使用我的 DFS 计算连接块的错误结果
Wrong results on counting connected blocks with my DFS
我有一个随机生成的 5 x 5 矩阵,由 1 和 0 组成,我应该找到 1 的块。
一个1
本身就是一个方块,如果它从任何方向连接到另一个1
,我的算法应该找到所有连接的1并将它们全部算作一个方块。
问题是我无法获得一致的结果,有时计数器甚至会达到 -1
。
这是主要的,我已经删除了随机化函数,只是放置了一个程序有问题的数组。
struct parent {
int row;
int col;
}
int ar[5][5] = { 0,0,0,0,1,
1,0,0,0,1,
0,0,0,0,0,
0,0,0,0,0,
0,0,0,0,0 };
int countblock=0,num=0;
struct parent pos[30] = { 0 };
srand(time(NULL));
printf(" ---Find the connected blocks--- \n\n");
printf("Creating a random array:\n");
//random5x5(ar);
for (int row = 0, col; row < 5; row++)
{
for (col = 0; col < 5; col++)
{
if (ar[row][col] == 1)
{
countblock++;
ar[row][col] = -1;
flag = 0;
check(ar, row, col, pos, &num);
}
printf("%d ", ar[row][col]); // Had problems with this loop not checking 1's
} //So I printed them to try to make sure
printf("\n");
}
printf("\n\n Found the connected blocks! I see only %d of them.", countblock);
return 0;
而 DFS 函数应该标记它在任何方向上找到的 1
int flag=0; // declared in global
void check(int ar[5][5], int row, int col, struct parent pos[30],int *num){
if (col - 1 >= 0 && row - 1 >= 0) //Checking Upper-left
{
if (ar[row - 1][col - 1] == 1) {
ar[row - 1][col - 1] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //Store this position
check(ar, row - 1, col - 1, pos, num); //Jump to the found 1
}
}
if (col - 1 >= 0) //Checking Left
{
if (ar[row][col - 1] == 1) {
ar[row][col - 1] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //Store this position
check(ar, row, col - 1, pos, num); //Jump to the found 1
}
}
if (col - 1 >= 0 && row + 1 <= 4) //Checking Bottom-Left
{
if (ar[row + 1][col - 1] == 1) {
ar[row + 1][col - 1] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //
check(ar, row + 1, col - 1, pos, num); //
}
}
if (row + 1 <= 4) //Checking Bottom
{
if (ar[row + 1][col] == 1) {
ar[row + 1][col] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //
check(ar, row + 1, col, pos, num); //
}
}
if (col + 1 <= 4 && row + 1 <= 4) //Checking Bottom-Right
{
if (ar[row + 1][col + 1] == 1) {
ar[row + 1][col + 1] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //
check(ar, row + 1, col + 1, pos, num); //
}
}
if (col + 1 <= 4) //Checking Right
{
if (ar[row][col + 1] == 1) {
ar[row][col + 1] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //
check(ar, row, col + 1, pos, num); //
}
}
if (col + 1 <= 4 && row - 1 >= 0) //Checking Upper-Right
{
if (ar[row - 1][col + 1] == 1) {
ar[row - 1][col + 1] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //
check(ar, row - 1, col + 1, pos, num); //
}
}
if (row - 1 >= 0) //Checking Up
{
if (ar[row - 1][col] == 1) {
ar[row - 1][col] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //
check(ar, row - 1, col, pos, num); //
}
}
if (*num == 0) //There were no 1's around
return;
else //Reached the end of the path, call back
{
flag++;
if (flag >= 2) //On first call back the pos is still on parent call
(*num)-=1; // On second call it should go back
check(ar, pos[*num].row, pos[*num].col, pos, num);
}
编辑:这是错误
flag++;
if (flag >= 2) //On first call back the pos is still on parent call
(*num)-=1; // On second call it should go back
check(ar, pos[*num].row, pos[*num].col, pos, num);
通过制作修复它
flag++;
if (flag >= 2)
(*num)-=1;
if (*num == 0)
return;
check(ar, pos[*num].row, pos[*num].col, pos, num);
基本上 *num 会在 else
中变为零并调用错误的父位置。
进行另一次检查使它起作用。
您所做的范围检查与您检查的邻居之间存在一些不匹配。例如(这不是唯一的)看这个:
if (row + 1 <= 4) //Checking Bottom
{
if (ar[row][col + 1] == 1) {
ar[row][col + 1] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //
check(ar, row, col + 1, pos, num); //
}
}
您检查 row+1
在范围内...但随后您递归 col+1
。
这种错误会导致内存损坏(因为您最终写入了您不希望写入的位置),这几乎会导致任何事情发生:-)。
如果您仔细检查这种不匹配,修正您发现的任何问题,然后重试,会发生什么情况?
