0.9poj2083(递归绘图好题)

最新推荐文章于 2019-12-01 18:57:01 发布

Melicent1

最新推荐文章于 2019-12-01 18:57:01 发布

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分类专栏：搜索、回溯、遍历

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本文详细阐述了如何使用递归算法绘制盒形分形，包括输入解析、绘图过程和输出展示，提供了一个深入理解分形艺术和递归算法应用的实例。

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http://poj.org/problem?id=2083

Fractal

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 5892		Accepted: 2975

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：就是绘图。。。思维一定要活不能死啊
#include <stdio.h>
#include <math.h>
int n;
char map[1000][1000];
void dfs(int n,int x,int y)
{//dfs函数并不绘制图形，而是根据坐标改变map[x][y]的值
int size;
if(n==1)//递归边界
{
map[x][y] = 'X';
return;
}
size = pow(3.0,n-2);
dfs(n-1,x,y);//递归绘制左上角的图形
dfs(n-1,x+2*size,y);//递归绘制右上角的图形
dfs(n-1,x+size,y+size);//递归绘制中间的图形
dfs(n-1,x,y+2*size); //递归绘制左下角的图形
dfs(n-1,x+2*size,y+2*size);//递归绘制右下角的图形
}
int main()
{
int i,j,size;
while(scanf("%d",&n)!=EOF)
{
if(n==-1)break;
size=pow(3.0,n-1);
for(i=1;i<=size;i++)
for(j=1;j<=size;j++)
map[i][j]=' ';
dfs(n,1,1);
for(i=1;i<=size;i++)
{
for(j=1;j<=size;j++)
printf("%c",map[i][j]);
printf("\n");
}
printf("-\n");
}
return 0;
}