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OpenCv 大图中找小图

作者：shj 发布于：2014-1-22 17:03

为了实现大图中找小图，之前是考虑使用OpenCv中的模板匹配。。。但发现速度太慢了。

前几天实在受不了，自己动手写了个特定情况下使用的。

把代码贴上来有兴趣的看了。

    clock_t starttime = clock();
    clock_t endtime = 0;
    int found = 0;
     
    IplImage *source1 = cvLoadImage("G:\\vctest\\OpenCv_Test\\图片\\030.png");
    IplImage *source2 = cvLoadImage("G:\\vctest\\OpenCv_Test\\图片\\0311.png");
    cvShowImage("source1",source1);
    cvShowImage("source2",source2);
     
    //FILE *fp = fopen("G:\\vctest\\OpenCv_Test\\图片\\030.tes","wb");
    //FILE *fp1 = fopen("G:\\vctest\\OpenCv_Test\\图片\\031.tes","wb");
    //fwrite(source1->imageData,source1->imageSize,1,fp);
    //fwrite(source2->imageData,source2->imageSize,1,fp1);
    //fclose(fp);
    //fclose(fp1);
     
     
     
    char *source1_data = source1->imageData;
    char *source2_data = source2->imageData;
     
    for(int h=0;h<source1->height - source2->height-1;h++)
    {
    for (int w=0;w<source1->width-source2->width-1;w++)
    {
    if (1 == contrast((unsigned char *)&source1_data[h*source1->widthStep+w*source1->nChannels],(unsigned char *)&source2_data[source2->width*source2->nChannels],source2->width*source2->nChannels,1))
    {
    found = 1;
    for (int ih = 1;ih < source2->height-1;ih++)
    {
    if (0 == contrast((unsigned char *)&source1_data[(h+ih)*source1->widthStep+w*source1->nChannels],(unsigned char *)&source2_data[ih*source2->widthStep+source2->width*source2->nChannels],source2->width*source2->nChannels,1))
    {
    found = 0;
    continue;
    }
    }
    if (found == 0)
    {
    break;
    }
    AfxOutputDebugString("find 1 line at:%d %d\n",w,h);
    endtime = clock();
    AfxOutputDebugString("time:%d\n",endtime-starttime);
    break;
    }
     
     
    }
    // AfxOutputDebugString("h::%d\n",h);
    }
     
     
     
    cvReleaseImage(&source1);
    cvReleaseImage(&source2);
    endtime = clock();
    AfxOutputDebugString("time:%d\n",endtime-starttime);

    int Cimage_contrast_01Dlg::contrast(unsigned char *source1,unsigned char *source2,int len,int a)
    {
    int i = 0;
    for (i=0;i<len-1;i++)
    {
    //if(source1[i] <= source2[i] - a || source1[i] >= source2[i] + a)
    if((abs(source1[i] - source2[i]) >= a))
    {
    return 0;
    }
    }
     
    return 1;
    }

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