UVa 297 - Quadtrees
2014-07-27 10:52
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Quadtrees |
by a parent node, while the four quadrants are represented by four child nodes, in a predetermined order.
Of course, if the whole image is a single color, it can be represented by a quadtree consisting of a single node. In general, a quadrant needs only to be subdivided if it consists of pixels of different colors. As a result, the quadtree need not be of uniform
depth.
A modern computer artist works with black-and-white images of
units, for a total of 1024 pixels per image. One of the operations
he performs is adding two images together, to form a new image. In the resulting image a pixel is black if it was black in at least one of the component images, otherwise it is white.
This particular artist believes in what he calls the preferred fullness: for an image to be interesting (i.e. to sell for big bucks) the most important property is the number of filled (black) pixels in the image. So, before adding two images together,
he would like to know how many pixels will be black in the resulting image. Your job is to write a program that, given the quadtree representation of two images, calculates the number of pixels that are black in the image, which is the result of adding the
two images together.
In the figure, the first example is shown (from top to bottom) as image, quadtree, pre-order string (defined below) and number of pixels. The quadrant numbering is shown at the top of the figure.
Input Specification
The first line of input specifies the number of test cases (N) your program has to process.The input for each test case is two strings, each string on its own line. The string is the pre-order representation of a quadtree, in which the letter 'p' indicates a parent node, the letter 'f' (full) a black quadrant and the letter 'e'
(empty) a white quadrant. It is guaranteed that each string represents a valid quadtree, while the depth of the tree is not more than 5 (because each pixel has only one color).
Output Specification
For each test case, print on one line the text 'There are X black pixels.', whereX is the number of black pixels in the resulting image.
Example Input
3 ppeeefpffeefe pefepeefe peeef peefe peeef peepefefe
Example Output
There are 640 black pixels. There are 512 black pixels. There are 384 black pixels.
题目:
利用四叉树处理图片,给你两张黑白图片的四叉树,问两张图片叠加后黑色的面积。
分析:
搜索、数据结构。把图片分成1024块1*1的小正方形,建立一位数组记录对应小正方形的颜色。
用一个数组存颜色0表示白色1表示黑色,然后两次递归填充一下数组,落在相同位置的就算一个黑色,最后统计一下黑色就可以了
#include <iostream> #include <stdio.h> #include <string> #include <string.h> using namespace std; const int N = 32; int cnt; int map ; //把字符串str[p..]导出到以(r,c)为坐上角,边长为w的缓冲区中 void draw(string str,int &p,int r,int c,int w) { char ch = str[p++]; if(ch == 'p') { draw(str,p,r,c+w/2,w/2); draw(str,p,r,c,w/2); draw(str,p,r+w/2,c,w/2); draw(str,p,r+w/2,c+w/2,w/2); } else if(ch == 'f') { for(int i = r; i < r+w; i++) { for(int j = c; j < c+w; j++) { if(map[i][j] == 0) { map[i][j] = 1; cnt++; } } } } } int main() { int t; int p; string str1,str2; while(cin >> t) { getchar(); while(t--) { memset(map,0,sizeof(map)); cnt = 0; cin >> str1; cin >> str2; p = 0; draw(str1,p,0,0,N); p = 0; draw(str2,p,0,0,N); cout <<"There are "<< cnt <<" black pixels."<<endl; } } return 0; }
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