UVA - 297 Quadtrees（四分图）

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

3
ppeeefpffeefe
pefepeefe
peeef
peefe
peeef
peepefefe

Example Output

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

题目大意：如果某子节点对应的区域全黑或全白，则直接用一个黑节点或白节点表示，如果既有黑又有白则用灰节点表示。题中给出两棵四分树的先序遍历，求两者合并后（黑色部分合并）黑色像素的个数，p表示中间节点，f表示黑色，e表示白色。

注：题中给出树最多5层

思路：根据递归建树，然后将最下面的黑色节点标记，最后循环统计，（很暴力）。

#include<stdio.h>
#include<string.h>
char s[5400];
int book[1400],k;
int judge(int x)
{
    if(x>1364)
        return 1;
    for(int i=x; i<x+4; i++)//同父的四个节点
    {
        if(judge(4*i+1))//判断为最后一层的节点然后标记
        {
            book[i]=1;
        }
    }
    return 0;
}
void solve(int x)//编号
{
    if(s[k]=='p')//p为中间节点，需要跳过向后遍历
    {
        k++;
        solve(x*4+1);
        solve(x*4+2);
        solve(x*4+3);
        solve(x*4+4);
        return ;
    }
    else if(s[k]=='f')
    {
        if(judge(4*x+1)&&x<=1364)//f为它的子节点都为黑（向下递归）
            book[x]=1;
    }
    k++;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int cnt=0;
        memset(book,0,sizeof(book));//book数组标记
        memset(s,'\0',sizeof(s));
        scanf("%s",s);
        k=0;
        solve(0);//从0根节点开始

        scanf("%s",s);
        k=0;
        solve(0);//
        for(int i=0;i<=1364;i++)//一共1364个节点
        {
            if(book[i]==1)
                cnt++;
        }
        printf("There are %d black pixels.\n",cnt);
    }
    return 0;
}