UVa 297 Quadtrees -SilverN
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.', where X 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.
觉得这题挺好玩，于是写篇博客留个纪念
1 /**/ 2 #include<iostream> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 using namespace std; 7 int T; 8 int mp[200][200]; 9 char s[2000]; 10 int ans=0; 11 void draw(char s[],int &p,int r,int c,int w){//在mp上的一个正方形区域中操作，其中(r,c)为区域左上角坐标，w为区域边长 12 char ch=s[p++];//p为目前处理到的位数，用&实现传值 13 if(ch=='p')//处理中间节点 14 { 15 draw(s,p,r,c+w/2,w/2); 16 draw(s,p,r,c,w/2); 17 draw(s,p,r+w/2,c,w/2); 18 draw(s,p,r+w/2,c+w/2,w/2); 19 }else if(ch=='f'){ 20 for(int i=r,j;i<r+w;i++) 21 for(j=c;j<c+w;j++) 22 if(mp[i][j]==0){ 23 mp[i][j]=1; 24 ans++; 25 } 26 } 27 return; 28 } 29 int main(){ 30 scanf("%d",&T); 31 while(T--){ 32 memset(mp,0,sizeof(mp)); 33 ans=0; 34 for(int i=1;i<=2;i++){//先建第一棵树，再把第二棵树覆盖上去 35 scanf("%s",s); 36 int p=0; 37 draw(s,p,0,0,32); 38 39 } 40 printf("There are %d black pixels.\n",ans); 41 } 42 return 0; 43 }
