UVA 297 Quadtrees(四叉樹）

Question：
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 32 × 32 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
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
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.
Sample Input
3
ppeeefpffeefe
pefepeefe
peeef
peefe
peeef
peepefefe
Sample Output
There are 640 black pixels.
There are 512 black pixels.
There are 384 black pixels.
題目大意：有一個1024個像素的格子，按照從左到右從上到下2，1，3，4這樣建樹，p表示這個節點的子節點有黑有白，f（full）表示充滿了黑色（子節點全爲黑，不許表示子節點了），e（empty）表示其中全爲白，合併兩顆二叉樹，爲黑色像素到底有多少（奇妙的是四叉樹只需要先序遍歷就可以得到整棵樹的建樹）
解題思路：建一個32的二維數組來保存節點狀態。並遍歷完成統計。建樹過程詳見代碼
(http://vjudge.net/contest/133965#problem/A)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
const int len=32;
const int maxn=32*32+10;
char s[maxn];
int ans,a[len][len];
void draw(char s[],int &p,int r,int c,int w)    //r,c,w表示頂點爲(r,c)邊長爲w的正方形
{
    char ch=s[p++];    //用p來表示s數組下標
    if(ch=='p')
    {                   //將原本的正方形四等分
        draw(s,p,r,c+w/2,w/2);       //建‘1’這一塊
        draw(s,p,r,c,w/2);           //建‘2’這一塊
        draw(s,p,r+w/2,c,w/2);       //建‘3’這一塊
        draw(s,p,r+w/2,c+w/2,w/2);  //建‘4’這一塊
    }
    else if(ch=='f'){      //遍歷這個正方形中所有的像素
        for(int i=r;i<r+w;i++){
            for(int j=c;j<c+w;j++){
                if(!a[i][j]){
                    a[i][j]=1;
                    ans++;
                }
            }
        }
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        memset(a,0,sizeof(a));
        ans=0;
        for(int i=0;i<2;i++)
        {
            scanf("%s",s);
            int p=0;
            draw(s,p,0,0,len);
        }
        printf("There are %d black pixels.\n",ans);
    }
    return 0;
}

體會：四叉樹題型很少見，不過還是挺有趣的