我有一个随机生成的 5 x 5 矩阵,由 1 和 0 组成,我应该找到 1 的块。
一个1
本身就是一个方块,如果它从任何方向连接到另一个1
,我的算法应该找到所有连接的1并将它们全部算作一个方块。
问题是我无法获得一致的结果,有时计数器甚至会达到 -1
。
这是主要的,我已经删除了随机化函数,只是放置了一个程序有问题的数组。
struct parent {
int row;
int col;
}
int ar[5][5] = { 0,0,0,0,1,
1,0,0,0,1,
0,0,0,0,0,
0,0,0,0,0,
0,0,0,0,0 };
int countblock=0,num=0;
struct parent pos[30] = { 0 };
srand(time(NULL));
printf(" ---Find the connected blocks--- \n\n");
printf("Creating a random array:\n");
//random5x5(ar);
for (int row = 0, col; row < 5; row++)
{
for (col = 0; col < 5; col++)
{
if (ar[row][col] == 1)
{
countblock++;
ar[row][col] = -1;
flag = 0;
check(ar, row, col, pos, &num);
}
printf("%d ", ar[row][col]); // Had problems with this loop not checking 1's
} //So I printed them to try to make sure
printf("\n");
}
printf("\n\n Found the connected blocks! I see only %d of them.", countblock);
return 0;
而 DFS 函数应该标记它在任何方向上找到的 1
int flag=0; // declared in global
void check(int ar[5][5], int row, int col, struct parent pos[30],int *num){
if (col - 1 >= 0 && row - 1 >= 0) //Checking Upper-left
{
if (ar[row - 1][col - 1] == 1) {
ar[row - 1][col - 1] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //Store this position
check(ar, row - 1, col - 1, pos, num); //Jump to the found 1
}
}
if (col - 1 >= 0) //Checking Left
{
if (ar[row][col - 1] == 1) {
ar[row][col - 1] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //Store this position
check(ar, row, col - 1, pos, num); //Jump to the found 1
}
}
if (col - 1 >= 0 && row + 1 <= 4) //Checking Bottom-Left
{
if (ar[row + 1][col - 1] == 1) {
ar[row + 1][col - 1] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //
check(ar, row + 1, col - 1, pos, num); //
}
}
if (row + 1 <= 4) //Checking Bottom
{
if (ar[row + 1][col] == 1) {
ar[row + 1][col] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //
check(ar, row + 1, col, pos, num); //
}
}
if (col + 1 <= 4 && row + 1 <= 4) //Checking Bottom-Right
{
if (ar[row + 1][col + 1] == 1) {
ar[row + 1][col + 1] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //
check(ar, row + 1, col + 1, pos, num); //
}
}
if (col + 1 <= 4) //Checking Right
{
if (ar[row][col + 1] == 1) {
ar[row][col + 1] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //
check(ar, row, col + 1, pos, num); //
}
}
if (col + 1 <= 4 && row - 1 >= 0) //Checking Upper-Right
{
if (ar[row - 1][col + 1] == 1) {
ar[row - 1][col + 1] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //
check(ar, row - 1, col + 1, pos, num); //
}
}
if (row - 1 >= 0) //Checking Up
{
if (ar[row - 1][col] == 1) {
ar[row - 1][col] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //
check(ar, row - 1, col, pos, num); //
}
}
if (*num == 0) //There were no 1's around
return;
else //Reached the end of the path, call back
{
flag++;
if (flag >= 2) //On first call back the pos is still on parent call
(*num)-=1; // On second call it should go back
check(ar, pos[*num].row, pos[*num].col, pos, num);
}
编辑:这是错误
flag++;
if (flag >= 2) //On first call back the pos is still on parent call
(*num)-=1; // On second call it should go back
check(ar, pos[*num].row, pos[*num].col, pos, num);
通过制作修复它
flag++;
if (flag >= 2)
(*num)-=1;
if (*num == 0)
return;
check(ar, pos[*num].row, pos[*num].col, pos, num);
基本上 *num 会在 else
中变为零并调用错误的父位置。
进行另一次检查使它起作用。
您所做的范围检查与您检查的邻居之间存在一些不匹配。例如(这不是唯一的)看这个:
if (row + 1 <= 4) //Checking Bottom
{
if (ar[row][col + 1] == 1) {
ar[row][col + 1] = -1;
(*num)+=1;
pos[*num].row = row; pos[*num].col = col; //
check(ar, row, col + 1, pos, num); //
}
}
您检查 row+1
在范围内...但随后您递归 col+1
。
这种错误会导致内存损坏(因为您最终写入了您不希望写入的位置),这几乎会导致任何事情发生:-)。
如果您仔细检查这种不匹配,修正您发现的任何问题,然后重试,会发生什么情况?