数学建模97年B题 截断切割问题 源程序
Mathematical Modeling 97 B title truncated mutilation source (2016-07-18, Visual C++, 1KB, 下载1次)
四种路径选择算法，DF,BF,DJ以及A*
Four kinds of path selection algorithm, DF, BF, DJ and A* (2016-05-28, Visual C++, 5547KB, 下载1次)
矩阵链乘： P=a1×a2×a3×……×an，依据乘法结合律，不改变其顺序，只用括号表示成对的乘积，试问有几种括号化的方案？(h(n)种)
出栈次序问题。
Matrix chain multiplication: P=a1* A2* A3*......* an, according to the multiplication with law, does not change the order, the brackets only representation of the pairwise product, ask several brackets scheme? (H (n) species)
出栈次序问题。
Sequence problem.
(2015-09-16, Visual C++, 18KB, 下载1次)
对初学者很实用的BF算法，值得一看。希望能帮助到大家
Very useful for beginners BF algorithm, worth a visit. We are hoping to help everyone (2015-06-03, Visual C++, 553KB, 下载2次)
矩阵类模板Matrix<Type>是专门为线性代数中矩阵而设计的一个模板类，具体的函数声明与定义可以参见“matrix.h。见测试文件 matrix_test.cpp，功能如下：
Matrix<Type> m 创建一个空矩阵
Matrix<Type> m2(m1) 创建矩阵 m2的拷贝 m1
Matrix<Type> m(r,c,x) 创建常数矩阵
Matrix<Type> m(r,c,a) 通过数组创建矩阵
m.~Matrix<Type> () 销毁矩阵并释放空间
矩阵运算：
A += x 矩阵自身加常数
A -= x 矩阵自身减常数
A *= x 矩阵自身乘常数
A /= x 矩阵自身除以常数
A1 += A2 矩阵自身加矩阵
A1 -= A2 矩阵自身减矩阵
A1 *= A2 矩阵自身乘矩阵(逐元素)
A1 /= A2 矩阵自身除以矩阵(逐元素)
x + A 常数与矩阵之和
A + x 矩阵与常数之和
A1 + A2 矩阵与矩阵之和
A - x 矩阵与常数之差
x - A 常数与矩阵之差
A1 - A2 矩阵与矩阵之差
Matrix Matrix class template <Type> Is a template class specifically for linear algebra and matrix design, the specific function declarations and definitions can be found in " matrix.h see test file matrix_test.cpp, functions as follows:. Matrix <Type> Create an empty matrix Matrix m <Type> m2 (m1) to create a copy of the matrix m2 m1 Matrix <Type> m (r, c, x) to create a constant matrix Matrix <Type> m (r, c, a) create a matrix m through the array. ~ Matrix <Type> () Destroy the matrix and release space matrix calculation: A+ = x matrix itself plus the constant A- = x matrix itself less constant A* = x matrix itself by a constant A/= x matrix itself is divided by the constant A1+ = A2 matrix itself plus Matrix A1- = A2 matrix itself reduced matrix A1* = A2 matrix multiplication matrix itself (by elements) A1/= A2 matrix itself divided by the matrix (by elements) x+ A matrix of constants and A+ x matrix with constants and A1+ A2 matrix and matrix (2014-04-24, Visual C++, 4KB, 下载5次)
这是一段双边滤波的快速实现算法，利用了下采用上采用的算法
fast fast BF (2014-03-10, Visual C++, 3059KB, 下载7次)
给定含有n 个元素的多重集合S = { a1，a2 ，……，an} ，1≤ai≤n，1≤i≤n。
每个元素在S中出现的次数称为该元素的重数
多重集S中重数大于n/2 的元素称为主元素
设计一个O(n)时间算法，计算S 的主元素
Given the multiple of n elements containing the set S = {a1, a2, ......, an}, 1 ≤ ai ≤ n, 1 ≤ i ≤ n. Each element in S is called the number of occurrences of the element multiplicity of heavy multi-set S is greater than n/2 elements is called the primary element design an O (n) time algorithm to calculate the main elements of S (2013-09-07, Visual C++, 1KB, 下载1次)
矩阵（二）
编写C++程序完成以下功能：
（1）假定矩阵大小为4×5（整型）；
（2）矩阵空间采用new动态申请，保存在指针中；
（3）定义矩阵初始化函数，可以从cin中输入矩阵元素；
（4）定义矩阵输出函数，将矩阵格式化输出到cout；
（5）定义矩阵相加的函数，实现两个矩阵相加的功能,结果保存在另一个矩阵中；
（6）定义矩阵相减的函数，实现两个矩阵相减的功能,结果保存在另一个矩阵中；
（7）动态申请三个矩阵：A1、A2、A3；
（8）初始化A1、A2；
（9）计算并输出A3 = A1加A2，A3 = A1减A2；
（10）释放矩阵空间。
Matrix (two) write C++ program performs the following functions: (1) assuming matrix size of 45 (integer) (2) matrix spaces using new dynamic applications, stored in the pointer (3) define the matrix initialization function, you can Enter the matrix elements from cin (4) defines the output function matrix, the matrix format the output to cout (5) defines a function of matrix addition, to achieve the function of two matrices and stores the result in another matrix (6) Define the matrix subtraction function, the realization of the function of two matrices subtraction result is stored in another matrix (7) for the three matrices Dynamic: A1, A2, A3 (8) initialization A1, A2 (9) calculate and output A3 = A1 plus A2, A3 = A1 minus A2 (10) release matrix space. (2013-09-06, Visual C++, 1KB, 下载4次)
设V1=<{a1, a2}, max >和V2=<{b1, b2}, min >是两个数系统, 其中max (x, y)表示x 与y中较大的数；min (x, y)表示 x 与y 中较小的数。max 和min 可以看作二元运算。输入a1,a2，b1, b2 为整数, 考虑积代数V1 × V2.
