Flood Fill

Resources

Introduction

Flood fill is an algorithm that identifies and labels the connected component that a particular cell belongs to in a multidimensional array.

For example, suppose that we want to split the following grid into components of connected cells with the same number.

Let's start the floodfill from the top-left cell. The color scheme will be red for the node currently being processed, blue for nodes already visited, and uncolored for nodes not yet visited.

As opposed to an explicit graph where the edges are given, a grid is an implicit graph. This means that the neighbors are just the nodes directly adjacent in the four cardinal directions.

Usually, grids given in problems will be $N$ by $M$, so the first line of the input contains the numbers $N$ and $M$. In this example, we will use an two-dimensional integer array to store the grid, but depending on the problem, a two-dimensional character array or a two-dimensional boolean array may be more appropriate. Then, there are $N$ rows, each with $M$ numbers containing the contents of each square in the grid. Example input might look like the following (varies between problems):

3 4
1 1 2 1
2 3 2 1
1 3 3 3

And we’ll want to input the grid as follows:

int grid[MAXN][MAXM];
int n, m;

int main(){
	cin >> n >> m;
	for(int i = 0; i < n; i++){
		for(int j = 0; j < m; j++){
			cin >> grid[i][j];
		}
	}
	return 0;
}

static int[][] grid;
static int n, m;

public static void main(String[] args){
	int n = r.nextInt();
	int m = r.nextInt();
	grid = new int[n][m];
	for(int i = 0; i < n; i++){
		for(int j = 0; j < m; j++){
			grid[i][j] = r.nextInt();
		}
	}
}

Implementation

When doing floodfill, we will maintain an $N\times M$ array of booleans to keep track of which squares have been visited, and a global variable to maintain the size of the current component we are visiting. Make sure to store the grid, the visited array, dimensions, and the current size variable globally.

This means that we want to recursively call the search function from the squares above, below, and to the left and right of our current square. The algorithm to find the size of a connected component in a grid using floodfill is as follows (we’ll also maintain a 2D visited array).

The code below shows the global/static variables we need to maintain while doing floodfill, and the floodfill algorithm itself.

int grid[MAXN][MAXM]; // the grid itself
int n, m; // grid dimensions, rows and columns
bool visited[MAXN][MAXM]; // keeps track of which nodes have been visited
int currentSize = 0; // reset to 0 each time we start a new component

void floodfill(int r, int c, int color){
	if(r < 0 || r >= n || c < 0 || c >= m) return; // if outside grid
	if(grid[r][c] != color) return; // wrong color
	if(visited[r][c]) return; // already visited this square
	visited[r][c] = true; // mark current square as visited

static int[][] grid; // the grid itself
static int n, m; // grid dimensions, rows and columns
static boolean[][] visited; // keeps track of which nodes have been visited
static int currentSize = 0; // reset to 0 each time we start a new component

public static void main(String[] args){
	/**
	* input code and other problem-specific stuff here
	*/
	for(int i = 0; i < n; i++){

Example - Counting Rooms

Implementation

#include <bits/stdc++.h>
using namespace std;

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)

const int xd[4] = {0,1,0,-1}, yd[4] = {1,0,-1,0};

int n,m;
string g[2500];

import java.io.*;
import java.util.*;

public class test{
	public static boolean[][] v;
	public static char[][] g;
	public static final int xd[] = {0,1,0,-1}, yd[] = {1,0,-1,0};
	public static int N, M;
	static class Pair{
		public int a, b;

Problems