浙大版《数据结构》习题3.10 汉诺塔的非递归实现（25 分）

题目：https://pintia.cn/problem-sets/434/problems/5892

借助堆栈以非递归（循环）方式求解汉诺塔的问题（n, a, b, c），即将N个盘子从起始柱（标记为“a”）通过借助柱（标记为“b”）移动到目标柱（标记为“c”），并保证每个移动符合汉诺塔问题的要求。
输入格式:
输入为一个正整数N，即起始柱上的盘数。
输出格式:
每个操作（移动）占一行，按柱1 -> 柱2的格式输出。
输入样例:

3

输出样例:

a -> c
a -> b
c -> b
a -> c
b -> a
b -> c
a -> c

代码：Inspired by (https://blog.csdn.net/royzdr/article/details/79032032)

#include <stdio.h>
#include <stdlib.h>
#define MAX_STACK_SIZE 100

typedef struct node{
    int pieces;
    char start, medium, dst;
}ElemType;
ElemType ERROR = {0, 'a', 'b', 'c'};
typedef struct SNode{
    ElemType Data[MAX_STACK_SIZE]; /*结构体数组*/
    int Top;
    int MaxSize;
}SNode, *Stack;
typedef enum{false, true} bool;


bool IsFull(Stack S);
bool IsEmpty(Stack S);
bool Push(Stack S, ElemType e);
ElemType Pop(Stack S);
void Hanoi(int N);


int main(void){
    int N;
    scanf("%d", &N);
    Hanoi(N);

    return 0;
}



bool IsFull(Stack S){
    return (S->Top == S->MaxSize-1);
}
bool IsEmpty(Stack S){
    return (S->Top == -1);
}

bool Push(Stack S, ElemType e){
    if(IsFull(S)){
        return false;
    }
    else{
        S->Data[++(S->Top)] = e;
        return true;
    }
}

ElemType Pop(Stack S){
    if(IsEmpty(S)){
        return ERROR;
    }
    else{
        return S->Data[(S->Top)--];
    }
}

void Hanoi(int N){
    Stack S = (Stack)malloc(sizeof(SNode));
    S->Top = -1;
    S->MaxSize = MAX_STACK_SIZE;

    ElemType init={N, 'a', 'b', 'c'};
    ElemType cur, temp;
    Push(S, init);
    while(!IsEmpty(S)){
        cur = Pop(S);
        if(cur.pieces>1){ /*先进后出*/
            temp = cur;
            temp.pieces--;
            temp.start = cur.medium;
            temp.medium = cur.start;
            Push(S, temp); /*后：{N-1, 'b', 'a', 'c'}*/

            temp = cur;
            temp.pieces = 1;
            Push(S, temp); /*再：{1, 'a', 'b', 'c'}*/

            temp =cur;
            temp.pieces--;
            temp.medium = cur.dst;
            temp.dst = cur.medium;
            Push(S, temp); /*先：{N-1, 'a', 'c', 'b'}*/
        }
        else if(cur.pieces==1){
            printf("%c -> %c\n", cur.start, cur.dst);
        }
    }
}