One matrix is given; the matrix is representing the one screen. Each element (i, j) of the screen is denoted as a pixel, the color of that pixel is marked with different numbers. In this algorithm, the pixels will be filled with new color when it is already in selected previous color. If the previous color is not the previous color, that pixel will not be filled. After filling a pixel, it will check for its up, down, left and right pixels to do the same.
The idea is really simple, first, we check whether the selected position is colored with the previous color or not, of not, the algorithm will not work. Otherwise, it will fill that pixel with new color and recur for its four neighbors.
Input and Output
Input: The screen matrix: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 0 0 1 1 0 1 1 1 2 2 2 2 0 1 0 1 1 1 2 2 0 1 0 1 1 1 2 2 2 2 0 1 1 1 1 1 2 1 1 1 1 1 1 1 2 2 1 Output: Screen matrix after flood fill 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 0 0 1 1 0 1 1 1 3 3 3 3 0 1 0 1 1 1 3 3 0 1 0 1 1 1 3 3 3 3 0 1 1 1 1 1 3 1 1 1 1 1 1 1 3 3 1
Algorithm
fillScreen(x, y, prevColor, newColor)
Input: The (x,y) coordinate to start, previous color, and new color.
Output − Screen after changing the color from previous to new, if possible.
Begin if (x, y) not in the screen range, then return if color of (x, y) ≠ prevColor, then return screen[x, y] := newColor fillScreen(x+1, y, prevColor, newColor) fillScreen(x-1, y, prevColor, newColor) fillScreen(x, y+1, prevColor, newColor) fillScreen(x, y-1, prevColor, newColor) End
Example
#include<iostream>
#define M 8
#define N 8
using namespace std;
int screen[M][N] = { //the screen dimention and colors
{1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 0, 0},
{1, 0, 0, 1, 1, 0, 1, 1},
{1, 2, 2, 2, 2, 0, 1, 0},
{1, 1, 1, 2, 2, 0, 1, 0},
{1, 1, 1, 2, 2, 2, 2, 0},
{1, 1, 1, 1, 1, 2, 1, 1},
{1, 1, 1, 1, 1, 2, 2, 1}
};
void fillScreen(int x, int y, int prevColor, int newColor) { //replace previous color of (x,y), with new color
if (x < 0 || x >= M || y < 0 || y >= N) //when point exceeds the screen
return;
if (screen[x][y] != prevColor) //if the point(x,y) are not containing prevColor, do nothing
return;
screen[x][y] = newColor; //update the color
fillScreen(x+1, y, prevColor, newColor); //for the right of (x,y)
fillScreen(x-1, y, prevColor, newColor); //for the left of (x,y)
fillScreen(x, y+1, prevColor, newColor); //for the top of (x,y)
fillScreen(x, y-1, prevColor, newColor); //for the bottom of (x,y)
}
void floodFill(int x, int y, int newColor) {
int prevColor = screen[x][y]; //take the color before replacing with new color
fillScreen(x, y, prevColor, newColor);
}
int main() {
int x = 4, y = 4, newColor = 3;
cout << "Previous screen: "<< endl;
for (int i=0; i<M; i++) {
for (int j=0; j<N; j++)
cout << screen[i][j] << " ";
cout << endl;
}
cout << endl;
floodFill(x, y, newColor); //start from (4, 4), with new color 3
cout << "Updated screen: "<< endl;
for (int i=0; i<M; i++) {
for (int j=0; j<N; j++)
cout << screen[i][j] << " ";
cout << endl;
}
}Output
Previous screen 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 0 0 1 1 0 1 1 1 2 2 2 2 0 1 0 1 1 1 2 2 0 1 0 1 1 1 2 2 2 2 0 1 1 1 1 1 2 1 1 1 1 1 1 1 2 2 1 Updated screen: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 0 0 1 1 0 1 1 1 3 3 3 3 0 1 0 1 1 1 3 3 0 1 0 1 1 1 3 3 3 3 0 1 1 1 1 1 3 1 1 1 1 1 1 1 3 3 1