UVa 四叉树

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。

 #include<iostream>
 #include<cstring>
 using namespace std;

 const int len = ;
 const int maxn =  + ;
 char s[maxn];
 char buf[maxn][maxn];
 int number;

 void solve(char *s, int &p,int r,int c,int w)
 {
     char q = s[p++];
     if (q == 'p')
     {
         solve(s, p, r, c + w / , w / );
         solve(s, p, r, c, w / );
         solve(s, p, r + w / , c, w / );
         solve(s, p, r + w / , c + w / , w / );
     }
     else if (q=='f')
     for (int i = r; i < r + w;i++)
     for (int j = c; j < c + w;j++)
     if (buf[i][j] == )
     {
         buf[i][j] = ;
         number++;
     }
 }

 int main()
 {
     int t;
     cin >> t;
     while (t--)
     {
         memset(buf, , sizeof(buf));
         number = ;
         cin >> s;
         int p = ;
         solve(s,p,,,len);
         cin >> s;
         p = ;
         solve(s,p,,,len);
         cout << "There are " << number << " black pixels." << endl;
     }
     return ;
 }