#c #algorithm #graph-algorithm
#c #алгоритм #граф-алгоритм
Вопрос:
Я нашел здесь реализацию алгоритма Хошена-Копельмана, но он проверяет соседей только вверх и влево, что означает, что диагональное соединение не считается соединением.
Как я могу улучшить этот код, чтобы даже диагональное соединение считалось соединением?
В следующем примере я ожидаю 1 объект, а не 7 объектов:
4 5
1 0 1 0 1
0 1 0 1 0
1 0 1 0 0
0 0 1 0 0
--input--
1 0 1 0 1
0 1 0 1 0
1 0 1 0 0
0 0 1 0 0
--output--
1 0 2 0 3
0 4 0 5 0
6 0 7 0 0
0 0 7 0 0
HK reports 7 clusters found
Это реализация (полный код можно найти здесь):
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
/* Implementation of Union-Find Algorithm */
/* The 'labels' array has the meaning that labels[x] is an alias for the label x; by
following this chain until x == labels[x], you can find the canonical name of an
equivalence class. The labels start at one; labels[0] is a special value indicating
the highest label already used. */
int* labels;
int n_labels = 0; /* length of the labels array */
/* uf_find returns the canonical label for the equivalence class containing x */
int uf_find(int x)
{
int y = x;
while (labels[y] != y)
y = labels[y];
while (labels[x] != x)
{
int z = labels[x];
labels[x] = y;
x = z;
}
return y;
}
/* uf_union joins two equivalence classes and returns the canonical label of the resulting class. */
int uf_union(int x, int y)
{
return labels[uf_find(x)] = uf_find(y);
}
/* uf_make_set creates a new equivalence class and returns its label */
int uf_make_set(void)
{
labels[0] ;
assert(labels[0] < n_labels);
labels[labels[0]] = labels[0];
return labels[0];
}
/* uf_intitialize sets up the data structures needed by the union-find implementation. */
void uf_initialize(int max_labels)
{
n_labels = max_labels;
labels = calloc(sizeof(int), n_labels);
labels[0] = 0;
}
/* uf_done frees the memory used by the union-find data structures */
void uf_done(void)
{
n_labels = 0;
free(labels);
labels = 0;
}
/* End Union-Find implementation */
#define max(a,b) (a>b?a:b)
#define min(a,b) (a>b?b:a)
/* print_matrix prints out a matrix that is set up in the "pointer to pointers" scheme
(aka, an array of arrays); this is incompatible with C's usual representation of 2D
arrays, but allows for 2D arrays with dimensions determined at run-time */
void print_matrix(int** matrix, int m, int n)
{
for (int i = 0; i < m; i )
{
for (int j = 0; j < n; j )
printf("= ", matrix[i][j]);
printf("n");
}
}
/* Label the clusters in "matrix". Return the total number of clusters found. */
int hoshen_kopelman(int** matrix, int m, int n)
{
uf_initialize(m * n / 2);
/* scan the matrix */
for (int y = 0; y < m; y )
{
for (int x = 0; x < n; x )
{
if (matrix[y][x])
{ // if occupied ...
int up = (y == 0 ? 0 : matrix[y - 1][x]); // look up
int left = (x == 0 ? 0 : matrix[y][x - 1]); // look left
switch (!!up !!left)
{
case 0:
matrix[y][x] = uf_make_set(); // a new cluster
break;
case 1: // part of an existing cluster
matrix[y][x] = max(up, left); // whichever is nonzero is labelled
break;
case 2: // this site binds two clusters
matrix[y][x] = uf_union(up, left);
break;
}
}
}
}
/* apply the relabeling to the matrix */
/* This is a little bit sneaky.. we create a mapping from the canonical labels
determined by union/find into a new set of canonical labels, which are
guaranteed to be sequential. */
int* new_labels = calloc(sizeof(int), n_labels); // allocate array, initialized to zero
for (int i = 0; i < m; i )
for (int j = 0; j < n; j )
if (matrix[i][j])
{
int x = uf_find(matrix[i][j]);
if (new_labels[x] == 0)
{
new_labels[0] ;
new_labels[x] = new_labels[0];
}
matrix[i][j] = new_labels[x];
}
int total_clusters = new_labels[0];
free(new_labels);
uf_done();
return total_clusters;
}
/* This procedure checks to see that any occupied neighbors of an occupied site
have the same label. */
void check_labelling(int** matrix, int m, int n)
{
int N, S, E, W;
for (int i = 0; i < m; i )
for (int j = 0; j < n; j )
if (matrix[i][j])
{
N = (i == 0 ? 0 : matrix[i - 1][j]);
S = (i == m - 1 ? 0 : matrix[i 1][j]);
E = (j == n - 1 ? 0 : matrix[i][j 1]);
W = (j == 0 ? 0 : matrix[i][j - 1]);
assert(N == 0 || matrix[i][j] == N);
assert(S == 0 || matrix[i][j] == S);
assert(E == 0 || matrix[i][j] == E);
assert(W == 0 || matrix[i][j] == W);
}
}
/* The sample program reads in a matrix from standard input, runs the HK algorithm on
it, and prints out the results. The form of the input is two integers giving the
dimensions of the matrix, followed by the matrix elements (with data separated by
whitespace).
