由题意可得，最终校园的边长是 $n^k$。取极限数据时，$n^k=2401$，所以可以按照题意，暴力构建校园，最后用 DFS 求连通块数目即可。

#include<bits/stdc++.h>
using namespace std;
#define endl '\n' 
const int N = 2500;
int a[N][N], nw[N][N], res[N][N];
int siz;
bool vis[N][N];
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
void dfs(int x,int y)
{
    vis[x][y] = 1;
    for (int i = 0; i < 4; i++)
    {
        int xx = x + dx[i], yy = y + dy[i];
        if (xx >= 1 && xx <= siz && yy >= 1 && yy <= siz && res[xx][yy] && !vis[xx][yy])
            dfs(xx, yy);
    }
}
int main()
{
    freopen("exp.in","r",stdin);
    freopen("exp.out","w",stdout);
    
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    int n, k, c;
    cin >> n >> k >> c;
    for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) cin >> a[i][j];
    siz = 1, res[1][1] = 1;
    while(k--)
    {
        for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) // 枚举当前是被复制的第几块
        {
            for(int x = 1; x <= siz; x++) for(int y = 1; y <= siz; y++) // 枚举这一块中的所有点
            {
                if(a[i][j] == 1)
                    nw[(i - 1) * siz + x][(j - 1) * siz + y] = res[x][y];
                else
                    nw[(i - 1) * siz + x][(j - 1) * siz + y] = 0;
            }
        }
        siz *= n;
        for(int i = 1; i <= siz; i++) for(int j = 1; j <= siz; j++) res[i][j] = nw[i][j];
    }
    int ans = 0;
    for(int i = 1; i <= siz; i++) for(int j = 1; j <= siz; j++) if(res[i][j] && !vis[i][j])
    {
        ans++;
        dfs(i, j);
    }
    cout << ans;
    return 0;
}