264 lines
6.3 KiB
C++
264 lines
6.3 KiB
C++
/* 程序名：outline.c
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功能：输入文字另外由于轮廓图像。返回相应的轮廓特征值
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*/
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#pragma once
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#include <cv.h>
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#include <highgui.h>
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#include <stdlib.h>
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#include <stdio.h>
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#include "Point.h"
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int outlinefeature(IplImage* imglk,int feature[ ][50]){
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/*定义变量*/
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int i,j;
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int height,width,step,channels;
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uchar *data;
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int feat[50][50]={0}; //特征值初始化
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Point featblk[32]; //标记相同H的黑点坐标
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int featk; //标记相同H的黑点数目
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int m; //for 里面的变量
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/* 获取图像信息*/
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height = imglk->height;
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width = imglk->width;
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step = imglk->widthStep;
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channels = imglk->nChannels;
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data = (uchar *)imglk->imageData;
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//初始化特征矩阵 最大值为47 ，非空的特征字有1081个
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int outllab[9][9]={\
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{3,37,10,36,2,35,9,34,1},{38,3,21,20,2,19,18,1,33},\
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{11,22,3,10,2,9,1,17,8},{39,23,11,3,2,1,8,16,32},\
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{4,4,4,4,0,0,0,0,0},{40,24,12,5,6,7,15,31,47},\
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{12,25,5,13,6,14,7,30,15},{41,5,26,27,6,28,29,7,46},\
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{5,42,13,43,6,44,14,45,7}};
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// for(i=0;i<9;i++) //输出测试
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// {
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// for(j=0;j<9;j++)
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// printf("%d*",outllab[i][j]);
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// printf("\n");
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// }
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for(i=4;i<=width-5;i++)
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{
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for(j=4;j<=height-5;j++)
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{
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if(data[j*step+i*channels]==0)
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{
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//**************H=1
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memset(featblk, 0, sizeof(Point)*32); //归零
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featk=0;
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if(data[j*step+(i+1)*channels]==0) //右侧点
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{
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featblk[featk].x=i+1;
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featblk[featk].y=j;
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featk++;
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}
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for(m=i+1;m>=i-1;m--) //上排点
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{
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if(data[(j-1)*step+m*channels]==0)
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{
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featblk[featk].x=m;
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featblk[featk].y=j-1;
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featk++;
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}
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}
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if(data[j*step+(i-1)*channels]==0) //左侧点
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{
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featblk[featk].x=i-1;
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featblk[featk].y=j;
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featk++;
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}
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for(m=i-1;m<=i+1;m++) //下排点
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{
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if(data[(j+1)*step+m*channels]==0)
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{
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featblk[featk].x=m;
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featblk[featk].y=j+1;
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featk++;
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}
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}
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//统计特征点
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//****************************************************
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if(featk>=2)
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{
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for(m=1;m<=featk-1;m++)
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{
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feat[outllab[featblk[m-1].x-i+4][featblk[m-1].y-j+4]][outllab[featblk[m].x-i+4][featblk[m].y-j+4]]++;
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}
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}
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//H=1
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//H=2
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memset(featblk, 0, sizeof(Point)*32); //归零
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featk=0;
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for(m=j+1;m>=j-2;m--)
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{
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if(data[m*step+(i+2)*channels]==0) //右排点
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{
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featblk[featk].x=i+2;
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featblk[featk].y=m;
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featk++;
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}
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}
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for(m=i+1;m>=i-2;m--) //上排点
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{
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if(data[(j-2)*step+m*channels]==0)
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{
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featblk[featk].x=m;
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featblk[featk].y=j-2;
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featk++;
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}
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}
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for(m=j-1;m<=j+2;m++) //左侧点
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{
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if(data[m*step+(i-2)*channels]==0)
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{
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featblk[featk].x=i-2;
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featblk[featk].y=m;
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featk++;
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}
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}
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for(m=i-1;m<=i+2;m++) //下排点
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{
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if(data[(j+2)*step+m*channels]==0)
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{
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featblk[featk].x=m;
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featblk[featk].y=j+2;
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featk++;
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}
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}
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//统计特征点
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//****************************************************
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if(featk>=2)
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{
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for(m=1;m<=featk-1;m++)
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{
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feat[outllab[featblk[m-1].x-i+4][featblk[m-1].y-j+4]][outllab[featblk[m].x-i+4][featblk[m].y-j+4]]++;
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}
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}
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//H=2
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//H=3
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memset(featblk, 0, sizeof(Point)*32); //归零
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featk=0;
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for(m=j+2;m>=j-3;m--)
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{
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if(data[m*step+(i+3)*channels]==0) //右排点
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{
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featblk[featk].x=i+3;
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featblk[featk].y=m;
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featk++;
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}
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}
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for(m=i+2;m>=i-3;m--) //上排点
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{
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if(data[(j-3)*step+m*channels]==0)
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{
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featblk[featk].x=m;
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featblk[featk].y=j-3;
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featk++;
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}
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}
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for(m=j-2;m<=j+3;m++) //左侧点
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{
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if(data[m*step+(i-3)*channels]==0)
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{
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featblk[featk].x=i-3;
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featblk[featk].y=m;
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featk++;
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}
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}
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for(m=i-2;m<=i+3;m++) //下排点
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{
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if(data[(j+3)*step+m*channels]==0)
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{
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featblk[featk].x=m;
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featblk[featk].y=j+3;
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featk++;
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}
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}
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//统计特征点
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//******************************************
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if(featk>=2)
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{
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for(m=1;m<=featk-1;m++)
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{
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feat[outllab[featblk[m-1].x-i+4][featblk[m-1].y-j+4]][outllab[featblk[m].x-i+4][featblk[m].y-j+4]]++;
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}
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}
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//H=3
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//H=4
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memset(featblk, 0, sizeof(Point)*32); //归零
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featk=0;
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for(m=j+3;m>=j-4;m--)
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{
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if(data[m*step+(i+4)*channels]==0) //右排点
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{
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featblk[featk].x=i+4;
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featblk[featk].y=m;
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featk++;
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}
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}
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for(m=i+3;m>=i-4;m--) //上排点
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{
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if(data[(j-4)*step+m*channels]==0)
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{
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featblk[featk].x=m;
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featblk[featk].y=j-4;
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featk++;
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}
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}
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for(m=j-3;m<=j+4;m++) //左侧点
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{
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if(data[m*step+(i-4)*channels]==0)
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{
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featblk[featk].x=i-4;
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featblk[featk].y=m;
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featk++;
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}
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}
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for(m=i-3;m<=i+4;m++) //下排点
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{
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if(data[(j+4)*step+m*channels]==0)
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{
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featblk[featk].x=m;
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featblk[featk].y=j+4;
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featk++;
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}
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}
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//统计特征点
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if(featk>=2)
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{
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for(m=1;m<=featk-1;m++)
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{
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feat[ outllab[featblk[m-1].x-i+4][featblk[m-1].y-j+4]] [outllab[featblk[m].x-i+4][featblk[m].y-j+4] ]++;
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}
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}
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//H=4***********************
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}
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}
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}
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//****注：最终特征值为feat(x,y)+feat(y,x)，放入feat(x,y)中，x<y
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for(i=1;i<50;i++)
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for(j=0;j<i;j++)
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{
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feat[j][i]=feat[i][j]+feat[j][i];
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feat[i][j]=0;
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}
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memcpy(feature,feat,2500*4); //int有四个字节
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return 0;
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} |