我能够想出一个算法(逻辑)来解决汉诺塔问题的k-peg解决方案,但是当我实现我的代码时,我得到了分段错误。
void move(int number_of_disks, int source, int dest, vector <int> free_peg, int pointer)
{
int p;
if (1 == number_of_disks)
{
moves++;
move_top_disk (source, dest);
}
if(free_peg.size() > 2)
p = number_of_disks/2;
else
p = number_of_disks - 1;
moves++;
//Move top "p" disks from peg 1 to peg i
move(p, source, free_peg.back(),free_peg, pointer);
//Move "n - p - 1" disks from peg 1 to another peg
move(number_of_disks - p - 1, source, free_peg[pointer--], free_peg, pointer++);
//Move the "last disk" from the source peg to the destination
move_top_disk(source, dest);
//Move "n - p - 1" disks from peg (i - 1) to the final peg
move(number_of_disks - p - 1, free_peg[pointer--], dest, free_peg, pointer++);
//Move "p" disks from peg i to the destination
move(p, free_peg.back(), dest, free_peg, pointer);
}这个想法很简单,我保留了一个空闲的钉子(或塔)的向量,当我移动我的磁盘时,我会更新它。因此,对于6个pegs和n个磁盘的情况,我有一个源,一个目的地和4个空闲pegs。其思想是将(n - p)处的p~ n/2从源移动到free_peg3。现在我的向量中只有3个空闲的peg,我使用这3个空闲的peg将(n -p- 1)个磁盘移动到free_peg2,然后将最后一个磁盘从源磁盘移动到目标磁盘。所以现在我有2个空闲的pegs和1个source =3个空闲的pegs。接下来,我需要使用3个空闲的peg(包括现在空闲的源)将(n -p- 1)个磁盘从peg2移动到目的地。最后,使用4个空闲的peg将p个磁盘从free_peg3移动到目的地。然而,当我在我的代码中实现这一点时,我得到了一个分段错误,有人能帮我解决这个问题吗?
发布于 2012-09-23 06:02:18
我能够解决一个k-peg通用解决方案,谢谢你的帮助。下面是算法的运行方式:
void move(int number_of_disks, int source, int dest, vector <int> free_peg)
{
int p, middle, g;
if (1 == number_of_disks)
{
moves++;
move_top_disk (source, dest);
}
else
{
moves++;
if(free_peg.size() >= 2)
p = number_of_disks/2;
else
p = number_of_disks - 1;
//Move top "p" disks from peg 1 to peg i
middle = free_peg.back();
free_peg.pop_back();
free_peg.push_back(dest);
move(p, source, middle,free_peg);
//Move "n - p " disks from peg 1 to another peg
free_peg.pop_back();
move(number_of_disks - p, source, dest, free_peg);
//Move p from current peg to the final peg
free_peg.push_back(source);
move(p, middle, dest, free_peg);
}
}https://stackoverflow.com/questions/12545231
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