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Description: 一个很好的cost算法对文件夹、文档、图片、视频等的加解密可以有效的保证文件的安全-cost of a good algorithm for folders, documents, pictures, video, the encryption and decryption can effectively guarantee the safety document
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Size: 2048 |
Author: JAKE |
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Description: 模仿cisco路由器，产生netflow，送到指定的收集器上，以相对低廉的软件代价来实现昂贵的路由器的功能 fprobe -i eth0 -l 3 host:port-imitate, have netflow sent to a designated collection device, a relatively low cost of the software to achieve expensive routers function fprobe-i eth0-l three host : port
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Size: 122880 |
Author: yatagan |
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Description: prolog 找路例子程序：
=== === ===
Part 1-Adding connections
Part 2-Simple Path
example
| ?- path1(a,b,P,T).
will produce the response:
T = 15
P = [a,b] ?
Part 3 - Non-repeating path
As an example, the query:
?- path2(a,h,P,T).
will succeed and may produce the bindings:
P = [a,depot,b,d,e,f,h]
T = 155
Part 4 - Generating a path below a cost threshold
As an example, the query:
?- path_below_cost(a,[a,b,c,d,e,f,g,h],RS,300).
returns:
RS = [a,b,depot,c,d,e,g,f,h] ?
RS = [a,c,depot,b,d,e,g,f,h] ?
no
==================================
-prolog to find a way examples : ==================================== Part 1-Adding connections Part 2- Simple Path example |- path1 (a, b, P, T). will produce the response : T = 15, P = [a, b] Part 3-Non-repeating path As an example, the query :-path2 (a, h, P, T). will succeed and may produce the bindings : P = [a, depot, b, d, e, f, h] T = 155 Part 4-Generating a path below a cost threshold As an example, the query :-path_below_cost (a, [a, b, c, d, e, f, g, h], RS, 300). returns : RS = [a, b, depot, c, d, e , g, f, h] RS = [a, c, depot, b, d, e, g, f, h] no =====================
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Size: 2048 |
Author: Fa |
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Description: 本文用Ｃ语言实现了优化算法－最小费用最大流算法，主要用于图论领域-Using C programming language, the optimization algorithm-minimum cost maximum flow algorithm, mainly for the field of graph theory
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Size: 3072 |
Author: 申辉 |
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Description: 最大最小流程序。包括最大流网络，最小费用最大流。算法的一些实现。-Max-min flow procedures. Including the maximum flow network, minimum cost maximum flow. Some of realize algorithm.
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Size: 110592 |
Author: HuiCong |
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Description: 最小费用流的最小费用路算法SuccessiveShortest-Minimum cost flow of minimum cost path algorithm SuccessiveShortest
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Size: 2048 |
Author: 叶博 |
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Description: 网络优化算法：最小费用流的消圈算法cyclecanceling-Network Optimization Algorithms: Minimum Cost Flow Algorithm circle of extinction cyclecanceling
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Size: 2048 |
Author: 叶博 |
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Description: For solving the following problem:
"There is No Free Lunch"
Time Limit: 1 Second Memory Limit: 32768 KB
One day, CYJJ found an interesting piece of commercial from newspaper: the Cyber-restaurant was offering a kind of "Lunch Special" which was said that one could "buy one get two for free". That is, if you buy one of the dishes on their menu, denoted by di with price pi , you may get the two neighboring dishes di-1 and di+1 for free! If you pick up d1, then you may get d2 and the last one dn for free, and if you choose the last one dn, you may get dn-1 and d1 for free.
However, after investigation CYJJ realized that there was no free lunch at all. The price pi of the i-th dish was actually calculated by adding up twice the cost ci of the dish and half of the costs of the two "free" dishes. Now given all the prices on the menu, you are asked to help CYJJ find the cost of each of the dishes.
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Size: 1024 |
Author: anthony chan |
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Description: 本人参加ACM竞赛使用的一些算法模板，包括二分图匹配，欧拉回路的构造以及网络流中的最大流与最小费用最大流等，可以说实战性非常强。-ACM competitions I take part in a number of algorithms used in templates, including two sub-graph matching, Euler circuit, as well as network flow structure of the maximum flow and minimum cost maximum flow, etc., it can be said of very strong combat.
