How should I implement a flood fill function to my c++ program? - c++

Question

Im currently looking to turn something like this:
.............
.............
..XXX.....X..
..XXX.....X..
..XXX........
..XXX........
..XXXXXXX....
..XXXXXXX....
..XXXXXXX....
.............
.............
into this:
.............
.............
..XXX.....O..
..XXX.....O..
..XXX........
..XXX........
..XXXXXXX....
..XXXXXXX....
..XXXXXXX....
.............
.............
with the user entering ./a.exe input4.txt floodfill 2 10 o
I believe im going to need to implement some recursion into the program to be able to only look at indexes that match that of the user pointer (including positions of up, down, left, and right) instead of reading the entire vector (which i wouldn't mind doing but dont see how i would begin doing so).
here is the code I have so far for the flood fill function:
void floodfilll(vector<vector<char>> &vec, int x, int y, char r, char dum)
{
int ii;
ii = 0;
int jj;
jj = 0;
for (int i = 0; i < vec.size(); i ++) {
for (int j = 0; j < vec[i].size(); j++) {
if (vec[i][j] == r) {
vec[i][j] == r;
if ((i + ii) > 0) {
if (vec[i-1][j] == r)
vec[i-1][j] = dum;
vec[i][j] == r;
ii--;
floodfilll(vec, x + ii, y, r, dum);
}
if ((j + jj) > 0) {
if(vec[i][j-1] != r)
vec[i][j-1] = dum;
vec[i][j] == r;
jj--;
floodfilll(vec, x, y + jj, r, dum);
}
if ((i + ii)<vec.size()) {
if (vec[i+1][j] != r)
vec[i+1][j] = dum;
vec[i][j] == r;
ii++;
floodfilll(vec, x + ii, y, r, dum);
}
if ((j + jj)<vec[i].size()) {
if (vec[i][j+1] != r)
vec[i][j+1] = dum;
vec[i][j] == r;
jj++;
floodfilll(vec, x, y + jj, r, dum);
}
}
}
replacee(vec, dum, r);
}
}
NOTE: Im using a function called replacee to replace Var dum, with Var R. Var dum is assigned 'i' and r is 'X'.
Also, the text file is parsed as a 2d vector of char's (char)**
Its just the way the rest of my program is based. Here is the replace function:
void replacee(vector<vector<char>> &vec, char oldd, char neww)
{
for (vector<char> &v : vec) // reference to innver vector
{
replace(v.begin(), v.end(), oldd, neww); // standard library algorithm
}
}
This is the int main file im using:
int main(int argc, char* argv[]) {
fstream fin; char ch;
string name (argv[1]); //File Name.
vector<vector<char>> data;
// 2D Vector.
vector<char> temp;
// Temporary vector to be pushed
// into vec, since its a vector of vectors.
fin.open(name.c_str(),ios::in);
// Assume name as an arbitary file.
string argument2 (argv[2]);
while(fin)
{
ch = fin.get();
if(ch!='\n') {
temp.push_back(ch);
}
else
{
data.push_back(temp);
temp.clear();
}
}
if (argument2 == "floodfill") {
string argument3 (argv[3]);
string argument4 (argv[4]);
string argument5 (argv[5]);
int x = 0;
int y = 0;
stringstream xx(argument3);
stringstream yy(argument4);
xx >> x;
yy >> y;
floodfilll(data, x, y, argument5[0], 'i');
for (int m = 0; m < data.size(); m ++) {
for (int n = 0; n < data[m].size(); n++) {
cout << data[m][n];
}
cout << endl;
}
}
fin.close();
}
Sorry if it looks like im just pasting code for grabs, this is incase anyone has a solution outside my mode of thought. The int main and replacee functions work as intended. I just need help coming up with a way to make floodfilll work correctly.
This is the output i get with my code:
$ ./a.exe input4.txt floodfill 2 10 o
.............
.............
..XXX.....X..
..XXX.....X..
..XXX........
..XXX........
..XXXXXXX....
..XXXXXXX....
..XXXXXXX....
.............

Answer 1

Why do you iterate over the whole field in each recursion?
Normally, flood filling works as follows:
You have a specific starting point.
You fill this starting point with the colour intended
You check for each of the four (or 8, if you consider diagonals as well) neighbours, if they have the same colour as the starting point originally had; if so, you continue with recursively.
So an implementation might look like this:
void floodfill
(
std::vector<std::vector<char>>& v,
unsigned int x, unsigned int y, char r
)
{
char p = v[x][y];
v[x][y] = r;
if(x > 0 && v[x - 1][y] == p)
floodfill(v, x - 1, y, r);
if(x + 1 < v.size() && v[x + 1][y] == p)
floodfill(v, x + 1, y, r);
if(y > 0 && v[x][y - 1] == p)
floodfill(v, x, y - 1, r);
if(y + 1 < v[x].size() && v[x][y + 1] == p)
floodfill(v, x, y + 1, r);
}
Note that I did not check for the case of colour to fill being the same as the one of the starting pixel, neither did I initially check the range check of x and y. For efficiency, I wouldn't add these checks in the recursive function, but in a specific entry function starting the recursion, so they are done only once when needed and not needlessly repeated.

Answer 2

One option is to use recursion, as suggested by the other answer. However, personally I prefer avoiding recursion where it is not necessary. An alternative, is a queue-based approach.
void floodfill (std::vector<std::vector<char>>& v, unsigned int x, unsigned int y, char r) {
char init = v[x][y];
if (init == r) return; //We don't want to get caught in an endless loop.
if (x >= v.size() || y >= v[x].size) return; //Index out of bounds.
std::queue<std::pair<unsigned int, unsigned int>> q;
q.push(std::make_pair(x, y)); //Push the initial position into the queue.
v[x][y] = r; //Replace the initial point.
while (!q.empty()) {//Keep looking at relevant neighbours so long as there is something in the queue.
auto pt = q.front(); //Collect the first entry.
q.pop(); //Remove it, we don't want to keep processing the same point.
//Now add neighbours if they match our initial point.
if(pt.first > 0 && v[pt.first - 1][pt.second] == init)
q.push(std::make_pair(pt.first - 1, pt.second);
v[pt.first - 1][pt.second] = r; //Replace the value here to avoid pushing the same point twice.
if(pt.first + 1 < v.size() && v[pt.first + 1][pt.second] == init)
q.push(std::make_pair(pt.first + 1, pt.second);
v[pt.first + 1][pt.second] = r;
if(pt.second > 0 && v[pt.first][pt.second - 1] == init)
q.push(std::make_pair(pt.first, pt.second - 1);
v[pt.first][pt.second - 1] = r;
if(pt.second + 1 < v[pt.first].size() && v[pt.first][pt.second + 1] == init)
q.push(std::make_pair(pt.first, pt.second + 1);
v[pt.first][pt.second + 1] = r;
}
}
This gives you a BFS-like flood-fill pattern without recursion. Alternatively you could also use a stack instead of the queue, then the flood-fill would behave more like a DFS (much more similar to what the recursive pattern will do). It might even perform a little better than the queue, given the data structure is a little simpler.