Implemented BFS, DFS and A* To compile this project, use the following command: g++ -o search main.cpp Then you can run it: ./search The input is loaded from a input file in.txt Here is the format of the input file: The first line of the input file shoud contain two chars indicate the source and destination city for breadth first and depth first algorithm. The second line of input file shoud be an integer m indicate the number of connections for the map. Following m lines describe the map, each line represents to one connection in this form: dist city1 city2, which means there is a connection between city1 and city2 with the distance dist. The following input are for A* The following line contains two chars indicate the source and destination city for A* algorithm. Then there is an integer h indicate the number of heuristic. The following h lines is in the form: city dist which means the straight-line distance from the city to B is dist.
标签： Implemented following compile command
上传时间： 2014-01-01
上传用户：lhc9102
功能说明： 次组件提供的功能有二： 1、解析出数学表达式中的参数 2、计算数学表达式 此组件不但可以解析和计算数学表达式，还提供了一系列的数学基本函数，可应用于表达式中一块处理。提供的数学基本函数有：将值舍入到最接近的整数 rount(x)、向下取整函数 int(x)、向上取整 ceiling(x)、求两个数中最大数的函数 max(a,b)、求两个数中最小数的函数 min(a,b)、幂函数 pow(x,n)、求平方根的函数 sqrt(x)。 下载地址：http://www.qiusuo365.com/qiusuo365/viewthread.php?tid=2422&extra=page％3D1
上传时间： 2014-01-18
上传用户：李梦晗
C# BigInteger class. BigInteger.cs is a csharp program. It is the BIgInteger class. It has methods: abs() , FermatLittleTest(int confidence) ,gcd(BigInteger bi) , genCoPrime(int bits, Random rand) , genPseudoPrime(int bits, int confidence, Random rand) , genRandomBits(int bits, Random rand) , isProbablePrime(int confidence) , isProbablePrime() , Jacobi(BigInteger a, BigInteger b) , LucasSequence(BigInteger P, BigInteger Q, BigInteger k, BigInteger n) ,max(BigInteger bi) , min(BigInteger bi) , modInverse(BigInteger modulus) , RabinMillerTest(int confidence) ,
标签： BigInteger class BIgInteger program
上传时间： 2013-12-23
上传用户：ynzfm
采用 DVB反向链路（DVB-RCS）标准中的双二元Turbo码matlab仿真程序和仿真结果。支持1/2和1/3码率。支持MAP，LOG-MAP解码算法。
上传时间： 2015-11-19
上传用户：xhz1993
基于DSP平台实现turbo码Max_Log_MAP算法，基于标准C 语言研究 了Turbo 码Max - Log - MAP 译码算法的软件编程与实现,为了提高程序的运行效率,结合TMS320C6000 系列DSP 芯片 的结构与特点采用循环展开、数据的存取优化设计、算法改进等措施进行了代码优化,给出了测试结果，非常值得一看
标签： Max_Log_MAP turbo DSP 算法
上传时间： 2014-01-11
上传用户：
％radon transform clear all ％ N=800 n=1:N fs=200 t=n/fs x1=exp(j*2*pi*(5*t+0.5*5*t.^2)) x2=exp(j*2*pi*(5*t+0.5*15*t.^2)) x=x1+x2 ％N=length(x) ％ ambifunb(x ) ％*****************************************RAT naf=ambifunb(x) htl(abs(naf)) ％ [wh,rho,theta]=htl(abs(naf)) colormap([0,0,0]) ％ xlabel( 极半径 ) ％ ylabel( 角度 ) ％**************************************％找出峰值点的坐标,计算初始频率和调频斜率(正确) ％找出峰值点的坐标 b=max(max(wh)) [u,a]=find(wh>=0.8*b)
上传时间： 2014-10-27
上传用户：Yukiseop
The task of clustering Web sessions is to group Web sessions based on similarity and consists of maximizing the intra- group similarity while minimizing the inter-group similarity． The first and foremost question needed to be considered in clustering W b sessions is how to measure the similarity between Web sessions．However．there are many shortcomings in traditiona1 measurements．This paper introduces a new method for measuring similarities between Web pages that takes into account not only the URL but also the viewing time of the visited web page．Yhen we give a new method to measure the similarity of Web sessions using sequence alignment and the similarity of W eb page access in detail Experiments have proved that our method is valid and e币cient．
标签： sessions clustering similarity Web
上传时间： 2014-01-11
上传用户：songrui
一个小的telnet程序与大家共享，/* 本程序支持如一些参数： * --host IP地址 或者 -H IP地址 * --port 端口 或者 -P 端口 * --back 监听数量 或者 -B 监听数量 * --dir 服务默认目录 或者 -D 服务默认目录 * --log 日志存放路径 或者 -L 日志存放路径 * --daemon 使程序进入后台运行模式 */
上传时间： 2016-11-20
上传用户：515414293
这个连接池是直接从JIVE中取出来的，进行了一下修改，使得连接参数直接在程序中设定而不是从属性文件中读取。 [b]用法：[/b] 先设定自己的连接参数，在DbConnectionDefaultPool.java文件的loadProperties方法中。注意你也需要设定连接池的log文件的存放位置。
上传时间： 2016-11-21
上传用户：TF2015
Instead of finding the longest common subsequence, let us try to determine the length of the LCS. Then tracking back to find the LCS. Consider a1a2…am and b1b2…bn. Case 1: am=bn. The LCS must contain am, we have to find the LCS of a1a2…am-1 and b1b2…bn-1. Case 2: am≠bn. Wehave to find the LCS of a1a2…am-1 and b1b2…bn, and a1a2…am and b b b b1b2…bn-1 Let A = a1 a2 … am and B = b1 b2 … bn Let Li j denote the length of the longest i,g g common subsequence of a1 a2 … ai and b1 b2 … bj. Li,j = Li-1,j-1 + 1 if ai=bj max{ L L } a≠b i-1,j, i,j-1 if ai≠j L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
标签： the subsequence determine Instead
上传时间： 2013-12-17
上传用户：evil
