图像细化算法大全

图像细化算法大全(1)

#include "StdAfx.h"

#include

#include

void beforethin(unsigned char *ip, unsigned char *jp,

unsigned long lx, unsigned long ly)

{

unsigned long i,j;

for(i=0; i

{

for(j=0; j

{

//这里要视前景是白点还是黑点而定，可以改动

//如果前景是白点，就是这样；反之反过来

if(ip[i*lx+j]>0)

jp[i*lx+j]=1;

else

jp[i*lx+j]=0;

}

}

}

/////////////////////////////////////////////////////////////////////////

//Hilditch细化算法

//功能：对图象进行细化

//参数：image：代表图象的一维数组

// lx：图象宽度

// ly：图象高度

// 无返回值

void ThinnerHilditch(void *image, unsigned long lx, unsigned long ly) {

char *f, *g;

char n[10];

unsigned int counter;

short k, shori, xx, nrn;

unsigned long i, j;

long kk, kk11, kk12, kk13, kk21, kk22, kk23, kk31, kk32, kk33, size; size = (long)lx * (long)ly;

g = (char *)malloc(size);

if(g == NULL)

{

printf("error in allocating memory!\n");

return;

}

f = (char *)image;

for(i=0; i

{

for(j=0; j

{

kk=i*ly+j;

if(f[kk]!=0)

{

f[kk]=1;

g[kk]=f[kk];

}

}

}

counter = 1;

do

{

printf("%4d*",counter); counter++;

shori = 0;

for(i=0; i

{

for(j=0; j

kk = i*ly+j;

if(f[kk]<0)

f[kk] = 0;

g[kk]= f[kk];

}

}

for(i=1; i

{

for(j=1; j

kk=i*ly+j;

if(f[kk]!=1)

continue;

kk11 = (i-1)*ly+j-1;

kk12 = kk11 + 1;

kk13 = kk12 + 1;

kk21 = i*ly+j-1;

kk22 = kk21 + 1;

kk23 = kk22 + 1;

kk31 = (i+1)*ly+j-1;

kk32 = kk31 + 1;

kk33 = kk32 + 1;

if((g[kk12]&&g[kk21]&&g[kk23]&&g[kk32])!=0)

continue;

nrn = g[kk11] + g[kk12] + g[kk13] + g[kk21] + g[kk23] + g[kk31] + g[kk32] + g[kk33];

if(nrn <= 1)

{

f[kk22] = 2;

continue;

}

n[4] = f[kk11];

n[3] = f[kk12];

n[2] = f[kk13];

n[5] = f[kk21];

n[1] = f[kk23];

n[6] = f[kk31];

n[7] = f[kk32];

n[8] = f[kk33];

n[9] = n[1];

xx = 0;

for(k=1; k<8; k=k+2)

{

if((!n[k])&&(n[k+1]||n[k+2]))

xx++;

}

if(xx!=1)

{

f[kk22] = 2;

continue;

if(f[kk12] == -1)

{

f[kk12] = 0;

n[3] = 0;

xx = 0;

for(k=1; k<8; k=k+2)

{

if((!n[k])&&(n[k+1]||n[k+2])) xx++;

}

if(xx != 1)

{

f[kk12] = -1;

continue;

}

f[kk12] = -1;

n[3] = -1;

}

if(f[kk21]!=-1)

{

f[kk22] = -1;

shori = 1;

continue;

}

f[kk21] = 0;

n[5] = 0;

xx = 0;

for(k=1; k<8; k=k+2)

{

if((!n[k])&&(n[k+1]||n[k+2])) {

xx++;

}

}

if(xx == 1)