启动web harvest的时候出现Caused by: java.io.IOException: Could not get shel

当使用java -jar命令的时候,出现如下异常,请各位看看,我还是新手:

d:\work\tec_crawlers>java -jar webharvest.jar
Exception in thread "AWT-EventQueue-0" java.lang.ExceptionInInitializerError
at org.webharvest.gui.Ide.(Unknown Source)
at CommandLine$1.run(Unknown Source)
at java.awt.event.InvocationEvent.dispatch(InvocationEvent.java:209)
at java.awt.EventQueue.dispatchEvent(EventQueue.java:597)
at java.awt.EventDispatchThread.pumpOneEventForFilters(EventDispatchThread.java:273)
at java.awt.EventDispatchThread.pumpEventsForFilter(EventDispatchThread.java:183)
at java.awt.EventDispatchThread.pumpEventsForHierarchy(EventDispatchThread.java:173)
at java.awt.EventDispatchThread.pumpEvents(EventDispatchThread.java:168)
at java.awt.EventDispatchThread.pumpEvents(EventDispatchThread.java:160)
at java.awt.EventDispatchThread.run(EventDispatchThread.java:121)
Caused by: java.lang.RuntimeException: java.io.IOException: Could not get shell folder ID list
at sun.awt.shell.Win32ShellFolder2$ComTask.execute(Win32ShellFolder2.java:1223)
at sun.awt.shell.Win32ShellFolder2.getFileSystemPath(Win32ShellFolder2.java:557)
at sun.awt.shell.Win32ShellFolder2.composePathForCsidl(Win32ShellFolder2.java:211)
at sun.awt.shell.Win32ShellFolder2.(Win32ShellFolder2.java:224)
at sun.awt.shell.Win32ShellFolderManager2.getNetwork(Win32ShellFolderManager2.java:123)
at sun.awt.shell.Win32ShellFolder2$7.call(Win32ShellFolder2.java:541)
at sun.awt.shell.Win32ShellFolder2$7.call(Win32ShellFolder2.java:538)
at sun.awt.shell.Win32ShellFolder2$ComTask.execute(Win32ShellFolder2.java:1214)
at sun.awt.shell.Win32ShellFolder2.getFileSystemPath(Win32ShellFolder2.java:538)
at sun.awt.shell.Win32ShellFolder2.access$400(Win32ShellFolder2.java:55)
at sun.awt.shell.Win32ShellFolder2$11.call(Win32ShellFolder2.java:711)
at sun.awt.shell.Win32ShellFolder2$11.call(Win32ShellFolder2.java:702)
at sun.awt.shell.Win32ShellFolder2$ComTask.execute(Win32ShellFolder2.java:1214)
at sun.awt.shell.Win32ShellFolder2.getChildByPath(Win32ShellFolder2.java:702)
at sun.awt.shell.Win32ShellFolderManager2.getPersonal(Win32ShellFolderManager2.java:137)
at sun.awt.shell.Win32ShellFolder2$10.call(Win32ShellFolder2.java:652)
at sun.awt.shell.Win32ShellFolder2$10.call(Win32ShellFolder2.java:639)
at java.util.concurrent.FutureTask$Sync.innerRun(FutureTask.java:303)
at java.util.concurrent.FutureTask.run(FutureTask.java:138)
at java.util.concurrent.ThreadPoolExecutor$Worker.runTask(ThreadPoolExecutor.java:885)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:907)
at sun.awt.shell.Win32ShellFolder2$ComTaskExecutor$2.run(Win32ShellFolder2.java:1180)
at java.lang.Thread.run(Thread.java:619)
Caused by: java.io.IOException: Could not get shell folder ID list
at sun.awt.shell.Win32ShellFolder2.getFileSystemPath0(Native Method)
at sun.awt.shell.Win32ShellFolder2.access$1000(Win32ShellFolder2.java:55)
at sun.awt.shell.Win32ShellFolder2$8.call(Win32ShellFolder2.java:559)
at sun.awt.shell.Win32ShellFolder2$8.call(Win32ShellFolder2.java:557)
at sun.awt.shell.Win32ShellFolder2$ComTask.execute(Win32ShellFolder2.java:1214)
... 22 more

1个回答

boshao122
波少love 哦,好的,多谢啦,我去下一个试试,等会搞不定的时候再找你。。。
接近 7 年之前 回复
jinnianshilongnian
jinnianshilongnian 是的
接近 7 年之前 回复
boshao122
波少love 什么意思,不懂,难道是要把我的jdk升级一下,对了,我的系统是win7的
接近 7 年之前 回复
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关于learning opencv这本书的源码
