uva297(四叉树建立)

思路：根据题意，每个字母都是代表一个正方形的方块，那么我们要表示这个正方形就只需要左上角的坐标和宽度，字母'P'表示这个方块里面有黑块，但是不是全部都是，所有要针对这个方块继续递归建树，‘f’表示这个方块全是黑色的，那么就可以开始统计数目了。‘e’就直接忽略丢就好了。

点击题目链接

/*****************************************
Author      :Crazy_AC(JamesQi)
Time        :2015
File Name   :
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
using namespace std;
#define MEM(a,b) memset(a,b,sizeof a)
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
inline int Readint(){
	char c = getchar();
	while(!isdigit(c)) c = getchar();
	int x = 0;
	while(isdigit(c)){
		x = x * 10 + c - '0';
		c = getchar();
	}
	return x;
}
char s[10000];
int col[100][100];
int cnt;
const int len = 32;
void solve(const char *s,int &p,int x,int y,int w){
	char tmp = s[p++];
	if (tmp == 'p'){
		solve(s,p,x,y + w / 2,w / 2);
		solve(s,p,x,y,w / 2);
		solve(s,p,x + w / 2,y,w / 2);
		solve(s,p,x + w / 2,y + w / 2,w / 2);
	}else if (tmp == 'f'){
		for (int i = x;i < x + w;i++){
			for (int j = y;j < y + w;j++){
				if(col[i][j] == 0){
					col[i][j] = 1;
				 	cnt++;
				}
			}
		}
	}
}
int main()
{	
	// freopen("in.txt","r",stdin);
	// freopen("out.txt","w",stdout);
	int t;
	scanf("%d",&t);
	while(t--){
		memset(col, 0,sizeof col);
		cnt = 0;
		for (int i = 1;i <=2;i++){
			scanf("%s",s);
			int p = 0;
			int len = 32;
			solve(s,p,0,0,len);
		}
		// printf("%d\n",cnt);
		printf("There are %d black pixels.\n",cnt);
	}
	return 0;
}