Unix网络编程 第十二章 IPv4 and IPv6 Interoperability

# Ethernet header contains a type fileld of 0x0800, which identifies the frame as an IPv4 frame.

# 若支持dual-stack的server既有IPv4又有IPv6。则IP层让server透明地既可处理IPv4又可处理IPv6.该server需要绑定到wildcard且未设置IPV6_V6ONLY socket option.

# UNP Page359 Figure 12.5 Summary of interoperability between IPv4 and IPv6 clients and servers.

# 尽可能用IPv6,  since an IPv6 client can still communicate with IPv4 servers, and vice versa.

Unix网络编程 第11章 Name and Address Conversions 笔记

# gethostbyname 和 gethostbyaddr 用来在 IPv4 地址和 hostname 之间转换. getservbyport 和 getservbyname 则是与服务相关。gethostbyname出错时不设errno而是设h_errno,并有hstrerror()函数。

# FQDN的全称是: Fully Qualified domain name. 技术上说必须以点号(period)终止.

# AAAA 被称为 "quad A" rcord, 给出了从hostname到Ipv6地址的映射。 PTR用来把IP地址到hostname.

# Entries in the DNS are known as Resource Records(RRs).

# 一个点分十进制(dotted-decimal)IPv4的地址前加 0::ffff:就是 IPv6的字符串形式。

# 与getpeername对应的函数不是gethostname而是getsockname.

# getaddrinfo函数的host参数指定为dotted-decimal IPv4或 IPv6 hex string,会使得只有IPv4或IPv6的addrinfo返回。

# 不给UDP套接字设置SO_REUSEADDR选项。We do not set the SO_REUSEADDR socket option for the UDP socket because this socket option can allow multiple sockets to bind the same UDP port on hosts that support multicasting. Since there is nothing like TCP's TIME_WAIT state for a UDP socket, there is no need to set this socket option when the server is started.

# 一般情况下,同端口的不同协议对应同样的服务。但也有例外。对于端口514,which is the rsh service with TCP, but the syslog service with UDP.

# gethostbyaddr的第一个参数是char* addr,而其实它并非指向一个char* 事实上指向in_addr结构体。

# getaddrinfo好复杂呀!hint的ai_flags设置了AI_CONONNAME成员得到host的canonical name.

# port 53 是domain service的端口号.

# 如果设置了IPV6_V6ONLY.那么一个来自ipv4 client的连接会被拒绝。

# POSIX says that specifying AF_UNSPEC will return addresses that can be used with any protocol family that can be used with the hostname and service name.

# POSIX specification also implies that if the AI_PASSIVE flag is specified without a hostname, then the IPv6 wildcard address(IN6ADDR_ANY_INIT or 0::0) should be returned as a sockaddr_in6 structure, along with the IPv4 wildcard address(INADDR_ANY or 0.0.0.0), which is returned as a sockaddr_in structure.

# An ipv6 server socket can handle both ipv4 and ipv6 on a dual-stack host. Refer to page319 in UNP for details.

Prof. Zixiang Xiong讲座 脏纸编码

作为主要兴趣在Linux方面的偏应用硕士,听了一个电信传输方面的学术讲座。所谓Dirty Paper Coding,我的理解是如果已知信道干扰,在发射机对发射信号预先减去该干扰,则发射信号经信道传输被接收机接收后,则信道干扰被自动抵消。很简单的Idea,但实际实施起来应该非常困难吧。Prof. Zixiang Xiong先介绍了他所工作的学校。呵。然后从Dirty-paper Coding的历史发展讲起。Instances of dirty paper coding include Costa precoding(1983) ,Tomlinson-Harashima precoding (1971) and the vector perturbation technique of Hochwald et al. (2005)。两位科学家GelfandPinsker开创了这个领域,到信息理论的实际设计准则如何应用到实践之中。Prof. Zixiang Xiong提出了两种有效的编码设计方法。理论回到实路,再介绍这两种编码设计方法在图形图象的数据隐藏,MIMO广播,抗噪预编码,无线网络传输协调方面的实际应用。Gaussian arbitrarily varying channel with a known interference signal at the encoder.

好复杂好深奥的数学变换。想起本科阶段的傅丽叶级数变换,卷积分就已经让我伤透脑筋。本次讲座的收获也就是让我一定程度上开拓了视野,接触到N多电信方面的术语。小波变换,脏纸编码,数字水印之类。术业有专功,我要专功我的Linux去嘞。。。

Sicily 1031 Campus 解题报告.

