https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=233

参考了刘汝佳的算法，写得太妙了。

因为最多是1024块，所以每行每列最多是32，利用先序遍历，一旦是'p'时，就访问第1块，如果第一块内还有细分，则继续递归下去。然后继续依次访问第2,3,4块空间。

1 #include
     
       2 #include
      
        3 using namespace std; 4  5 const int len = 32; 6 const int maxn = 1024 + 10; 7 char s[maxn]; 8 char buf[maxn][maxn]; 9 int number;10 11 void solve(char *s, int &p,int r,int c,int w)12 {13     char q = s[p++];14     if (q == 'p')15     {16         solve(s, p, r, c + w / 2, w / 2);17         solve(s, p, r, c, w / 2);18         solve(s, p, r + w / 2, c, w / 2);19         solve(s, p, r + w / 2, c + w / 2, w / 2);20     }21     else if (q=='f')22     for (int i = r; i < r + w;i++)23     for (int j = c; j < c + w;j++)24     if (buf[i][j] == 0)25     {26         buf[i][j] = 1;27         number++;28     }29 }30 31 int main()32 {33     int t;34     cin >> t;35     while (t--)36     {37         memset(buf, 0, sizeof(buf));38         number = 0;39         cin >> s;40         int p = 0;41         solve(s,p,0,0,len);42         cin >> s;43         p = 0;44         solve(s,p,0,0,len);45         cout << "There are " << number << " black pixels." << endl;46     }47     return 0;48 }