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Size: 14336 |
Author: 诗苇 |
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Description: a cost based algortihm for travelling salesman problem
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Size: 25600 |
Author: sadi |
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Description: Least cost method ---transportation model implementation in
c-Least cost method---transportation model implementation in
c++
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Size: 1024 |
Author: Akshay |
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Description: 问题描述：给定n个石子，其重量为a1,a2...,an,要求将其划分为m堆，每一份划分的费用定义为这堆石头中最大重量与最小重量的差的平方。总划分费用等于各堆费用之和。
输入：n m 及a1,a2...,an
输出：sum
问题描述：给定n个石子，其重量为a1,a2...,an,要求将其划分为m堆，每一份划分的费用定义为这堆石头中最大重量与最小重量的差的平方。总划分费用等于各堆费用之和。
输入：n m 及a1,a2...,an
输出：sum
-Description of the problem: Given n-stones, the weight of a1, a2 ..., an, requests that it be divided into m heap, the cost of each division is defined as the weight of this pile of stone, the largest and the smallest weight of the square of the difference . Divide the total cost is equal to the cost of each stack and the. Input: nm and a1, a2 ..., an output: sum description of the problem: Given n-stones, the weight of a1, a2 ..., an, requests that it be divided into m piles, each division of the cost of is defined as the weight of this pile of stone, the largest and the smallest weight of the square of the difference. Divide the total cost is equal to the cost of each stack and the. Input: nm and a1, a2 ..., an output: sum
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Size: 15360 |
Author: Adler.C |
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Description: 最小费用最大流
C语言编程，运筹学
C语言最小费用最大流-Minimum Cost Flow C language programming, operations research
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Size: 1024 |
Author: aoxue |
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Description: 线性分配算法的C++版本,在MATLAB调用mex命令编译后可以直接在MATLAB中运行,函数assianstment=lap(cost),cost为匹配的代价矩阵,assianstment为输出匹配矩阵-Linear assignment algorithm C++ version of the call in the MATLAB mex command can be compiled to run directly in the MATLAB function assianstment = lap (cost), cost for the match cost matrix, assianstment matrix for the output matching
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Size: 5120 |
Author: 123 |
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Description: 低成本AVR+LCD示波器，介绍用avr+LCD开发示波器-Low-cost AVR+ LCD oscilloscope
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Size: 25995264 |
Author: SHIGANG |
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Description: 给定一个N*N 的方形网格，设其左上角为起点◎，坐标为（1，1），X轴向右为正，Y轴向下为正，每个方格边长为1。一辆汽车从起点◎出发驶向右下角终点▲，其坐标为（N，N）。在若干个网格交叉点处，设置了油库，可供汽车在行驶途中加油。汽车在行驶过程中应遵守如下规则：
（1）汽车只能沿网格边行驶，装满油后能行驶K 条网格边。出发时汽车已装满油，在起点与终点处不设油库。
（2）当汽车行驶经过一条网格边时，若其X 坐标或Y 坐标减小，则应付费用B，否则免付费用。
（3）汽车在行驶过程中遇油库则应加满油并付加油费用A。
（4）在需要时可在网格点处增设油库，并付增设油库费用C（不含加油费用A）。
（5）(1)～(4)中的各数N、K、A、B、C均为正整数。
编程任务：
求汽车从起点出发到达终点的一条所付费用最少的行驶路线。
数据输入：
第一行是N，K，A，B，C的值，2 <=N<=100，2<=K<=10。
第二行起是一个N*N 的0-1方阵，每行N 个值，至N+1行结束。方阵的第i
行第j 列处的值为1 表示在网格交叉点（i，j）处设置了一个油库，为0 时表示未设油库。各行相邻的2 个数以空格分隔。
结果输出:
第1 行中的数是最小费用值。
-Given an N* N square grid, set the upper left corner as a starting point ◎, coordinates (1,1), X-axis to the right is positive, Y axis, is positive, each square side length of 1. Starting a car approached from the lower right corner of the end point ◎ ▲, its coordinates (N, N). In a number of grid intersection points, set the oil tanks, fuel for vehicles in a moving way. Vehicle in motion the process should follow the following rules:
(1) car can only travel along the grid edges, filled with oil can travel after the K edge of the grid. Starting when the car has been filled with oil, not in the starting and ending at the depot is located.
(2) when the vehicle is traveling through a mesh edge, if the X coordinate or Y coordinate decreases, then the fees payable B, or toll-free use.
(3) the process of moving car in case of fuel oil tanks should fill up and pay the cost of A.
(4) when needed at additional grid point depot, depot and pay the additional cost of C (excluding fuel costs A)
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Size: 1024 |
Author: 张开 |
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Description: 生成最小生成树，利用克鲁斯卡尔和普莱姆算法实现图的最小生成树，程序用的是C++编程-to create a minimun-cost spanning tree
by Kruscal and Prim
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Size: 24576 |
Author: 齐枫 |
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Description: 2.4G 产品应用比较广泛，有些芯片性能也很不错，但价位都比较偏高，很难进入量产的产品。为降低成本JF24D模块采用裸片绑定，虽然性能指标略低于具有代表性的 nRF2401 CC2500 A7105但它的价格要比它们低很多，完全可以满足一般需要双向数据传输及双向遥控的短距离产品应用。
-2.4 G products are widely used, some chip performance is also very good, but prices are high, it is difficult to enter mass production of the product. JF24D module adopts the bare chip binding to reduce cost, although the performance index is slightly lower than the typical nRF2401 CC2500 A7105 but its price is much lower than they are, can meet the needs of the general short product application of two-way data transmission and remote
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Size: 71680 |
Author: 陈治升 |
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Description: A wireless communication project by nRF24L01 low cost 2.4Ghz wireless chip.
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Size: 123904 |
Author: mehram |
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Description: 在二分检索树中，为了把动态规划应用于得到一颗最优二分检索树的问题，需要把构造这样的一个数看成是一系列决策的结果，而且要能列出求取最优决策序列的递推式，可以帮助你了解最优成本二叉树的实现-In a binary search tree, dynamic programming is applied in order to get the best question a binary search tree, we need to construct a number such as a result of a series of decisions, but also to be able to strike the optimal decision sequence lists The recursive formula that can help you understand the optimal cost to achieve the minimum cost of a binary tree binary tree
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Size: 55296 |
Author: wanyang |
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