");
  For (int i = 1; i    StrTmp [i] = ss. Substring (10, 12) / / Country 
  If (strTmp.equals ( "CN")) 
  StrTmp = "" + strTmp + ""; 
  Str1 = removeAllTag (ss [i]); 
  Bw.write ( "" + i + "" + strTmp + "" + + str1 " 
");

  ) 
  Bw.write ( "</ td> </ tr> </ table>"); 
  ) 
  Bw.close (); 
  System.out.println ( "complete"); 
  System.exit (0); 

  ) 


  Private String getProxy1 () ( 
  Int i = 0; 
  Try ( 
  URL url = new URL ( "http://www.cybersyndrome.net/pla.html"); 
  BufferedReader br = new BufferedReader (new InputStreamReader (url.openStream ())); 
  String s = ""; 
  StringBuffer sb = new StringBuffer (""); 
  While ((s = br.readLine ())!= null) ( 
  I + +; 
  If (i> 80 & & i <89) ( 
  Sb.append (s + "\ r \ n"); 
  ) 
  ) 
  Br.close (); 
  Return sb.toString (); 
  ) Catch (Exception e) ( 
  Return "error open url" + e.toString (); 
  ) 
  ) 

  Public static String removeAllTag (String src) ( 
  Return src.replaceAll ("<[^>]*>", ""); 
  ) 
  ) 


  Then javac JavaProxy.java 

  Java JavaProxy 

  Waiting for a good result.    </ Td> </ tr> <tr> 

  ↑ Back 

Depth-first search and breadth-first search

  Abstract: The depth-first search and breadth-first search 

  </ Td> </ tr> <tr> <td height="35" valign="top" class="ArticleTeitle"> <table width = "100%" border = "0" cellspacing = "0" cellpadding = " 0 "> <tr> <td width="483" height="86" align="center" valign="top"> 

  First, depth-first search 
  Depth-first search is in the search tree each layer has expanded to only a subset of nodes, and continuing to advance the further progress until no longer (at leaf nodes or depth limit), only from the current node to return to the previous level nodes, along the other direction and continue to move forward.    This method is the search tree roots from the beginning of an evolving one. 

  Depth-first search is also known as the vertical search.    As the issue of a broken tree branch may contain infinite, if the depth-first search of potential infinite branches (that is, infinite depth), it is impossible to find the target node.    Therefore, the depth-first search strategy is not complete.    In addition, the application of this strategy are the solution is not necessarily optimal solution (Shortest Path). 

  Second, rearrangement Jiugong quiz 
  In a 3 x 3 Jiugong 1-8 in the eight and the number of randomly placed in a box which the grid yard.    As shown below left.    Now demands of the question: the Jiugong adjusted as follows in the form shown in Figure Right.    Adjustment rules: Each box will be (on the left or right) of the provisional figures-moved to a box.    Test programming. 

  | 2 | 8 | 3 | | 1 | 2 | 3 | 
  – 
  | 1 | | 4 | | 8 | | 4 | 

  | 7 | 6 | 5 | | 7 | 6 | 5 | 

  Depth-first search path diagram: 

  </ Td> <td width="201" valign="top"> </ td> </ tr> <tr> <td height="20" colspan="2"> 

  </ Td> </ tr> </ table> 

  Third, BFS 

  In the depth-first search algorithm, the greater the depth is the first node expanded.    If the search algorithm to the node according to the level of search, the layer nodes not dealing with the search, not the lower node processing, that is, the smaller the depth of the first node be expanded, that is to say The first node can be expanded to deal with, such as breadth-first search algorithm search method. 

  BFS path diagram: 

  Four flights (from "The Art of Java") 
  A customer to a scheduled from New York to Los Angeles flight tickets Below are the flight line, you look for a discount to customers programmes. 