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Calibrate the cameras and display the // rectified results along with the computed disparity images. // static void StereoCalib(const char* imageList, int nx, int ny, int useUncalibrated) { int displayCorners = 0; int showUndistorted = 1; bool isVerticalStereo = false;//OpenCV can handle left-right //or up-down camera arrangements const int maxScale = 1; const float squareSize = 1.f; //Set this to your actual square size FILE* f = fopen(imageList, "rt"); int i, j, lr, nframes, n = nx*ny, N = 0; vector<string> imageNames[2]; vector<CvPoint3D32f> objectPoints; vector<CvPoint2D32f> points[2]; vector<int> npoints; vector<uchar> active[2]; vector<CvPoint2D32f> temp(n); CvSize imageSize = {0,0}; // ARRAY AND VECTOR STORAGE: double M1[3][3], M2[3][3], D1[5], D2[5]; double R[3][3], T[3], E[3][3], F[3][3]; CvMat _M1 = cvMat(3, 3, CV_64F, M1 ); CvMat _M2 = cvMat(3, 3, CV_64F, M2 ); CvMat _D1 = cvMat(1, 5, CV_64F, D1 ); CvMat _D2 = cvMat(1, 5, CV_64F, D2 ); CvMat _R = cvMat(3, 3, CV_64F, R ); CvMat _T = cvMat(3, 1, CV_64F, T ); CvMat _E = cvMat(3, 3, CV_64F, E ); CvMat _F = cvMat(3, 3, CV_64F, F ); if( displayCorners ) cvNamedWindow( "corners", 1 ); // READ IN THE LIST OF CHESSBOARDS: if( !f ) { fprintf(stderr, "can not open file %s\n", imageList ); return; } for(i=0;;i++) { char buf[1024]; int count = 0, result=0; lr = i % 2; vector<CvPoint2D32f>& pts = points[lr]; //points0和1分别是左右图像 if( !fgets( buf, sizeof(buf)-3, f )) break; size_t len = strlen(buf); while( len > 0 && isspace(buf[len-1])) buf[--len] = '\0'; if( buf[0] == '#') continue; IplImage* img = cvLoadImage( buf, 0 ); //打开图像到img if( !img ) break; imageSize = cvSize((*img).width, (*img).height); //imageSize = cvGetSize(img); imageNames[lr].push_back(buf); //FIND CHESSBOARDS AND CORNERS THEREIN: for( int s = 1; s <= maxScale; s++ ) //一张图像检测maxscale次 { IplImage* timg = img; if( s > 1 ) //检测不到怎么办,扩大图像使之更容易检测 { timg = cvCreateImage(cvSize(img->width*s,img->height*s), img->depth, img->nChannels ); cvResize( img, timg, CV_INTER_CUBIC ); } result = cvFindChessboardCorners( timg, cvSize(nx, ny), &temp[0], &count, CV_CALIB_CB_ADAPTIVE_THRESH | CV_CALIB_CB_NORMALIZE_IMAGE); if( timg != img ) cvReleaseImage( &timg ); if( result || s == maxScale ) //检测角点完成或者达到最大检测次数 for( j = 0; j < count; j++ ) { temp[j].x /= s; //temp储存的是精确坐标 temp[j].y /= s; } if( result ) break; } if( displayCorners ) { printf("%s\n", buf); IplImage* cimg = cvCreateImage( imageSize, 8, 3 ); cvCvtColor( img, cimg, CV_GRAY2BGR ); cvDrawChessboardCorners( cimg, cvSize(nx, ny), &temp[0], count, result ); cvShowImage( "corners", cimg ); cvWaitKey(); //注意不写这个无法显示图像 cvReleaseImage( &cimg ); if( cvWaitKey(0) == 27 ) //Allow ESC to quit exit(-1); } else putchar('.'); //一个.表示执行完一张图 N = pts.size(); // pts.resize(N + n, cvPoint2D32f(0,0)); //pts是个容器,将pts调整到能容纳原来的元素和角点个数 active[lr].push_back((uchar)result); //assert( result != 0 ); if( result ) { //Calibration will suffer without subpixel interpolation cvFindCornerSubPix( img, &temp[0], count, cvSize(11, 11), cvSize(-1,-1), cvTermCriteria(CV_TERMCRIT_ITER+CV_TERMCRIT_EPS, 30, 0.01) ); copy( temp.begin(), temp.end(), pts.begin() + N ); } cvReleaseImage( &img ); } fclose(f); printf("\n"); // HARVEST CHESSBOARD 3D OBJECT POINT LIST: nframes = active[0].size();//Number of good left chessboads picture found objectPoints.resize(nframes*n); //使objectPoints能够包含全部图像的角点个数 for( i = 0; i < ny; i++ ) for( j = 0; j < nx; j++ ) objectPoints[i*nx + j] = cvPoint3D32f(i*squareSize, j*squareSize, 0); for( i = 1; i < nframes; i++ ) copy( objectPoints.begin(), objectPoints.begin() + n, objectPoints.begin() + i*n ); npoints.resize(nframes,n); N = nframes*n; CvMat _objectPoints = cvMat(1, N, CV_32FC3, &objectPoints[0] ); CvMat _imagePoints1 = cvMat(1, N, CV_32FC2, &points[0][0] ); CvMat _imagePoints2 = cvMat(1, N, CV_32FC2, &points[1][0] ); //points的第一个索引,0是left,1是right CvMat _npoints = cvMat(1, npoints.size(), CV_32S, &npoints[0] ); //npoints的第i个元素指的是第i个图片有多少个角点 cvSetIdentity(&_M1); cvSetIdentity(&_M2); cvZero(&_D1); cvZero(&_D2); // CALIBRATE THE STEREO CAMERAS printf("Running stereo calibration ..."); fflush(stdout); cvStereoCalibrate( &_objectPoints, &_imagePoints1, &_imagePoints2, &_npoints, &_M1, &_D1, &_M2, &_D2, imageSize, &_R, &_T, &_E, &_F, cvTermCriteria(CV_TERMCRIT_ITER+ CV_TERMCRIT_EPS, 100, 1e-5), CV_CALIB_FIX_ASPECT_RATIO + CV_CALIB_ZERO_TANGENT_DIST + CV_CALIB_SAME_FOCAL_LENGTH ); } int main(void) { StereoCalib("ch12_list.txt", 9, 6, 1); return 0; } ``` 在调试过程中会出现![图片说明](https://img-ask.csdn.net/upload/201711/18/1511003216_595462.png) 初步调试应该是cvStereoCalibrate函数出了问题,c++基础不好,求大神帮忙看看啊。。。
