# 为Java运行环境导入根证书解决Eclipse的TFS插件的"PKIX path building failed"错误

sun.security.validator.ValidatorException: PKIX path building failed: sun.security.provider.certpath.SunCertPathBuilderException: unable to find valid certification path to requested target
...
Caused by: sun.security.provider.certpath.SunCertPathBuilderException: unable to find valid certification path to requested target
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For Linux:

sudo keytool -importkeystore -srckeystore ~/mykeystore -destkeystore JAVA_HOME/jre/lib/security/cacerts

For Mac:

$sudo keytool -importkeystore -srckeystore ~/mykeystore -destkeystore /Library/Java/Home/lib/security/cacerts 第二个命令将之前建好的keystore中的证书导入jre自带的keystore文件中。过程中会要求输入目标keystore（也就是jre自带的keystore）文件的密码，这个密码默认是 changeit （linux和mac下），或者是 changeme 。 之后，重启Eclipse，再次连结TFS服务器，即可成功通过服务器验证。此外，可以下载一个小工具SSLPoke验证证书是否导入正常，下载之后在文件所在目录下，命令行运行$ java SSLPoke tfs.yourserver.com 4343
Successfully connected

# ant 中通过重新定义 project.all.jars.path 在 classpath 中引入外部 jar 文件

<path id="project.javac.classpath">
<path refid="project.all.jars.path"></path>
<path refid="tested.project.classpath"></path>
</path>

