我正在尝试使用迭代方法来解决Hanoi问题。我试图通过使用两个嵌套循环来实现这一点，以便在每个循环中重复n-1个步骤，其中n是移动次数。我认为我用两个来解决这个问题已经很好了，但我不明白如何在每次传球时改变塔楼的顺序。有人能帮我完成这项任务吗？
尼齐奥开始了，好就是结束支持
#include <stdlib.h>
#include <stdio.h>
void tower_Organizer(int *inizio, int *fine, int *supp);
void hanoi_Iter(int n, int inizio, int fine, int supp);
int main(void){
int n;
printf("%sn" ,"inserisci un nuero positivo");
scanf("%d", &n);
hanoi_Iter(n, 1, 2, 3);
return 0;
}
void hanoi_Iter(int n, int inizio, int fine, int supp){
for(int i = 1 ; i <= n ; i++){
for(int j = 1 ; j <= i - 1 ; j++){
printf("%d -> %dn", inizio, fine);
tower_Organizer(&inizio, &fine, &supp);
}
printf("%d -> %dn", inizio, fine);
tower_Organizer(&inizio, &fine, &supp);
}
}
void tower_Organizer(int *inizio, int *fine, int *supp){
static int count = 1;
int c = count % 6;
switch( c ){
case 0:
*inizio = 1;
*fine = 2;
*supp = 3;
break;
case 1:
*inizio = 1;
*fine = 3;
*supp = 2;
break;
case 2:
*inizio = 2;
*fine = 3;
*supp = 1;
break;
case 3:
*inizio = 1;
*fine = 2;
*supp = 3;
break;
case 4:
*inizio = 3;
*fine = 1;
*supp = 2;
break;
case 5:
*inizio = 3;
*fine = 2;
*supp = 1;
break;
}
count++;
}
移动次数为2𝑛−1，因此嵌套循环无法实现这一点，因为它只会产生𝑛(𝑛+1(/2移动。
移动哪个光盘的逻辑也不像6次不同移动的循环那么简单。
有几种方法可以实现迭代解决方案，其中一些方法在维基百科上有解释。然而，上面列出的那些要求你保持每个桩的状态，这样你就可以检查什么是有效的移动。
您还可以查看移动序列号的位模式，并从中得出正确的移动。
以下是后一种想法的实现：
void hanoi(int n, int start, int end, int helper) {
// Create a converter for a pile index (0, 1 or 2) to the identifiers
// that are given as arguments:
int map[] = {start, end, helper};
// Perform as many iterations as there are moves to perform:
for (int last = 1<<n, i = 1; i < last; i++) {
// Use bit pattern of i to determine the source peg.
// First remove and count the trailing 0 bits in i:
int j = i;
int zeroCount = 0;
while (j % 2 == 0) {
j >>= 1;
zeroCount++;
}
// Derive the source pile from that zero-stripped number
int source = (j >> 1) % 3;
// Get next pile as destination
int target = (source + 1) % 3;
// Depending on parity, the source/target pile should be mirrored
if ((n + zeroCount) % 2 == 0) {
source = (3 - source) % 3;
target = (3 - target) % 3;
}
printf("Move from %d to %dn", map[source], map[target]);
}
}
int main(void) {
int n;
printf("%sn" ,"Enter a positive number (of discs)");
scanf("%d", &n);
hanoi(n, 1, 2, 3);
return 0;
}