POJ 2083 Fractal

链接：http://poj.org/problem?id=2083

Fractal

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 8119		Accepted: 3890

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

1
2
3
4
-1

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

大意——给你一个数n，显示对应的一个分形图形,B(n)由5个B(n-1)的图形组成，如下所示：

B(n-1)          B(n-1)

B(n-1)

B(n-1)          B(n-1)

比如n=1时，图形是
X
n=2时，图形是
X    X
   X
X    X
n=3时，图形是
X    X           X    X
   X                 X
X    X           X    X
         X      X
             X
         X      X
X    X           X    X
   X                 X
X    X           X    X
n最大为7。

思路——因为n比较小，所以我们可以把n=7时的图形打印到2维数组中，然后读到n时，输出对应的图形即可。打印图形到2维数组中，使用递归是最自然的做法。因为这个分形图形本身定义就是递归的。因此我们只需要确定5个n-1规模问题的起始坐标和图形宽度就可以了。首先图形是一个正方形，n=1时，图形宽度是1；n=2时，图形宽度是3；显然，n时，图形宽度是3^(n-1)。接下来考虑起始坐标问题：假设w为n-1规模问题的图形宽度，n规模图形的起始坐标为(x,y)，其被表示成5个n-1规模问题的组合。那么5个n-1规模图形的起始坐标依次是(x,y),(x,y+2w),(x+w,y+w),(x+2w,
y),(x+2w,y+2w)。解决了这两个问题，我们就可以递归打印图形了。

复杂度分析——时间复杂度:O(1),空间复杂度:O(1)

附上AC代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
#include <stack>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
typedef long long ll;
const double pi = acos(-1.0);
const double e = exp(1.0);
const double eps = 1e-8;
const short maxlen = 730;
short num[7] = {1, 3, 9, 27, 81, 243, 729}; // 图形宽度
char mat[maxlen][maxlen];
short n;

void print(short n); // 打印图形
void init(short n, short x, short y); // 预处理图形，递归实现

int main()
{
	ios::sync_with_stdio(false);
	memset(mat, ' ', sizeof(mat));
	init(7, 0, 0);
	while (~scanf("%hd", &n) && n!=-1)
		print(n-1);
	return 0;
}

void print(short n)
{
	for (short i=0; i<num[n]; ++i)
	{
		for (short j=0; j<num[n]; ++j)
			putchar(mat[i][j]);
		putchar('\n');
	}
	printf("-\n");
}

void init(short n, short x, short y)
{
	if (n == 1)
		mat[x][y] = 'X';
	else
	{
		short k = num[n-2];
		init(n-1, x, y); // 五个部分图形起始坐标关系
		init(n-1, x, y+2*k);
		init(n-1, x+k, y+k);
		init(n-1, x+2*k, y);
		init(n-1, x+2*k, y+2*k);
	}
}