 

1 原题中文大意;
 
Sicily 1031 Campus实际上就是求一个图中给出两点的最短路径的问题。
 
2 算法思想及解题用到的主要数据结构;
 
用 dijkstra 算法求给定两点的最短路径.
 
按路径长度递增次序产生最短路径算法:   把V分成两组:   (1)S:已求出最短路径的顶点的集合   (2)V-S=T:尚未确定最短路径的顶点集合   将T中顶点按最短路径递增的次序加入到S中,   保证:(1)从源点V0到S中各顶点的最短路径长度都不大于   从V0到T中任何顶点的最短路径长度   (2)每个顶点对应一个距离值   S中顶点:从V0到此顶点的最短路径长度   T中顶点:从V0到此顶点的只包括S中顶点作中间   顶点的最短路径长度   依据:可以证明V0到T中顶点Vk的最短路径,或是从V0到Vk的   直接路径的权值;或是从V0经S中顶点到Vk的路径权值之和。
 
3  逐步求精算法描述
 
用邻接矩阵存放图.1. 初使时令 S={V0},T={其余顶点},T中顶点对应的距离值   若存在<V0,Vi>,d(V0,Vi)为<V0,Vi>弧上的权值   若不存在<V0,Vi>,d(V0,Vi)为∝   2. 从T中选取一个其距离值为最小的顶点W且不在S中,加入S   3. 对T中顶点的距离值进行修改:若加进W作中间顶点,从V0到Vi的   距离值比不加W的路径要短,则修改此距离值   重复上述步骤2、3,直到S中包含所有顶点,即S=T为止
 
3 程序注释清单 // 这份代码通不过sicily。有case错误。我实在找不到原因。
 
#include <map>

#include <vector>

#include <string>

#include <iostream>

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <cassert>



#define INFINITE 0x0FFFFFFF



// 这个类的函数几乎不做参数有效性的检查. 

class graph_t

{

public:

// 构建一个 num x num 的邻接矩阵

graph_t(int num)

{

m_matrix = new int[num*num];

m_sunit = new bool[num];

m_spath_table = new int[num];

// memset(m_matrix, INFINITE, num*num);  // INFINITE 表示无穷远. 

for ( int i = 0; i < num*num; i++ )

{

m_matrix[i] = INFINITE;

}

m_num = num;

}



// row,column 取 [0, m_num)

int& operator()(int row, int column)

{

return m_matrix[row * m_num + column];

}



// vtx_start, vtx_end 取 [vtx_start, vtx_end)

int shortestpath_dij(int vtx_start, int vtx_end)

{

init_dij(vtx_start);  // 初始化dijkstra算法需要的一些数据. 



if ( vtx_end == vtx_start )

{

return this->operator()(vtx_start, vtx_end);

}



// 还剩 m_num - 1 个点. 

for ( int i = 1; i < m_num; i++ )

{

// 找下一个最近的点. 

int vtx = -1, min = INFINITE;

for ( int j = 0; j < m_num; j++ )

{

if ( m_sunit[j] )  // 这个点已经在已经求得最短路径的点的集合中了. 

continue;



if ( m_spath_table[j] < min )

{

vtx = j;

min = m_spath_table[j];

}

}

// 已经没有连通的顶点了. 

if ( vtx == -1 )

{

return -1;

}

m_sunit[vtx] = true;  // 把这个点加入集合中. 



// 这个点是终点.  http://blog.ykyi.net

if ( vtx == vtx_end )

{

return min;

}



// 因为找到了一个新的点加入了集合,更新最短路径表. 

for ( int k = 0; k < m_num; k++ )

{

if ( m_sunit[k] )

continue;



if (  min + this->operator()(vtx, k) < m_spath_table[k] )

{

m_spath_table[k] = min + this->operator()(vtx, k);

}

}

}



assert(0); // it's not suppossed to reach here.

return -1;

}



~graph_t()

{

delete m_matrix;

delete m_sunit;

delete m_spath_table;

}



private:

void init_dij(int vtx_start)

{

memset(m_sunit, 0, m_num);   // 初始化为没有点加入这个已求得最短路径的终点集合.

for ( int i = 0; i < m_num; i++ )

{

m_spath_table[i] = this->operator()(vtx_start, i);

}

assert( 0 == m_spath_table[vtx_start] );

m_sunit[vtx_start] = true;

}



private:

int   m_num;

int*  m_matrix;  // 存邻接矩阵

bool* m_sunit;   // 在dijkstra计算过程中用来存某点是否已经是最短路径中的点.

int*  m_spath_table;  // 在dijkstra计算过程中存到某点是的最短路径是多少.

