题解

dfs递归，3个数组，vis存是否被找过，dx、dy反向数组，a存地图
代码AC后点个赞吧！
参考代码：

#include <bits/stdc++.h>
using namespace std;
int n,m,cnt = 1;
int dx[4] = {-1,1,0,0};
int dy[4] = {0,0,-1,1};
char a[25][25];
bool vis[25][25] = {0};
void dfs(int x,int y){
    int nx,ny;
    vis[x][y] = true;
    for (int l = 0;l < 4;l++){
        nx = x + dx[l];
        ny = y + dy[l];
        if (nx > n || ny < 1 || ny > m || nx < 1 || vis[nx][ny] || a[nx][ny] == '#') continue;
        cnt++;
        dfs(nx,ny);
    }
}
int main(){
    cin >> m >> n;
    for (int i = 1;i <= n;i++){
        for (int j = 1;j <= m;j++){
            cin >> a[i][j];
        }
    }
    for (int i = 1;i <= n;i++){
        for (int j = 1;j <= m;j++){
            if (a[i][j] == '@'){
                a[i][j] = '.';
                dfs(i,j);
                cout << cnt;
            }
        }
    }
    return 0;
}