（1）设积代数中的二元运算为*运算，输出它的运算表
（2）输出积代数的单位元和零元
Let V1 = < {a1, a2}, max> and V2 = < {b1, b2}, min> is the two number system, where max (x, y) represents the x and y, the larger number min (x , y) represents the x and y in the smaller number. max and min can be seen as a binary operator. Enter a1, a2, b1, b2 is an integer, consider the product algebra V1 V2. (1) Let plot algebra binary operation is* operation, the output of its operation table (2) units of output product algebras and $ nil (2013-06-19, Visual C++, 45KB, 下载1次)
假设所有新修建的公路都是直接连接两个村庄，除村庄以外没有交汇点，并且所有公路均为双向通行，不存在单行道。请你设计一个算法，解决以下问题：
1. 给出一个造价最低的修建方案（若存在）；
2. 判断造价最低的方案是否唯一。如果唯一，给出一个除该方案外造价最低的备选方案（若存在）；否则，给出另一个造价最低的备选方案。
Assuming that all new construction of roads are directly connecting two villages, in addition to the meeting point outside the village, and all roads are two-way traffic, there is no one-way street. You design an algorithm to solve the following problems: 1. Give a minimum cost of construction of the program (if it exists) 2. Judgment lowest cost program is unique. If only given an addition to the program outside the lowest cost option (if it exists) otherwise, given another lowest cost options. (2013-03-30, Visual C++, 556KB, 下载7次)
符号平衡问题
给定n（1<=n<=3,000,000）个整数a1,a2,…，an组成的序列，试设计一个O(n) 时间算法，计算其最长符号平衡段的长度
Symbol balance problems given n (1 < = n < = 3,000,000) integers a1, a2, ..., aN sequence of trial design of an O (n) time algorithm to calculate the equilibrium stage the length of the longest symbol (2012-11-21, Visual C++, 62KB, 下载10次)
多项式计算（单向链表的应用）
设计要求：设计处理n次一元多项式P(x) = a0 + a1x + a2x2 + … + anxn,多项式实际上包含n+1项，由n+1个系数唯一确定。为了节省存储空间，只存储多项式中系数非0 的项。运用链表中的每一个结点存放多项式的一个系数非0项，它包含三个域，分别存放该项的系数、指数以及指向下一项结点的指针，来对其进行相应的处理（某一多项式根据x求值、多个同元多项式处理等等）。
Polynomial calculations (one-way linked list application) design requirements: designed to handle n I. polynomial P (x) = a0+ a1x+ a2x2+ ...+ anxn, polynomials actually contains n+1 entries, the n+1 coefficients uniquely determined. To save storage space, storing only the coefficients of polynomials in non-0 entry. The use of the list in the store each node of a polynomial coefficient non 0, it contains three domains, namely the coefficient store, index and point to a pointer to the next node to its corresponding processing (x evaluated according to a polynomial, a number of the same polynomial processing, etc.). (2011-05-30, Visual C++, 8KB, 下载5次)
用BF、KMP、BM算法实现字符串的匹配
BF, KMP, BM algorithm for string matching (2011-05-14, Visual C++, 5KB, 下载3次)
能帮你算出给的N次的X多项式之和，并输入相应的系数a0,a1,a2....an。式子形为：a0+a1*x+a2*x1+a3*x2+...+a(n+1)*xn=sum
Can help you figure out X to the N times and polynomials, and enter the corresponding coefficients a0, a1, a2 .... an. Formula of the form: a0+ a1* x+ a2* x1+ a3* x2+...+ a (n+1)* xn = sum (2010-07-22, Visual C++, 131KB, 下载2次)
词典顺序的排列生成 实现计算机程序设计艺术中的算法L 给定n个元素已按升序排好，算法生成{a1,a2,a3,...,an}的所有排列，并以词典顺序访问他们
The dictionary is arranged in order generated in the art of realizing computer programming algorithm is given n L elements and algorithm by ascending generation {a1, a2, a3,... , an arrangement, and all} in order to visit them. The dictionary (2010-04-26, Visual C++, 444KB, 下载5次)