  <table Width="581" border="1"> <tr> <td width="262"> flights </ td> <td width="303"> distance </ td> </ tr> <tr> < td> New York to Chicago </ td> <td> 900 miles </ td> </ tr> <tr> <td> Chicago to Denver </ td> <td> 1,000 miles </ td> </ tr> < tr> <td> New York to Toronto </ td> <td> 500 miles </ td> </ tr> <tr> <td> New York to Denver </ td> <td> 1,800 miles </ td> < / tr> <tr> <td> Toronto to Calgary </ td> <td> 1,700 miles </ td> </ tr> <tr> <td> Toronto to Los Angeles </ td> <td> 2,500 miles </ td> </ tr> <tr> <td> Toronto to Chicago </ td> <td> 500 miles </ td> </ tr> <tr> <td> Denver to Urbana </ td> <td> 1,000 miles </ td> </ tr> <tr> <td> Denver to Houston </ td> <td> 1,000 miles </ td> </ tr> <tr> <td> Houston to Los Angeles </ td> <td > 1,500 miles </ td> </ tr> <tr> <td> Denver to Los Angeles </ td> <td> 1,000 miles </ td> </ tr> </ table> 

  Below are using depth-first search for the procedure: 

  / / Find connections using a depth-first search. Import java.util .*; import java.io. *; / / Flight information. Class FlightInfo (String from; String to; int distance; boolean skip; / / used in backtracking FlightInfo (String f, String t, int d) (f = from; to = t; distance = d; skip = false;)) (class Depth final int MAX = 100; / / This array holds the flight information. FlightInfo flights [] = new FlightInfo [MAX]; int numFlights = 0 / / number of entries in flight array Stack btStack = new Stack (); / / backtrack stack public static void main (String args []) (String to and from; Depth ob = new Depth (); BufferedReader br = new BufferedReader (new InputStreamReader (System.in)); ob.setup (); try (System.out.print ( "From?"); from br.readLine = () ; System.out.print ( "To?"); to br.readLine = (); ob.isflight (from, to); if (ob.btStack.size ()! = 0) ob.route (to); ) catch (IOException exc) (System.out.println ( "Error on input.");)) / / Initialize the flight database. void setup () (addFlight ( "New York" and "Chicago", 900); addFlight ( "Chicago" and "Denver", 1000); addFlight ( "New York" and "Toronto", 500); addFlight ( "New York", "Denver", 1800); addFlight ( "Toronto", "Calgary" 1700); addFlight ( "Toronto", "Los Angeles", 2500); addFlight ( "Toronto", "Chicago", 500); addFlight ( "Denver", "Urbana," 1000); addFlight ( "Denver," " Houston, "1000); addFlight (" Houston, "" Los Angeles ", 1500); addFlight (" Denver "," Los Angeles ", 1000);) / / Put flights into the database. void addFlight (String from, String to, int dist) (if (numFlights <MAX) (flights [numFlights] = new FlightInfo (from, to, dist); numFlights + +;) else System.out.println ( "Flight database full \ n");) / / Show the route and total distance. void route (String to) (Stack rev = new Stack (); int dist = 0; FlightInfo f; btStack.size int num = () / / Reverse the stack to display route. for (int i = 0; i <num; i + +) rev.push (btStack.pop ()); for (int i = 0; i <num; i + +) (f = (FlightInfo) rev.pop (); System. out.print (f.from + "to"); dist + = f.distance;) System.out.println (to); System.out.println ( "Distance is" + dist);) / * If there is a flight between from and to, and return the distance of flight; otherwise, return 0 * / int match (String from, String to) (for (int i = numFlights-1; i> 1; i -) (if (flights [i]. from.equals (from) & & flights [i]. to.equals (to) & &! flights [i]. skip) (flights [i]. skip = true; / / prevent reuse return flights [ i]. distance;)) return 0; / / not found) / / Given from, find any connection. FlightInfo find (String from) (for (int i = 0; i <numFlights; i + +) (if (flights [i ]. from.equals (from) & &! flights [i]. skip) (FlightInfo f = new FlightInfo (flights [i]. from, flights [i]. to, flights [i]. distance); flights [i] . skip = true; / / prevent reuse return f;)) return null;) / / Determine if there is a route between from and to. void isflight (String from, String to) (int dist; FlightInfo f; / / See if at destination. dist = match (from, to); if (dist! = 0) (btStack.push (new FlightInfo (from, to, dist)) return;) / / Try another connection. f = find (from ); if (f! = null) (btStack.push (new FlightInfo (from, to, f.distance)); isflight (f.to, to);) else if (btStack.size ()> 0) (/ / Backtrack and try another connection. f = (FlightInfo) btStack.pop (); isflight (f.from, f.to);))) 

