四分樹(Quadtrees)__java

Quadtrees

Time Limit:3000MS

Memory Limit:Unknown

64bit IO Format:%lld & %llu

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Description

A quadtree is a representation format used to encode images. The fundamental idea behind the quadtree is that any image can be split into four quadrants. Each quadrant may again be split in four sub quadrants, etc. In the quadtree, the image is represented 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.', where X is the number of black pixels in the resulting image. Example Input

3ppeeefpffeefepefepeefepeeefpeefepeeefpeepefefe

Example Output

There are 640 black pixels.There are 512 black pixels.There are 384 black pixels.

【分析】

       由於當結點的顏色確定時，我們就可以通過判斷它是否有子結點，所以根據所給的先序遍曆我們就可以確定整棵樹了。我們可以通過遞迴還原“畫圖”時的先序操作，在遞迴的過程中，統計黑色結點的個數，當該結點已被訪問過，則不用再統計該結點（實際上該結點就是兩個圖合并後重疊的黑色結點）。

用C++語言編寫程式，代碼如下：

#include<iostream>#include<cstring>using namespace std;const int maxn = 1024 + 10;char s[maxn];//string s;const int len = 32;int buf[len][len], cnt;//把字串s[p..]匯出到以(r, c)為左上方，邊長為w的緩衝區中//2 1//3 4void draw(const char* s, int& p, int r, int c, int w) {char ch = s[p++];if (ch == 'p') {draw(s, p, r, c + w / 2, w / 2);              //1draw(s, p, r, c, w / 2);                           //2draw(s, p, r + w / 2, c, w / 2);              //3draw(s, p, r + w / 2, c + w / 2, w / 2); //4}else if (ch == 'f') { //畫黑像素（白像素不畫）for (int i = r; i < r + w; i++)for (int j = c; j < c + w; j++)if (buf[i][j] == 0) {buf[i][j] = 1;cnt++;}}}int main() {int T;cin >> T;while (T--) {cnt = 0;memset(buf, 0, sizeof(buf));for (int i = 0; i < 2; i++) {int p = 0;cin >> s;draw(s, p, 0, 0, len);}cout << "There are " << cnt << " black pixels." << endl;}return 0;}

用java語言編寫程式，代碼如下：

import java.io.BufferedInputStream;import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner input = new Scanner(new BufferedInputStream(System.in));final int len = 32;int T = input.nextInt();for(int i = 0; i < T; i++) {int[] cnt = new int[1];cnt[0] = 0;int[][] buf = new int[len][len];String s;for(int j = 0; j < 2; j++) {s = input.next();int[] p = new int[1];p[0] = 0;draw(s, p, 0, 0, len, buf, cnt);}System.out.println("There are " + cnt[0] + " black pixels.");}}public static void draw(String s, int[] p, int r, int c, int w, int[][] buf, int[] cnt) {char ch = s.charAt(p[0]++);if(ch == 'p') {draw(s, p, r, c + w / 2, w / 2, buf, cnt);draw(s, p, r, c, w / 2, buf, cnt);draw(s, p, r + w / 2, c, w / 2, buf, cnt);draw(s, p, r + w / 2, c + w / 2, w / 2, buf, cnt);}else if(ch == 'f') {for(int i = r; i < r + w; i++)for(int j = c; j < c + w; j++)if(buf[i][j] == 0) {buf[i][j] = 1;cnt[0]++;}}}}