Farm Irrigation
Benny has a spacious farm land to irrigate. The farm land is a rectangle, and is divided into a lot of samll squares. Water pipes are placed in these squares. Different square has a different type of pipe. There are 11 types of pipes, which is marked from A to K, as Figure 1 shows. Figure 1 Benny has a map of his farm, which is an array of marks denoting the distribution of water pipes over the whole farm. For example, if he has a map ADC FJK IHE then the water pipes are distributed like Figure 2 Several wellsprings are found in the center of some squares, so water can flow along the pipes from one square to another. If water flow crosses one square, the whole farm land in this square is irrigated and will have a good harvest in autumn. Now Benny wants to know at least how many wellsprings should be found to have the whole farm land irrigated. Can you help him? Note: In the above example, at least 3 wellsprings are needed, as those red points in Figure 2 show. Input There are several test cases! In each test case, the first line contains 2 integers M and N, then M lines follow. In each of these lines, there are N characters, in the range of 'A' to 'K', denoting the type of water pipe over the corresponding square. A negative M or N denotes the end of input, else you can assume 1 <= M, N <= 50. Output For each test case, output in one line the least number of wellsprings needed. Sample Input 2 2 DK HF 3 3 ADC FJK IHE -1 -1 Sample Output 2 3
Imperishable Night
It's the eve of the Harvest Moon Festival in Gensokyo when Flandre Scarlet senses that something is wrong with the moon. It appears that the moon has been replaced by a fake moon. Someone must freeze time and find the real moon to ensure a full moon on the night of the festival. Remilia Scarlet tells her that she must go into the Bamboo Forest of the Lost (迷いの竹林) to investigate and gain enough points to remedy this. 二小姐 Flandre sets off immediately. When she arrives at the entry of the Bamboo Forest of the Lost, she finds that there are n different parts. According to Remilia, there are two kinds of mysterious halidom, ITEM and TIME, that will increase Flandre's points: When Flandre is at the entry, she can choose any path she hasn't choosed before and collects all ITEM and TIME there, then goes back to the entry. There are two values, IV (ITEM Value) and TV (TIME Value) relating to how many points Flandre can gain when she collect a piece of ITEM or a piece of TIME. The initial IV, TV and point of Flandre are all 0. When she chooses a new part, her IV and TV and point keep the same as the end of he previous collection. In parti, there are xi pieces of ITEM with value ai and yi pieces of TIME with value bi. When she collects a piece of ITEM, her point will increase IV (point += IV), and her TV will increase ai (TV += ai) immediately as well. when she collects a piece of TIME, her point will increase TV (point += TV), and her IV will increase bi (IV += bi) immediately as well. Obviously, Flandre's IV and TV will be changing when she is collecting ITEM and TIME, thus the order of collection is important. Morever, the order of the parts she chooses is crucial too. Flandre should gain as many points as possible. Your task is to calculate the maximal point Flandre can gain. Input The input contains multiple test cases (no more than 100). The first line of each case is an integer n (1 ≤ n ≤ 15), indicating the number of parts of the forest. Then follows n lines, the i-th (1 ≤ i ≤ n) of them contains 4 integers ai, bi, xi, yi (0 ≤ ai, bi ≤ 20, 0 ≤ xi, yi ≤ 30), as described above. Output For each case, output the maximal point Flandre can gain in a single line. Sample Input 1 1 2 1 1 3 2 2 2 2 2 2 2 2 2 2 2 2 Sample Output 2 72
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