P1162 填涂颜色

# 填涂颜色
0 0 0 0 0 0
0 0 0 1 1 1
0 1 1 0 0 1
1 1 0 0 0 1
1 0 0 1 0 1
1 1 1 1 1 1
```
```plain
0 0 0 0 0 0
0 0 0 1 1 1
0 1 1 2 2 1
1 1 2 2 2 1
1 2 2 1 2 1
1 1 1 1 1 1
```

## 样例 #1

### 样例输入 #1

```
6
0 0 0 0 0 0
0 0 1 1 1 1
0 1 1 0 0 1
1 1 0 0 0 1
1 0 0 0 0 1
1 1 1 1 1 1
```

### 样例输出 #1

```
0 0 0 0 0 0
0 0 1 1 1 1
0 1 1 2 2 1
1 1 2 2 2 1
1 2 2 2 2 1
1 1 1 1 1 1
```

 解题思路：扫外面零变成3，最后在遍历                                                                      

#include<iostream>
using namespace std;
const int N = 30;
int g[N][N], n;
bool vis[N][N];
void dfs(int x, int y)
{
    if (x >= 0 && y >= 0 && x <= n+1&& y <= n+1)
    {
        if (g[x][y] == 1 || g[x][y] == 3)
        {
            return;
        }
        else
        {
            g[x][y] = 3;
            dfs(x + 1, y);
            dfs(x - 1, y);
            dfs(x, y + 1);
            dfs(x, y - 1);
        }
    }
}
int main()
{
    cin >> n;
    for (int i = 1; i<=n; i++)
    {
        for (int j = 1; j<=n; j++)
        {
            cin >> g[i][j];
        }
    }
    dfs(0, 0);
    for (int i = 1; i<=n; i++) {
        for (int j = 1; j<=n; j++)
        {
            if (g[i][j] == 3)
            {
                g[i][j] = 0;
            }
            else if (g[i][j] == 0)
            {
                g[i][j] = 2;
            }
        }
    }
    for (int i = 1; i<=n; i++) {
        for (int j = 1;j<=n; j++)
        {
            cout << g[i][j] << " ";
        }
        cout << endl;
    }
    return 0;
}