UVa 297 四分树

感觉特别像那个分治的日程表问题。是f的话就填，否则就不填，然后同一个表填两次。那么就是最后的结果。

 #include <iostream>
 #include <cstring>
 #include <string>
 #include <map>
 #include <set>
 #include <algorithm>
 #include <fstream>
 #include <cstdio>
 #include <cmath>
 #include <stack>
 #include <queue>
 using namespace std;
 const double Pi=3.14159265358979323846;
 typedef long long ll;
 const int MAXN=+;
 const int dx[]={,,,,-};
 const int dy[]={,-,,,};
 const int INF = 0x3f3f3f3f;
 const int NINF = 0xc0c0c0c0;
 const ll mod=1e9+;
 int a[MAXN][MAXN];
 void build(string str,int &cnt,int r1,int r2,int c1,int c2)
  {
      char s=str[cnt++];
      if(s=='p')
      {
          int midr=(r1+r2)/,midc=(c1+c2)/;
          build(str,cnt,r1,midr,midc+,c2);
          build(str,cnt,r1,midr,c1,midc);
          build(str,cnt,midr+,r2,c1,midc);
          build(str,cnt,midr+,r2,midc+,c2);
     }
     else if(s=='f')
     {
         //wcout <<"in";
         for(int i=r1;i<=r2;i++)
             for(int j=c1;j<=c2;j++)
                 a[i][j]=;
     }

  }

 int main()
 {
     int t;cin>>t;
     while(t--)
     {
         int cnt=;
         string str1,str2;
         cin>>str1>>str2;
         memset(a,,sizeof(a));
         build(str1,cnt,,,,);
         /*for(int i=1;i<=32;i++)
         {
             for(int j=1;j<=32;j++)
             {
                 cout <<a[i][j]<<" ";
             }
             cout <<endl;
         }*/
         cnt=;
         build(str2,cnt,,,,);
         int ans=;
         for(int i=;i<=;i++)
         {
             for(int j=;j<=;j++)
             {
                 if(a[i][j]) ans++;
             } 

         }

         printf("There are %d black pixels.\n",ans);
     }
     return ;
 }