 |
Author: gouyabin |
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Description:
Introduction
In this project, we will design a programming solution to the bounded-buffer problem using the producer and consumer processes . The solution uses three semaphores: empty and full, which count the number of empty and full slots in the buffer, and mutex, which is a binary (or mutual exclusive) semaphore that protects the actual insertion or removal of items in the buffer. For this project, standard counting semaphores will be used for empty and full, and, rather than a binary semaphore, a mutex lock will be used to represent mutex. The producer and consumer – running as separate threads – will move items to and from a buffer that is synchronized with these empty, full, and mutex structures. You are required to use the pthread package to solve this problem in this project.
And there can be several producers/consumers as many as you want.
-Introduction In this project, we will design a programming solution to the bounded-buffer problem using the producer and consumer processes. The solution uses three semaphores: empty and full, which count the number of empty and full slots in the buffer, and mutex, which is a binary (or mutual exclusive) semaphore that protects the actual insertion or removal of items in the buffer. For this project, standard counting semaphores will be used for empty and full, and, rather than a binary semaphore, a mutex lock will be used to represent mutex. The producer and consumer- running as separate threads- will move items to and from a buffer that is synchronized with these empty, full, and mutex structures. You are required to use the pthread package to solve this problem in this project.And there can be several producers/consumers as many as you want.
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Size: 96256 |
Author: MidnightD |
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Description: 模拟退火解决0-1背包问题，初学者可以借鉴-Simulated annealing to solve 0-1 knapsack problem, beginners can learn from
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Size: 4096 |
Author: 马贺 |
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Description: 基于粒子群优化算法（PSO）的50个城市TSP问题的求解，可推广至类似NP-hard问题。-Based on Particle Swarm Optimization (PSO) of the 50 cities TSP problem solving can be extended to a similar NP-hard problem.
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Size: 2048 |
Author: 孙岩 |
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Description: 基于遗传算法的50个城市TSP问题的求解，可推广至类似NP-hard问题。-Genetic Algorithm Based on 50 cities TSP problem solving can be extended to a similar NP-hard problem.
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Size: 1024 |
Author: 孙岩 |
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Description: 这是一篇关于多旅行商问题的学术论文，非常有新意，非常有利于我们深入了解mtsp问题的本质-This is a multi-traveling salesman problem on the papers, very fresh, very beneficial to our in-depth understanding of the nature of the problem mtsp
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Size: 49152 |
Author: 王斌 |
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Description: 中国邮递员问题的DNA计算，从学校的数据资源库里找着的，感觉不错，在此向大家支持一下-Chinese postman problem for DNA computing, data resources from the school library found, an excuse, and feel good, look at this opportunity to Members to support the
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Size: 228352 |
Author: 贺波 |
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Description: partition problem to divide a group in to 2 equal sets
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Size: 1117184 |
Author: Abdosebo |
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Description: mechine scheduling problem
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Size: 1024 |
Author: 张震 |
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Description: Data Abstraction and Problem Solving with Java - code1
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Size: 72704 |
Author: Sanch |
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Description: 迷宫问题
以一个m*n的长方阵表示迷宫，0和1分别表示迷宫中的通路和障碍。设计一个程序，对任意设定的迷宫，求出一条从入口到出口的通路，或得出没有通路的结论。
基本要求：
（1）首先实现一个以链表作存储结构的栈类型，然后编写一个求解迷宫的非递归程序。求得的通路以三元组(i,j,d)的形式输出，其中(i,j)指示迷宫中的一个坐标，d表示走到下一坐标的方向。
（2）测试几组数据，数据的规模由小变大，即网格越来越小，障碍越来越复杂。
拓展要求：
实现该问题的可视化界面，用鼠标点击即可一步步走出迷宫。
-Mazing Problem:use a m*n long square to indicate maze,0 or 1 represents maze s road and obstacle,Find out a way from entry to export access by programming
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Size: 44032 |
Author: panbowen |
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Description: 约瑟夫斯（Josephus）问题的一种描述是：编号为1，2，…，n的n个人按顺时针方向围坐一圈，每人持有一个密码（正整数）。一开始任选一个正整数作为报数上限值m，从第一个人开始按顺时针方向自1开始报数，报到m时停止报数。报m的人出列，将他的密码作为新的m值，从他在顺时针方向下一个人开始重新从1报数，如此下去，直至所有的人全部出列为止。试设计一个程序，按出列顺序印出各人编号。-Josephus (Josephus) a description of the problem are: number 1,2, ..., n of n individuals sitting around a circle clockwise, each holding a password (positive integer). Choose a positive integer beginning as a limit on the number of reported m, starting from the first person to start a clockwise direction from a report number, report the number of reported m stop. Who reported m out of line, his password as the new m value, in a clockwise direction from the next person he began to re-reported from a number, it goes on until all the people all of the columns so far. Design a program, according to the column order prints each number.
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Size: 57344 |
Author: 冯柳鑫 |
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Description: 分别通过顺序表和链表的方式，利用C++实现约瑟夫斯问题(C++ implementation of the Josephus problem)
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Size: 1533952 |
Author: 嘉树
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