a sample input file is the following:
8 8
1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 1
1 0 0 0 0 1 0 1
1 0 0 1 0 1 0 1
1 0 0 1 0 1 0 1
1 0 0 1 1 1 0 1
1 1 1 1 0 0 0 1
0 0 0 1 1 1 0 1
this sample input gives the following output:
--input--
1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 1
1 0 0 0 0 1 0 1
1 0 0 1 0 1 0 1
1 0 0 1 0 1 0 1
1 0 0 1 1 1 0 1
1 1 1 1 0 0 0 1
0 0 0 1 1 1 0 1
--output--
1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 1
2 0 0 0 0 2 0 1
2 0 0 2 0 2 0 1
2 0 0 2 0 2 0 1
2 0 0 2 2 2 0 1
2 2 2 2 0 0 0 1
0 0 0 2 2 2 0 1
HK reports 2 clusters found
*/
int main(int argc, char** argv)
{
int m, n;
int** matrix;
/* Read in the matrix from standard input
The whitespace-deliminated matrix input is preceeded
by the number of rows and number of columns */
while (2 == scanf_s("%d %d", amp;m, amp;n))
{ // m = rows, n = columns
matrix = (int**)calloc(m, sizeof(int*));
for (int i = 0; i < m; i )
{
matrix[i] = (int*)calloc(n, sizeof(int));
for (int j = 0; j < n; j )
scanf_s("%d", amp;(matrix[i][j]));
}
printf_s(" --input-- n");
print_matrix(matrix, m, n);
printf(" --output-- n");
/* Process the matrix */
int clusters = hoshen_kopelman(matrix, m, n);
/* Output the result */
print_matrix(matrix, m, n);
check_labelling(matrix, m, n);
printf("HK reports %d clusters foundn", clusters);
for (int i = 0; i < m; i )
free(matrix[i]);
free(matrix);
}
return 0;
}
Я попытался изменить функцию hoshen_kopelman , как описано ниже, но я по-прежнему получаю 2 объекта вместо 1:
int hoshen_kopelman(int** matrix, int m, int n)
{
uf_initialize(m * n / 2);
/* scan the matrix */
for (int y = 0; y < m; y )
{
for (int x = 0; x < n; x )
{
if (matrix[y][x])
{ // if occupied ...
int up = (y == 0 ? 0 : matrix[y - 1][x]); // look up
int left = (x == 0 ? 0 : matrix[y][x - 1]); // look left
// ----------- THE NEW CODE -------------
if (x > 0)
{
if (up == 0 amp;amp; y > 0) // left up
up = matrix[y - 1][x - 1];
if (left == 0 amp;amp; y < m - 1) // left down
left = matrix[y 1][x - 1];
}
// ---------- END NEW CODE --------------
switch (!!up !!left)
{
case 0:
matrix[y][x] = uf_make_set(); // a new cluster
break;
case 1: // part of an existing cluster
matrix[y][x] = max(up, left); // whichever is nonzero is labelled
break;
case 2: // this site binds two clusters
matrix[y][x] = uf_union(up, left);
break;
}
}
}
}
/* apply the relabeling to the matrix */
/* This is a little bit sneaky.. we create a mapping from the canonical labels
determined by union/find into a new set of canonical labels, which are
guaranteed to be sequential. */
int* new_labels = calloc(sizeof(int), n_labels); // allocate array, initialized to zero
for (int i = 0; i < m; i )
for (int j = 0; j < n; j )
if (matrix[i][j])
{
int x = uf_find(matrix[i][j]);
if (new_labels[x] == 0)
{
new_labels[0] ;
new_labels[x] = new_labels[0];
}
matrix[i][j] = new_labels[x];
}
int total_clusters = new_labels[0];
free(new_labels);
uf_done();
return total_clusters;
}
Теперь получен следующий результат (я ожидаю 1 и получил 2):
4 5
1 0 1 0 1
0 1 0 1 0
1 0 1 0 0
0 0 1 0 0
--input--
1 0 1 0 1
0 1 0 1 0
1 0 1 0 0
0 0 1 0 0
--output--
1 0 1 0 1
0 1 0 1 0
2 0 1 0 0
0 0 1 0 0
HK reports 2 clusters found
Как правильно исправить код, чтобы проверить все 8 соседей?
Ответ №1:
Я ввел вас в заблуждение, сказав проверить вниз-налево. Алгоритм полагается на то, что текущий узел, который он проверяет, находится после всех соседей, которых он проверяет. Итак, вам нужно проверить влево, вверх, вверх-влево и вверх-вправо. Вы можете использовать это вместо своего нового кода:
if (y > 0)
{
if (left == 0 amp;amp; x > 0) // left up
left = matrix[y - 1][x - 1];
if (up == 0 amp;amp; x < n-1) // right up
up = matrix[y - 1][x 1];
}