搜索资源列表
Tug og war.rar
- Tug of War A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of people on the two teams must not differ by more than
jzclass
- 改进的矩阵操作类,增加了诸如对矩阵任意行/列的存取,矩阵的积分图像的计算等-improved matrix operations, such as the right to increase the matrix arbitrary line / column access, image matrix of the integral calculation
1-wire_search
- 这是关于DALLAS的一线的ROMID搜索程序,很有用,可以找到ROMID,-This is on the front line of Dallas ROMID search procedures, useful and can be found ROMID.
分支限界算法求解TSP问题
- 本程序用分支界限方法解决ＴＳＰ问题，即旅行商问题．-the procedures used branch line solution to TSP, the traveling salesman problem.
货郎担分枝限界图形演示
- 货郎担分枝限界图形演示 问题描述:某售货员要到若干城市去推销商品，已知各城市之间的路程（或旅费）。他要选定一条从驻地出发，经过每个城市一遍，最后回到驻地的路线，使总的路程（或旅费）最小。-traveling salesman Branch and Bound graphic demonstration Problem descr iption : A salesman to a number of cities to sell commodities, the known distance
r8q
- 八皇后问题的C版本的解决方案 包括如何处理两个皇后不能在同一列、同一行以及同以斜行的判断问题-eight of the Queen's version of the C solutions, including how to deal with two Queen's not in the same series, and with the same line oblique to the question of judgment
magic_1
- 魔方的算法,即每行每列和对角线上的数值和相等,算法值得研究研究-Magic Square algorithm, which means that every line and every out on the diagonal and the same numerical algorithm worth studies
文档编辑器
- 实现对输入文件的操作:保存文档 , 回到起始行 ,改变当前段的部分内容 , 删除当前, 查找字符串,段跳转, 帮助, 字符数 , 跳至下一行 , 跳至上一行 ,结束任务 ,读文档,改变当前段的全部内容, 阅览文档,写文档, 改为小写 等多种功能-realization of the right to operate the importation documents : to preserve documents, to return to the starting line, change t
chejiandiaodu
- 有m台不同的机器，n个不同的工件。每个工件有多道工序，每道工序由指定的机器在固定的时间内完成。一道工序一旦开始处理，就不能中断。每台机器一次只能处理一道工序。一个调度就是决定每台机器上工序的处理顺序，使得机器完成所有工件的时间最短。具体的，该问题就是要求在满足（1）、（2）两个约束条件的前提下，确定每台机器上工序的顺序，使加工的时间跨度（从开始加工到全部工件都加工完所需要的时间）达到最小。其中，（1）表示工件约束条件：对每个工件而言，机器对它的加工路线是事先确定的；（2）表示机器约束条件：对每台
dul_list
- 双向链表的vc++实现 是在命令行下的双向链表-bidirectional Chain vc realization of the command line of two-way linked list
nearpiont
- 最接近点对问题是求二维坐标中的点对问题，该算法是为了将平面上点集S线性分割为大小大致相等的2个子集S1和S2，我们选取一垂直线l:x=m来作为分割直线。其中m为S中各点x坐标的中位数。由此将S分割为S1={p∈S|px≤m}和S2={p∈S|px>m}。从而使S1和S2分别位于直线l的左侧和右侧，且S=S1∪S2 。由于m是S中各点x坐标值的中位数，因此S1和S2中的点数大致相等。 递归地在S1和S2上解最接近点对问题，我们分别得到S1和S2中的最小距离δ1和δ2。现设δ=min(δ
accept
- 数字运算，判断一个数是否接近素数 A Niven number is a number such that the sum of its digits divides itself. For example, 111 is a Niven number because the sum of its digits is 3, which divides 111. We can also specify a number in another base b, and a number in
acjanchan
- 2^x mod n = 1 acm竞赛题 Give a number n, find the minimum x that satisfies 2^x mod n = 1. Input One positive integer on each line, the value of n. Output If the minimum x exists, print a line with 2^x mod n = 1. Print 2^?
fjdjf
- 用分支界限法解决的几个问题:包括0-1背包问题,最大团问题,电路布线问题,最大装载问题.作业最优处理问韪.-branch line with the law to solve several problems : including 0-1 knapsack problem, the largest group, circuit wiring problem, the greatest problem loading. Optimal handling operations Q premises
kchengji1
- k乘积问题的分支界限法实现, 并且给出了划分结果 ,很好理解-k branch of the product line Method, and the division is a result, a good understanding
AVDOS
- 演示了AVL的删除与插入算法，通过模拟dos命令行实现。-demonstration of the AVL delete and insert algorithm simulation dos command line to achieve.
assgin02
- 本程序解决下列问题：利用动态规划算法实现装配线调度问题。 要求测试数据以文本文件的形式存储， 即所有的数据由文本文件读入。-This program addresses the following issues: the use of dynamic programming algorithm assembly line scheduling problem. requires test data to a text file is st
THE-LENGTH-OF-LINE
- 关于枚举中求绳子长度的算法，希望对学习算法的同志有用-the algorithm about the length of line
line-is-interact
- 算法导论里面的关于线段是否相交以及点是否在多边形内的判断的源代码，另包含一个说明文档。-Introduction to Algorithms inside as well as on line intersects the point is within the polygon to determine the source code, and the other contains a descr iption of the document.
Max-Points-on-a-Line
- 给定的n个点在二维平面上,发现的最大数量点躺在同一直线。-Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.