  Explained: isflight () method using recursive method of depth-first search, which calls first match () method checks the flight database, and to judge from in between flights up there.    If so, the goal of obtaining information, and the line pressure Ruzhan, and then return (to find a programme).    Otherwise, we will find call () method View from arbitrary and other cities between the lines, if found return to a description on the line FlightInfo object, or return null.    If the existence of such a line, then put the lines kept in f, and the current flight information pressure to the top of the stack, and then recursive calls isflight () method, then stored in f.to in a new city the starting city. otherwise a regression, the first pop-up Zhanding nodes, and then recursive calls isflight () method.    The process will continue till the objective to find. 

  Running Results: 

  C: \ java> java Depth 
  From? New York 
  To? Los Angeles 
  New York to Chicago to Denver to Los Angeles 
  Distance is 2900 

  C: \ java> 

  Depth-first search to find a solution, at the same time, above this particular issue, depth-first search had not been rolled back, and once found a solution, but if the data organized in a manner different from when the solution is likely to find a number of regression.    Therefore, the example of the output is not universal.    Moreover, in a very long search, but which do not have a branch of the solution, depth-first search performance will be poor, in such circumstances, not only in the depth-first search of the search path, a waste of time, but also Back to the goals of a waste of time. 

  Look at the example of the use of this breadth-first search procedures: 