<javac encoding="{java.encoding}"         source="{java.source}" target="{java.target}"         debug="true" extdirs="" includeantruntime="false"         destdir="{out.classes.absolute.dir}"
bootclasspathref="project.target.class.path"
verbose="{verbose}"         classpathref="project.javac.classpath"         fork="{need.javac.fork}">
<src path="{source.absolute.dir}">     
 其中 project.all.jars.path 包含了所有的jar文件，我们可以通过在工程目录下的buildxml中重新定义这个变量来引入其他的jar文件。例如在我的工程中，引用了ormlite这个ORM库，为了能够在开发中使用“attach source”察看源码，该jar文件不能放在libs目录中，因为Eclipse不允许对libs目录中的jar文件“attach source”。因此我将此文件放到了libs/ormlite目录中，为了能够将这两个jar文件加入到classpath中，就要重新定义 project.all.jars.path 这个元素。 基本思路是，重新定义-pre-compile这个target，在其中重新定义 project.all.jars.path 的值。 <target name="-pre-compile"> <echo message="JARPATH={toString:project.all.jars.path}"> {toString:project.all.jars.path}"> {ormlite.dir}/ormlite-core-4.41.jar"> 
 分类： Android, Java | 2013/01/21 | 278 个字 | 评论 
 ThoughtWorks 的一道笔试题 PROBLEM ONE: TRAINS Problem: The local commuter railroad services a number of towns in Kiwiland. Because of monetary concerns, all of the tracks are 'one-way.' That is, a route from Kaitaia to Invercargill does not imply the existence of a route from Invercargill to Kaitaia. In fact, even if both of these routes do happen to exist, they are distinct and are not necessarily the same distance! The purpose of this problem is to help the railroad provide its customers with information about the routes. In particular, you will compute the distance along a certain route, the number of different routes between two towns, and the shortest route between two towns. Input: A directed graph where a node represents a town and an edge represents a route between two towns. The weighting of the edge represents the distance between the two towns. A given route will never appear more than once, and for a given route, the starting and ending town will not be the same town. Output: For test input 1 through 5, if no such route exists, output 'NO SUCH ROUTE'. Otherwise, follow the route as given; do not make any extra stops! For example, the first problem means to start at city A, then travel directly to city B (a distance of 5), then directly to city C (a distance of 4). 1. The distance of the route A-B-C. 2. The distance of the route A-D. 3. The distance of the route A-D-C. 4. The distance of the route A-E-B-C-D. 5. The distance of the route A-E-D. 6. The number of trips starting at C and ending at C with a maximum of 3 stops. In the sample data below, there are two such trips: C-D-C (2 stops). and C-E-B-C (3 stops). 7. The number of trips starting at A and ending at C with exactly 4 stops. In the sample data below, there are three such trips: A to C (via B,C,D); A to C (via D,C,D); and A to C (via D,E,B). 8. The length of the shortest route (in terms of distance to travel) from A to C. 9. The length of the shortest route (in terms of distance to travel) from B to B. 10. The number of different routes from C to C with a distance of less than 30. In the sample data, the trips are: CDC, CEBC, CEBCDC, CDCEBC, CDEBC, CEBCEBC, CEBCEBCEBC. Test Input: For the test input, the towns are named using the first few letters of the alphabet from A to D. A route between two towns (A to B) with a distance of 5 is represented as AB5. Graph: AB5, BC4, CD8, DC8, DE6, AD5, CE2, EB3, AE7 Expected Output: Output #1: 9 Output #2: 5 Output #3: 13 Output #4: 22 Output #5: NO SUCH ROUTE Output #6: 2 Output #7: 3 Output #8: 9 Output #9: 9 Output #10: 7 ========== 用一个带权重的有向图表示节点之间的关系，使用深度优先搜索。 1-5: package com.log4think.code; public class TrainMap { public int[][] map = { // A B C D E {-1, 5, -1, 5, 7}, {-1, -1, 4, -1, -1}, {-1, -1, -1, 8, 2}, {-1, -1, 8, -1, 6}, {-1, 3, -1, -1, -1}, }; public int getDistance(String path) { int distance = 0; for (int i=0; i 0) System.out.println(distance); else System.out.println("NO SUCH ROUTE"); // g.findDistance("AD"); // g.findDistance("ADC"); // g.findDistance("AEBCD"); // g.findDistance("AED"); } } 6: public class TrainMap { public int[][] map = { {-1, 5, -1, 5, 7}, {-1, -1, 4, -1, -1}, {-1, -1, -1, 8, 2}, {-1, -1, 8, -1, 6}, {-1, 3, -1, -1, -1}, }; // used to record the trip round count public int tripCount = 0; /*** * * @param end , the end node. if we reach it we found a path * @param path , the current path we already have * @param maxLength, the maximum stops */ public void dfs(String end, String path, int maxLength) { // this is for debug and trace // System.out.println(";; " + path); // if the path reach the maximum stops, then cancel search if (path.length() - 1 > maxLength) return; // check if we have reach the "end" node if ( path.length() > 1 && path.endsWith(end) ) { System.out.println(path + ", " + tripCount); tripCount ++; } // caculate the lastest node index in map char lastChar = path.charAt(path.length()-1); int lastNodeIndex = lastChar - 'A'; // loop all nodes in map which connected to lastNode, and try it for ( int i=0; i 0) { dfs(end, path + newChar, maxLength); } } } public static void main(String[] args) { TrainMap g = new TrainMap(); g.dfs("C", "C", 3); } } 7: public class TrainMap { public int[][] map = { { -1, 5, -1, 5, 7 }, { -1, -1, 4, -1, -1 }, { -1, -1, -1, 8, 2 }, { -1, -1, 8, -1, 6 }, { -1, 3, -1, -1, -1 } }; public void bfs(String start, String end, int hops) { String lastRoute = start; for (int hop = 0; hop < hops; hop++) { String route = ""; for (int i = 0; i < lastRoute.length(); i++) { char c = lastRoute.charAt(i); int id = c - 'A'; for (int j = 0; j < map[id].length; j++) { if (map[id][j] > 0) route = route + (char) (j + 'A'); } } // System.out.println(lastRoute); lastRoute = route; } // System.out.println(lastRoute); System.out.println(lastRoute.split(end).length - 1); } public static void main(String[] args) { TrainMap g = new TrainMap(); g.bfs("A", "C", 4); } } 8/9: public class TrainMap { public int[][] map = { {-1, 5, -1, 5, 7}, {-1, -1, 4, -1, -1}, {-1, -1, -1, 8, 2}, {-1, -1, 8, -1, 6}, {-1, 3, -1, -1, -1}, }; public void dfs(String end, String path, int cost) { if (path.endsWith(end) && cost < bestCost && cost > 0) { bestPath = path; bestCost = cost; return; } char lastChar = path.charAt(path.length() - 1); int lastNodeIndex = lastChar - 'A'; for (int i = 0; i < map[lastNodeIndex].length; i++) { char newChar = (char) (i + 'A'); int newCost = map[lastNodeIndex][i]; if (newCost > 0) { if (path.indexOf(newChar) > 0) continue; dfs(end, path + newChar, cost + newCost); } } } public String bestPath = ""; public int bestCost = Integer.MAX_VALUE; public static void main(String[] args) { TrainMap g = new TrainMap(); g.dfs("C", "A", 0); // 8 // g.dfs("B", "B", 0); // 9 System.out.println("Best Path: " + g.bestPath + "\nCost: " + g.bestCost); } } 10: public class TrainMap { public int[][] map = { {-1, 5, -1, 5, 7}, {-1, -1, 4, -1, -1}, {-1, -1, -1, 8, 2}, {-1, -1, 8, -1, 6}, {-1, 3, -1, -1, -1}, }; public void dfs(String end, String path, int cost) { if (cost >= 30) return; if (cost > 0 && path.endsWith(end)) { System.out.println(path + ", " + cost); } char lastChar = path.charAt(path.length() - 1); int lastNodeIndex = lastChar - 'A'; for (int i = 0; i < map[lastNodeIndex].length; i++) { char newChar = (char) (i + 'A'); int newCost = map[lastNodeIndex][i]; if (newCost > 0) { dfs(end, path + newChar, cost + newCost); } } } public static void main(String[] args) { TrainMap g = new TrainMap(); g.dfs("C", "C", 0); } } 分类： 写代码(coding) | 2008/12/11 | 1,014 个字 | 评论