KMP 算法是由Knuth，Morris和Pratt等人共同提出的，所以成为Knuth－Morris－Pratt算法，简称KMP算法。KMP算法是字符串模式匹配中的经典算法。和BF算法相比，KMP算法的不同点是匹配过程中，主串的位置指针不会回溯，这样的结果使得算法时间复杂度只为O（n＋m）。
采用VC++开发，实现KMP字符串匹配算法
KMP algorithm by Knuth, Morris and Pratt, who co-sponsored, so as Knuth-Morris-Pratt algorithm, called KMP algorithm. KMP algorithm is a classical string pattern matching algorithm. And the BF algorithm is compared, KMP algorithm is different from the matching process, the location of the main string pointer will not go back, this result makes the algorithm time complexity only O (n+ m). Using VC++ development, implementation KMP string matching algorithm (2010-04-15, Visual C++, 596KB, 下载15次)
问题描述：
给定n个石子，其重量分别a1,a2,a3...an,要求将其划分成m份，每一份的划分费 义为这份石了中最大重量与最小重量的差的平方。总划分费用等丁m份划分费用之和。
编程任务
对于给定的n个石子，求一种划分方案，使得总划分费用最小。
Description of the problem: Given n-stones, its weight, respectively a1, a2, a3 ... an, asked that it be divided into m copies, each of the division of costs defined as the weight of the stone of the largest and the smallest weight difference the square. Small m the total costs were divided into the sum of the cost. Programming tasks for a given n-stones, seek a kind of division, making the total cost of the smallest division. (2009-12-13, Visual C++, 739KB, 下载4次)
给定n个矩阵构成的一个链<A1,A2,…,An>其中i=1，2，…，n，矩阵Ai的维数为pi-1pi,对乘积A1A2…An以一种最小化标量乘法次数的方式进行加全部括号。该问题应用动态规划算法
N matrix of a given composition of a chain <A1,A2,…,An> Which i = 1,2, ..., n, the dimension of the matrix Ai is pi-1pi, on the product A1A2 ... An order to minimize a scalar multiplication method increases the number of all brackets. The problem by using dynamic programming algorithm (2009-04-09, Visual C++, 3KB, 下载29次)
给定n个矩阵{A1,A2,…,An}，其中Ai与Ai+1是可乘的，i=1,2,…,n-1。考察这n个矩阵的连乘积A1A2…An。由于矩阵乘法满足结合律，故计算矩阵的连乘积可以有许多不同的计算次序，这种计算次序可以用加括号的方式来确定。若一个矩阵连乘积的计算次序完全确定，则可以依此次序反复调用2个矩阵相乘的标准算法（有改进的方法，这里不考虑）计算出矩阵连乘积。若A是一个p×q矩阵，B是一个q×r矩阵，则计算其乘积C=AB的标准算法中，需要进行pqr次数乘。
Given n matrices (A1, A2, ..., An), which Ai and Ai+ 1 is the multiplicative, i = 1,2, ..., n-1. Visit this link n-matrix product A1A2 ... An. As a result of matrix multiplication to meet the combination of law, it (2008-05-10, Visual C++, 1KB, 下载6次)
问题描述
给定n个矩阵A1，A2，…，An，其中，Ai与Aj+1是可乘的，i=1，2，…，n-1。
你的任务是要确定矩阵连乘的运算次序，使计算这n个矩阵的连乘积A1A2…An时总的元素乘法次数达到最少。
例如：3个矩阵A1，A2，A3，阶分别为10×100、100×5、5×50，计算连乘积A1A2A3时按（A1A2）A3所需的元素乘法次数达到最少，为7500次。
Description of the problem given n matrices A1, A2, ..., An, which, Ai and Aj+ 1 is the multiplicative, i = 1,2, ..., n-1. Your mission is to identify the matrix even by computing the order, so that calculation of this n-matrix product A1A2 ... An even when the total number of elements to achieve a minimum of multiplication. For example: three matrices A1, A2, A3, bands were 10 × 100,100 × 5,5 × 50, the calculation even when the product A1A2A3 by (A1A2) A3 multiply the number of elements required to achieve at least for the 7500 meeting. (2007-10-19, Visual C++, 6KB, 下载15次)