  / / Find connections using a breadth-first search. Import java.util .*; import java.io. *; / / Flight information. Class FlightInfo (String from; String to; int distance; boolean skip; / / used in backtracking FlightInfo (String f, String t, int d) (f = from; to = t; distance = d; skip = false;)) (class Breadth final int MAX = 100; / / This array holds the flight information. FlightInfo flights [] = new FlightInfo [MAX]; int numFlights = 0 / / number of entries in flight array Stack btStack = new Stack (); / / backtrack stack public static void main (String args []) (String to and from; Breadth ob = new Breadth (); BufferedReader br = new BufferedReader (new InputStreamReader (System.in)); ob.setup (); try (System.out.print ( "From?"); from br.readLine = () ; System.out.print ( "To?"); to br.readLine = (); ob.isflight (from, to); if (ob.btStack.size ()! = 0) ob.route (to); ) catch (IOException exc) (System.out.println ( "Error on input.");)) / / Initialize the flight database. void setup () (addFlight ( "New York" and "Chicago", 900); addFlight ( "Chicago" and "Denver", 1000); addFlight ( "New York" and "Toronto", 500); addFlight ( "New York", "Denver", 1800); addFlight ( "Toronto", "Calgary" 1700); addFlight ( "Toronto", "Los Angeles", 2500); addFlight ( "Toronto", "Chicago", 500); addFlight ( "Denver", "Urbana," 1000); addFlight ( "Denver," " Houston, "1000); addFlight (" Houston, "" Los Angeles ", 1500); addFlight (" Denver "," Los Angeles ", 1000);) / / Put flights into the database. void addFlight (String from, String to, int dist) (if (numFlights <MAX) (flights [numFlights] = new FlightInfo (from, to, dist); numFlights + +;) else System.out.println ( "Flight database full \ n");) / / Show the route and total distance. void route (String to) (Stack rev = new Stack (); int dist = 0; FlightInfo f; btStack.size int num = () / / Reverse the stack to display route. for (int i = 0; i <num; i + +) rev.push (btStack.pop ()); for (int i = 0; i <num; i + +) (f = (FlightInfo) rev.pop (); System. out.print (f.from + "to"); dist + = f.distance;) System.out.println (to); System.out.println ( "Distance is" + dist);) / * If there is a flight between from and to, and return the distance of flight; otherwise, return 0 * / int match (String from, String to) (for (int i = numFlights-1; i> 1; i -) (if (flights [i]. from.equals (from) & & flights [i]. to.equals (to) & &! flights [i]. skip) (flights [i]. skip = true; / / prevent reuse return flights [ i]. distance;)) return 0; / / not found) / / Given from, find any connection. FlightInfo find (String from) (for (int i = 0; i <numFlights; i + +) (if (flights [i ]. from.equals (from) & &! flights [i]. skip) (FlightInfo f = new FlightInfo (flights [i]. from, flights [i]. to, flights [i]. distance); flights [i] . skip = true; / / prevent reuse return f;)) return null;) / * Determine if there is a route between from and to using breadth-first search * / void isflight (String from, String to) (int dist , dist2; FlightInfo f; / / This stack is needed by the breadth-first search. Stack resetStck = new Stack (); / / See if at destination. dist = match (from, to); if (dist! = 0) (btStack.push (new FlightInfo (from, to, dist)) return;) / * Following is the first part of the breadth-first modification. It checks all connecting flights from a specified node * / while ((f = find (from))! = null) (resetStck.push (f) if ((dist = match (f.to, to))! = 0) (resetStck.push (f.to); btStack.push (new FlightInfo (from, f.to, f.distance)); btStack.push (new FlightInfo (f.to, to, dist)) return;)) / * The following code resets the skip fields set by preceding while loop. This is also part of the breadth-first modifiction * / int i = resetStck.size (); for (; i! = 0; i -) resetSkip ((FlightInfo) resetStck.pop ()) / / Try another connection. f = find (from); if (f! = null) (btStack.push (new FlightInfo (from, to, f.distance)); isflight (f.to, to);) else if (btStack.size ()> 0) (/ / Backtrack and try another connection. f = (FlightInfo) btStack.pop (); isflight (f.from, f.to);)) / / Reset skip field of specified flight. void resetSkip ( FlightInfo f) (for (int i = 0; i <numFlights; i + +) if (flights [i]. from.equals (f.from) & & flights [i]. to.equals (f.to)) flights [i ]. skip = false;)) 

  Running Results: 

  C: \ java> java Breadth 
  From? New York 
  To? Los Angeles 
  New York to Toronto to Los Angeles 
  Distance is 3000 

  C: \ java> 

  It found a reasonable solution, but it does not have general.    Because the first of the path to find information on the physical forms of organization. 

  If the target in the search space is not too deep in hiding, then BFS performance will be very good. 

  </ Td> </ tr> <tr> 

  ↑ Back 

Google AJAX Search API + = delicious TAG site?

  Abstract: Google AJAX Search API + = delicious TAG site? 

  : 

  Foreword 

  Del.icio.us provides a wide range of reusable data format.    And it provides a common API to PHP and similar output format so that these data can also be JavaScript developers as JSON format will be used. 

  Google AJAX Search API to provide some simple web objects: they implementation of the various Google services (page search, local search, video search, Blog search, news search) on embedded (inline) Search. 

  How these two technologies together?    Google AJAX Search API + = delicious TAG site?    As this paper for 25, the following summary only listed some of the paragraphs, read and download the original source code: 
  Google AJAX Search API + = delicious TAG site. Pdf Java, java, J2SE, j2se, J2EE, j2ee, J2ME, j2me, ejb, ejb3, JBOSS, jboss, spring, hibernate, jdo, struts, webwork, ajax, AJAX, mysql, MySQL, Oracle, Weblogic, Websphere, scjp, scjd