（ 递归 ）Fractal -- POJ -- 2083

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

 

Fractal

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 8317		Accepted: 3957

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

 

递归这个神奇的东西， 然而我并没有会， 唉， 感觉就只是懂了皮毛， 稍微加点难度的就不会， 真失败， 以后好好用用

 

#include<stdio.h>
#include<string.h>

#define N 1100

int a[20];
bool G[N][N];

void DFS(int n, int x, int y)
{
    G[x][y] = true;

	if(n==7)
        return ;

    int s = a[n];

	DFS(n+1, x, y);
	DFS(n+1, x, y+s*2 );
	DFS(n+1, x+s, y+s);
	DFS(n+1, x+s*2, y);
	DFS(n+1, x+s*2, y+s*2);
}

int main()
{
    int i, j, n;

	memset(G, false, sizeof(G));

    a[1] = 1;
	for(i=2; i<=10; i++)
        a[i] = a[i-1]*3;

	DFS(1, 1, 1);

	while(scanf("%d", &n), n!=-1)
	{

		for(i=1; i<=a[n]; i++)
		{
			for(j=1; j<=a[n]; j++)
			{
				if(G[i][j]==true)
					printf("X");
				else
					printf(" ");
			}
			printf("
");
		}

		printf("-
");
	}
	return 0;
}

　

下面粘个错误代码， 自己刚开始写的， 仔细看看是我没有很好的注意分层的问题， 在每一层递归的时候，应该记录一下它有用的信息， 然而我并没有。这题， 错就错在没有很好的注意分层的问题

#include<stdio.h>
#include<string.h>

int n, G[N][N];

void DFS(int x, int y)
{
    G[x][y] = 1;

    if(x== 3 && y==3)
    {
        G[x][y] = 1;
        return ;
    }

    DFS(x, y+2);
    DFS(x+1, y+1);
    DFS(x+2, y);
    DFS(x+2, y+2);
}

int main()
{
    int n, a[20]={0,1};

    memset(G, 0, sizeof(G));

    for(i=2; i<=10; i++)
        a[i] = a[i-1]*3;

    DFS(1, 1);

    while(scanf("%d", &n), n!=-1)
    {

        int i, j;

        for(i=1; i<=a[n]; i++)
        {
            for(j=1; j<=a[n]; j++)
            {
                if(G[i][j]==1)
                    printf("X");
                else
                    printf(" ");
            }
            printf("
");
        }

        printf("-
");
    }
    return 0;
}