一道连通块型dfs,精髓在于0，0一定是不是内环，所以在0，0 dfs,剩余被包住的就涂色

#include <bits/stdc++.h>
using namespace std;
int a[50][50];
bool vis[50][50];
int n;
int fx[4] = {0,0,1,-1};
int fy[4] = {1,-1,0,0};
void dfs(int x,int y)
{
	vis[x][y] = true;
	int tx,ty;
	for(int i=0; i<4; i++)
	{
		tx = x + fx[i];
		ty = y + fy[i];
		if(tx>=0&&tx<=n+1&&ty>=0&&ty<=n+1
		&&vis[tx][ty]==false&&a[tx][ty]==0)
		{
			dfs(tx,ty);
		}
	}
} 
int main()
{
	cin>>n;
	for(int i=1; i<=n; i++)
	{
		for(int j=1; j<=n; j++)
		{
			cin>>a[i][j];
		}
	}
	dfs(0,0);
	for(int i=1; i<=n; i++)
	{
		for(int j=1; j<=n; j++)
		{
			if(vis[i][j]==true||a[i][j]==1)
			{
				cout<<a[i][j];
			}
			else
			{
				cout<<"2";
			}
			cout<<" ";
		}
		cout<<endl;
	}
	
	return 0;
 }

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

// 定义方向数组，表示上下左右四个方向
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};

void bfs(int x, int y, vector<vector<int>>& grid, int n) {
    queue<pair<int, int>> q;
    q.push({x, y});
    grid[x][y] = -1; // 标记为与边界相连的 0

    while (!q.empty()) {
        auto [cx, cy] = q.front();
        q.pop();

        // 遍历四个方向
        for (int i = 0; i < 4; ++i) {
            int nx = cx + dx[i];
            int ny = cy + dy[i];

            // 检查是否在方阵范围内且为 0
            if (nx >= 0 && nx < n && ny >= 0 && ny < n && grid[nx][ny] == 0) {
                grid[nx][ny] = -1; // 标记为与边界相连的 0
                q.push({nx, ny});
            }
        }
    }
}

int main() {
    int n;
    cin >> n;
    vector<vector<int>> grid(n, vector<int>(n));

    // 输入方阵
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            cin >> grid[i][j];
        }
    }

    // 遍历边界，找到所有边界上的 0，并标记为 -1
    for (int i = 0; i < n; ++i) {
        if (grid[i][0] == 0) bfs(i, 0, grid, n); // 左边界
        if (grid[i][n - 1] == 0) bfs(i, n - 1, grid, n); // 右边界
    }
    for (int j = 0; j < n; ++j) {
        if (grid[0][j] == 0) bfs(0, j, grid, n); // 上边界
        if (grid[n - 1][j] == 0) bfs(n - 1, j, grid, n); // 下边界
    }

    // 遍历整个方阵，填充被 1 包围的 0 为 2
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            if (grid[i][j] == 0) {
                grid[i][j] = 2; // 填充为 2
            } else if (grid[i][j] == -1) {
                grid[i][j] = 0; // 恢复为 0
            }
        }
    }

    // 输出结果
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            cout << grid[i][j] << " ";
        }
        cout << endl;
    }

    return 0;
}