};





int main(int argc, char* argv[])

{

int case_count, path_count, weight, next_ava_vtx;

int vtx_start, vtx_end;

std::map<std::string, int> loc2vtx;  // 存地点字符串到图的顶点号的映射. 

std::vector<int> spirit;

std::string line;



std::cin >> case_count;

for ( int i = 0; i < case_count; i++ )

{

loc2vtx.clear();

spirit.clear();

next_ava_vtx = 0;



std::cin >> path_count;

for ( int j = 0; j < path_count; j++ )

{

// 得到起点地址字符串. 

std::cin >> line;

std::map<std::string, int>::iterator ite = loc2vtx.find(line);

if ( loc2vtx.end() != ite )

{

vtx_start = ite->second;

}

else

{

vtx_start = next_ava_vtx;

loc2vtx[line] = vtx_start;

next_ava_vtx++;

}

spirit.push_back(vtx_start);



// 得到终点地址字符串. 

std::cin >> line;

ite = loc2vtx.find(line);

if ( loc2vtx.end() != ite )

{

vtx_end = ite->second;

}

else

{

vtx_end = next_ava_vtx;

loc2vtx[line] = vtx_end;

next_ava_vtx++;

}

spirit.push_back(vtx_end);



// 得到权重

std::cin >> weight;

spirit.push_back(weight);

}



// 至此 next_ava_vtx 中存放的实际上是邻接方阵的阶. 

graph_t graph(next_ava_vtx);

for ( int i = 0; i < spirit.size()/3; i++ )

{

int x = spirit[3*i];

int y = spirit[3*i+1];

int w = spirit[3*i+2];

graph(x, y) = w;

graph(y, x) = w;

}

for ( int i = 0; i < next_ava_vtx; i++ )

{

graph(i, i) = 0;

}



std::cin >> line;

std::map<std::string, int>::iterator ite = loc2vtx.find(line);

if ( ite == loc2vtx.end() )

{

std::cout << "-1\n";

continue;

}

vtx_start = loc2vtx[line];



std::cin >> line;

ite = loc2vtx.find(line);

if ( ite == loc2vtx.end() )

{

std::cout << "-1\n";

continue;

}

vtx_end = loc2vtx[line];



if ( vtx_start == vtx_end )

{

std::cout << "0\n";

continue;

}



std::cout << graph.shortestpath_dij(vtx_start, vtx_end) << std::endl;

}



return 0;

}

 

 
4 对时间复杂度,空间复杂度方面的分析、估算。
 
时间复杂度和空间复杂度都是O(n*n).   http://blog.ykyi.net
 
///////////// 原题
 
 
 
 
 
1031. Campus
 
 
 
 
 
Description
 
At present, Zhongshan University has 4 campuses with a total area of 6.17 square kilometers sitting respectively on both sides of the Pearl River or facing the South China Sea. The Guangzhou South Campus covers an area of 1.17 square kilometers, the North Campus covers an area of 0.39 square kilometers, the Guangzhou East Campus has an area of 1.13 square kilometers and the Zhuhai Campus covers an area of 3.48 square kilometers. All campuses have exuberance of green trees, abundance of lawns and beautiful sceneries, and are ideal for molding the temperaments, studying and doing research.
 
 
 
 
       Sometime, the professors and students have to go from one place to another place in one campus or between campuses. They want to find the shortest path between their source place S and target place T. Can you help them?
 
 
Input
 
The first line of the input is a positive integer C. C is the number of test cases followed. In each test case, the first line is a positive integer N (0<N<=100) that represents the number of roads. After that, N lines follow. The i-th(1<=i<=N) line contains two strings Si, Ti and one integer Di (0<=Di<=100). It means that there is a road whose length is Di between Si and Ti. Finally, there are two strings S and T, you have to find the shortest path between S and T. S, T, Si(1<=i<=N) and Ti(1<=i<=N) are all given in the following format: str_Campus.str_Place. str_Campus represents the name of the campus, and str_Place represents the place in str_Campus. str_Campus is "North", "South", "East" or "Zhuhai". str_Place is a string which has less than one hundred lowercase characters from "a-z". You can assume that there is at most one road directly between any two places.
 
Output
 
The output of the program should consist of C lines, one line for each test case. For each test case, the output is a single line containing one integer. If there is a path between S and T, output the length of the shortest path between them. Otherwise just output "-1" (without quotation mark). No redundant spaces are needed.