Planning garlic guest (outing) (DFS and BFS solution of dyeing block problem)

Garlic king and his friends are going to Summoner Canyon for an outing at the weekend . They found that the map of Summoner canyon was made up of squares , Some lattices are covered with grass , There is plenty of open space . The grass passes up, down, left and right 4 Expand other grasses in one direction to form a grassland , The grid in any piece of grass is grass , And all grids can be connected up, down, left and right . If you use ’#‘ Stands for grass ,’.' On behalf of the open space , There are... In the canyon below 2 A piece of grass .

​##..
..##

In the same grass 2 Individuals can see each other , The open space can't see the people in the grass . They found a friend missing , Now we need to find it separately , Everyone is responsible for a piece of grass , Garlic king wants to know how many people they need at least .
Input format
First line input n, m (1 <= n,m <= 100)(1≤n,m≤100) Indicates the size of the Canyon .

Next input n The line string represents the terrain of the Canyon .

Input format
At least how many people are needed to output .

The sample input

5 6
.#....
..#...
..#..#
...##.
.#....

Sample output

5

DFS solution

#include<bits/stdc++.h>
using namespace std;
int n,m;
char s[105][105];
bool vis[105][105];
int ans;
int dir[4][2]={
    {
    1,0},{
    -1,0},{
    0,1},{
    0,-1}};
void dfs(int x,int y){
    
    vis[x][y]=true;
    for(int i=0;i<4;i++){
    
        int tx=x+dir[i][0];
        int ty=y+dir[i][1];
        if(tx<n&&tx>=0&&ty<m&&ty>=0&&!vis[tx][ty]&&s[tx][ty]=='#'){
    
            dfs(tx,ty);
        }
    }
}
int main(){
    
    cin>>n>>m;
    for(int i=0;i<n;i++){
    
        scanf("%s",s[i]);
    }
    for(int i=0;i<n;i++){
    
        for(int j=0;j<m;j++){
    
            if(!vis[i][j]&&s[i][j]=='#'){
    
                dfs(i,j);
                ans++;
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}

BFS solution

#include<bits/stdc++.h>
using namespace std;
char s[105][105];
bool vis[105][105];
int dir[4][2]={
    {
    1,0},{
    -1,0},{
    0,-1},{
    0,1}};
int n,m,ans;
struct Node{
    
    int x,y;
};
void bfs(int x,int y){
    
    queue<Node> q;
    q.push({
    x,y});
    vis[x][y]=true;
    while(!q.empty()){
    
        Node now=q.front();
        q.pop();
        for(int i=0;i<4;i++){
    
            int tx=now.x+dir[i][0];
            int ty=now.y+dir[i][1];
            if(tx<n&&tx>=0&&ty<m&&ty>=0&&!vis[tx][ty]&&s[tx][ty]=='#'){
    
                q.push({
    tx,ty});
                vis[tx][ty]=true;
            }
        }
    }
}
int main(){
    
    cin>>n>>m;
    for(int i=0;i<n;i++){
    
        cin>>s[i];
    }
    for(int i=0;i<n;i++){
    
        for(int j=0;j<m;j++){
    
            if(!vis[i][j]&&s[i][j]=='#'){
    
                bfs(i,j);
                ans++;
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}