能走到边缘的都是在包围圈外面的，所以以边缘作为起点能搜到的就是包围圈里面的

用边缘作为起点

#include <bits/stdc++.h>
using namespace std;
int n;
int g[35][35];
queue<pair<int, int>> q;
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            cin >> g[i][j];
    // 起点入队
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            if (g[i][j] == 0 &&
                (i == 1 || i == n || j == 1 || j == n))
            {
                q.push({i, j});
                g[i][j] = 3;
            }
    // 只要队列非空就继续搜
    while (!q.empty())
    {
        // 取出队头
        pair<int, int> now = q.front();
        q.pop();
        // 考虑队头能扩展的位置
        for (int i = 0; i <= 3; i++)
        {
            int nx = now.first + dx[i];
            int ny = now.second + dy[i];
            // 只要能走到且没走过
            if (1 <= nx && nx <= n && 1 <= ny && ny <= n && g[nx][ny] == 0)
            {
                q.push({nx, ny});
                g[nx][ny] = 3;
            }
        }
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
            if (g[i][j] == 3)
                cout << 0 << " ";
            else if (g[i][j] == 0)
                cout << 2 << " ";
            else
                cout << g[i][j] << " ";
        cout << "\n";
    }
    return 0;
}

扩展第 0 行和第 n+1 行，把边缘连通起来

#include <bits/stdc++.h>
using namespace std;
int n;
int g[35][35];
queue<pair<int, int>> q;
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            cin >> g[i][j];
    // 起点入队
    q.push({0, 0});
    g[0][0] = 3;
    // 只要队列非空就继续搜
    while (!q.empty())
    {
        // 取出队头
        pair<int, int> now = q.front();
        q.pop();
        // 考虑队头能扩展的位置
        for (int i = 0; i <= 3; i++)
        {
            int nx = now.first + dx[i];
            int ny = now.second + dy[i];
            // 只要能走到且没走过
            if (0 <= nx && nx <= n + 1 && 0 <= ny && ny <= n + 1 && g[nx][ny] == 0)
            {
                q.push({nx, ny});
                g[nx][ny] = 3;
            }
        }
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
            if (g[i][j] == 3)
                cout << 0 << " ";
            else if (g[i][j] == 0)
                cout << 2 << " ";
            else
                cout << g[i][j] << " ";
        cout << "\n";
    }
    return 0;
}

真的修改原地图

#include <bits/stdc++.h>
using namespace std;
int n;
int a[35][35];
queue<pair<int, int>> q;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    // 边框（因为边框为 0,，所以不写也可以）
    for (int i = 0; i <= n + 1; i++)
        a[i][0] = a[0][i] = a[i][n + 1] = a[n + 1][i] = 0;
    // 原图
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            cin >> a[i][j];
    // 外面的 0 -> 3
    q.push(make_pair(0, 0));
    a[0][0] = 3;
    while (!q.empty())
    {
        pair<int, int> now = q.front();
        q.pop();
        for (int i = 0; i < 4; i++)
        {
            int nx = now.first + dx[i];
            int ny = now.second + dy[i];
            if (0 <= nx && nx <= n + 1 &&
                0 <= ny && ny <= n + 1 &&
                a[nx][ny] == 0)
            {
                a[nx][ny] = 3;
                q.push(make_pair(nx, ny));
            }
        }
    }
    // 0 -> 2
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            if (a[i][j] == 0)
                a[i][j] = 2;
    // 3 -> 0
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            if (a[i][j] == 3)
                a[i][j] = 0;
    // out
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
            cout << a[i][j] << " ";
        cout << "\n";
    }

    return 0;
}