2015年5月21日
C#实现验证码中粘连字符分割（六）
《C#实现验证码中粘连字符分割》发布了一系列文章，上一篇我们讲到了谷点的过滤。谷点过滤完成后，我们就可以根据最终确定的谷点来执行滴水算法，从而得到最终的分割路径。含四个字符的验证码图片可通过滴水算法得到3条分割路径，为方便程序统一处理，左侧与右侧边界会作为两条分割路径存储在路径集中。这样对于四个字符的图片而言，就有五条分割路径，每两条路径之间得到一个字符。
写文章时贴代码可能有所遗漏，这里提供完整的工程下载链接：链接: http://pan.baidu.com/s/1sjJoWKt 密码: xdeb
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粘连字符分割原则：
(1)分割后的图片大小与原图片一致，分割后的字符将从宽度上均分该图片，高度值遵循原字符所在高度。
(2)从滴水算法的原理可知，同一个高度坐标最多对应两个相同的宽度坐标。在根据路径分割图像时，必须要考虑到这一点。以高度值为循环基准，每个高度值都需要判断是否有两个相同的宽度坐标。左侧路径取靠左的坐标点作为左极限，右侧路径取靠右的坐标点作为右极限，将该高度值对应的左右极限宽度的中间坐标值提取出来。
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/// <summary> ///根据谷点获取图片的分割路径 /// </summary> /// <param name="valleyCollection">谷点列表</param> /// <param name="Boundary">图片中字符所在区域的边界</param> /// <param name="BinaryArray">二值化图片数组</param> /// <returns>返回粘粘图片的分割路径</returns> public static ArrayList getSegmentPath(ImgValley valleyCollection, ImgBoundary Boundary, Byte[,] BinaryArray) { ArrayList segPathList = new ArrayList(); int yIndex2 = Boundary.widthMax; int yIndex1 = Boundary.widthMin; int xIndex2 = Boundary.heightMax; int xIndex1 = Boundary.heightMin; int valleyNum = valleyCollection.upValley.Count; //将图像最左侧的点所在直线作为第一条分割路径 // int validHeight = xIndex2 - xIndex1; ArrayList pathList0 = new ArrayList(); for (int a = xIndex1; a < xIndex2; a++) { PixPos newPos = new PixPos(); newPos.widthPos = yIndex1; newPos.heightPos = a; pathList0.Add(newPos); } segPathList.Add(pathList0); for (int i = 0; i < valleyNum; i ++) { PixPos pos = (PixPos)valleyCollection.upValley[i]; ArrayList pathList = new ArrayList(); int iniPointY = pos.widthPos; int iniPointX = xIndex1;//任一分割路径的高度起始点均为有效字符出现的第一个点 byte leftRightFlg = 0;//该标志位置位时，表示情况5与情况6可能会循环出现 while ((iniPointX < xIndex2) && (iniPointY < yIndex2)) { Byte pointLeft = BinaryArray[iniPointX, (iniPointY - 1)]; Byte pointRight = BinaryArray[iniPointX, (iniPointY + 1)]; Byte pointDown = BinaryArray[(iniPointX + 1), iniPointY]; Byte pointDownLeft = BinaryArray[(iniPointX + 1), (iniPointY - 1)]; Byte pointDownRight = BinaryArray[(iniPointX + 1), (iniPointY + 1)]; PixPos newPos = new PixPos(); newPos.widthPos = iniPointY; newPos.heightPos = iniPointX; pathList.Add(newPos); if (((0 == pointLeft) && (0 == pointRight) && (0 == pointDown) && (0 == pointDownLeft) && ((0 == pointDownRight)))// all black || ((255 == pointLeft) && (255 == pointRight) && (255 == pointDown) && (255 == pointDownLeft) && ((255 == pointDownRight))//all white || ((0 == pointDownLeft) && (255 == pointDown))))//情况1与情况3 {//down iniPointX = iniPointX + 1; leftRightFlg = 0; } else if ((0 == pointLeft) && (0 == pointRight) && (0 == pointDown) && (255 == pointDownLeft) && (0 == pointDownRight)) {//left down //情况2 iniPointX = iniPointX + 1; iniPointY = iniPointY - 1; leftRightFlg = 0; } else if ((0 == pointDown) && (0 == pointDownLeft) && (255 == pointDownRight)) {//情况4 iniPointX = iniPointX + 1; iniPointY = iniPointY + 1; leftRightFlg = 0; } else if ((255 == pointRight) && (0 == pointDown) && (0 == pointDownLeft) && (0 == pointDownRight)) {//情况5 if (0 == leftRightFlg) { iniPointY = iniPointY + 1; leftRightFlg = 1; } else//标志位为1时说明上一个点出现在情况6，而本次循环的点出现在情况5，此种情况将垂直渗透 { iniPointX = iniPointX + 1; leftRightFlg = 0; } } else if ((255 == pointLeft) && (0 == pointDown) && (0 == pointDownLeft) && (0 == pointDownRight)) {//情况6 if (0 == leftRightFlg) { iniPointY = iniPointY - 1; leftRightFlg = 1; } else//标志位为1时说明上一个点出现在情况5，而本次循环的点出现在情况6，此种情况将垂直渗透 { iniPointX = iniPointX + 1; leftRightFlg = 0; } } else { iniPointX = iniPointX + 1; } } segPathList.Add(pathList); } //将图像最右侧的点所在直线作为最后一条分割路径 ArrayList pathListLast = new ArrayList(); for (int a = xIndex1; a < xIndex2; a++) { PixPos newPos = new PixPos(); newPos.widthPos = yIndex2; newPos.heightPos = a; pathListLast.Add(newPos); } segPathList.Add(pathListLast); return segPathList; } /// <summary> ///根据分割路径分割粘粘字符并另存为位图 /// </summary> /// <param name="valleyCollection">谷点列表</param> /// <param name="Boundary">图片中字符所在区域的边界</param> /// <param name="BinaryArray">二值化图片数组</param> /// <returns>返回粘粘图片的分割路径</returns> public static Bitmap getDivideImg(Byte[,] BinaryArray, ArrayList SegPathList) { int imageHeight = BinaryArray.GetLength(0); int imageWidth = BinaryArray.GetLength(1); int segCount = SegPathList.Count; int locationVal = imageWidth / (segCount - 1); int locationPos = 0; Byte[,] divideArray = new Byte[imageHeight, imageWidth]; for (Int32 x = 0; x < imageHeight; x++) { for (Int32 y = 0; y < imageWidth; y++) { divideArray[x, y] = 255; } } PixPos divPosRight = new PixPos(); PixPos divPosLeft = new PixPos(); ArrayList pathListLeft = new ArrayList(); ArrayList pathListRight = new ArrayList(); int indexWidthRight = 0; int indexWidthLeft = 0; int indexHeightRight = 0; int indexHeightLeft = 0; int pointIndexLeft = 0; int pointIndexRight = 0; int posCountLeft = 0; int posCountRight = 0; for (int i = 0; i < segCount - 1; i++) { //每条分割路径的起始点与终点均相同，且不会出现同一个高度值对应两个宽度值的状况，应此每条路径的点数均相同 pathListLeft = (ArrayList)SegPathList[i]; pathListRight = (ArrayList)SegPathList[i + 1]; posCountLeft = pathListLeft.Count; posCountRight = pathListRight.Count; /* int posCount = 0; if (posCountLeft < posCountRight) { posCount = posCountLeft; } else { posCount = posCountRight; } */ locationPos = 5 + locationVal * i; pointIndexLeft = 0; pointIndexRight = 0; //目前所用的滴水算法下，同一高度值最多同时对应两个宽度值 while ((pointIndexLeft < posCountLeft) && (pointIndexRight < posCountRight)) { divPosRight = (PixPos)pathListRight[pointIndexRight]; divPosLeft = (PixPos)pathListLeft[pointIndexLeft]; pointIndexLeft++; pointIndexRight++; indexWidthLeft = divPosLeft.widthPos; indexWidthRight = divPosRight.widthPos; indexHeightRight = divPosLeft.heightPos; indexHeightLeft = divPosRight.heightPos; if(pointIndexLeft < posCountLeft) { divPosLeft = (PixPos)pathListLeft[pointIndexLeft]; if (indexHeightLeft == divPosLeft.heightPos)//若下一点高度值相同，索引值加1，宽度值更新 { pointIndexLeft++; indexWidthLeft = divPosLeft.widthPos; } } if (pointIndexRight < posCountRight) { divPosRight = (PixPos)pathListRight[pointIndexRight]; if (indexHeightRight == divPosRight.heightPos)//若下一点高度值相同，索引值加1，宽度值更新 { pointIndexRight++; indexWidthRight = divPosRight.widthPos; } } for (Int32 y = indexWidthLeft; y < indexWidthRight; y++) { divideArray[indexHeightRight, (y - indexWidthLeft + locationPos)] = BinaryArray[indexHeightRight, y]; } } } Bitmap GrayBmp = ImageBinarization.ImageUtils.BinaryArrayToBinaryBitmap(divideArray); return GrayBmp; } |
