POJ 2083 fractal

最新推荐文章于 2020-09-02 10:12:07 发布

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算法竞赛宝典（第二部）—基础算法艺术同时被 3 个专栏收录

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本文介绍了一种使用递归算法绘制分形图的方法。通过分析不同级别分形图的规律，利用C语言实现了一个绘制箱型分形图的程序。文章详细解释了递归过程，并给出了完整的AC代码。

Fractal

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 11043		Accepted: 4941

Description

A fractal is an object or quantity that displays self-similarity, in a somewhat technical sense, on all scales. The object need not exhibit exactly the same structure at all scales, but the same "type" of structures must appear on all scales.
A box fractal is defined as below :

A box fractal of degree 1 is simply
X
A box fractal of degree 2 is
X X
X
X X
If using B(n - 1) to represent the box fractal of degree n - 1, then a box fractal of degree n is defined recursively as following
```
B(n - 1)        B(n - 1)

        B(n - 1)

B(n - 1)        B(n - 1)
```

Your task is to draw a box fractal of degree n.

Input

The input consists of several test cases. Each line of the input contains a positive integer n which is no greater than 7. The last line of input is a negative integer −1 indicating the end of input.

Output

For each test case, output the box fractal using the 'X' notation. Please notice that 'X' is an uppercase letter. Print a line with only a single dash after each test case.

Sample Input

Sample Output

X
-
X X
 X
X X
-
X X   X X
 X     X
X X   X X
   X X
    X
   X X
X X   X X
 X     X
X X   X X
-
X X   X X         X X   X X
 X     X           X     X
X X   X X         X X   X X
   X X               X X
    X                 X
   X X               X X
X X   X X         X X   X X
 X     X           X     X
X X   X X         X X   X X
         X X   X X
          X     X
         X X   X X
            X X
             X
            X X
         X X   X X
          X     X
         X X   X X
X X   X X         X X   X X
 X     X           X     X
X X   X X         X X   X X
   X X               X X
    X                 X
   X X               X X
X X   X X         X X   X X
 X     X           X     X
X X   X X         X X   X X
-

Source

Shanghai 2004 Preliminary

题意概括：画出分形图。

解题思路：递归。找出头2~3个图的规律，然后进行递归。

首先要找出第n个图形所占的行数与列数，即找出n与行数和列数的关系。然后将整个图形初始化：即把图形所占的行数与列数的每一个位置都赋初值为空格，同时在每一行的末尾用‘\0’来结束。

然后用dfs进行深搜，将n以及第一个图形的初始位置传入dfs函数，每一次n都减小1.

AC代码：

#include<stdio.h>
#include<math.h>

char map[730][730];

void dfs(int n, int x, int y)
{
    int size2;
    if(n == 1)
    {
        map[x][y] = 'X';   
        return;    //第二出错误，忘记了结束条件应该有return。
    }
    size2 = pow(3, n - 2);
    dfs(n - 1, x, y);
    dfs(n - 1, x + 2 * size2, y);
    dfs(n - 1, x, y + 2 * size2);
    dfs(n - 1, x + size2, y + size2);
    dfs(n - 1, x + 2 * size2, y + 2 * size2);
    return;
}

int main(void)
{
    int n, i, j, size1;
    while(scanf("%d", &n) != EOF)
    {
        if(n == -1)
        {
            break;
        }
        size1 = pow(3, n - 1);
        for(i = 1; i <= size1; i ++)  //第一处错误，没有对图形进行初始化
        {
            for(j = 1; j <= size1; j ++)
            {
                map[i][j] = ' ';
            }
            map[i][j] = '\0';
        }
        dfs(n, 1, 1);
        for(i = 1; i <= size1; i ++)  
        {
            printf("%s\n", map[i] + 1);  //第三处错误，map[i] + 1为从第一列开始
        }
        printf("-\n");
    }
    return 0;
}

错误原因：

1：不知道要对图形进行初始化，尤其是输出为%s时。

2：对于输出map[i]+1没概念，其实在使用%s对图形进行输出时，如果是从第i行j列输出的话，输出应为map[i] + j。

3：如果使用双重循环和%c输出图形超时的话，可以尝试改为单重循环和%s进行输出（有可能可以避免超时）！！

4：跳出递归时忘记了写return。敲打自己啊啊啊啊啊