05-树7 堆中的路径 (25 分) C++实现
#题目
将一系列给定数字插入一个初始为空的小顶堆H[]。随后对任意给定的下标i，打印从H[i]到根结点的路径。

输入格式:
每组测试第1行包含2个正整数N和M(≤1000)，分别是插入元素的个数、以及需要打印的路径条数。下一行给出区间[-10000, 10000]内的N个要被插入一个初始为空的小顶堆的整数。最后一行给出M个下标。

输出格式:
对输入中给出的每个下标i，在一行中输出从H[i]到根结点的路径上的数据。数字间以1个空格分隔，行末不得有多余空格。

输入样例:
5 3
46 23 26 24 10
5 4 3
输出样例:
24 23 10
46 23 10
26 10

#C++代码

#include<iostream>
using namespace std;
#define CAPACITY 10000
#define MINDATA -10000
typedef struct Stack Stack;
struct Stack
{
	int* H;
	int size;
	int capacity;
};

//Stack* creatstack(int N)
//{
//	Stack* stack = new Stack;
//	stack->capacity = CAPACITY;
//	stack->size = 0;
//	stack->H = new int[CAPACITY] {0};
//	stack->H[0] = MINDATA;
//	int num = 0;
//	for (int i = 1; i <= N; i++)
//	{
//		cin >> num;
//		stack->H[++stack->size] = num;
//	}
//	return stack;
//}
//int max(int a, int b)
//{
//	return a > b ? a : b;
//}
//void sort(Stack* stack)
//{
//	for (int i = stack->size / 2; i; i--)
//	{
//		int child = 2 * i;
//		if ((child + 1) <= stack->size && stack->H[child + 1] < stack->H[child])
//			child += 1;
//		if (stack->H[child] < stack->H[i])
//		{
//			int temp = stack->H[i];
//			stack->H[i] = stack->H[child];
//			stack->H[child] = temp;
//			if ((child * 2 <= stack->size && stack->H[child * 2] < temp) || ((child * 2 + 1) <= stack->size && stack->H[child * 2 + 1] < temp))
//				i = child + 1;
//		}
//	}
//}

//bool IsFull(Stack* stack)
//{
//	if (stack)
//	{
//		if (stack->size >= stack->capacity - 1)
//			return true;
//		else
//			return false;
//	}
//	else
//		return false;
//}
void insert(Stack* &stack,int num)
{
		int i = ++stack->size;
		for (; stack->H[i / 2] > num; i /= 2)
			stack->H[i] = stack->H[i / 2];
		stack->H[i] = num;
}
int main()
{
	int N = 0, M = 0;
	cin >> N >> M;
	int k = 0;
	Stack* stack = NULL;
	stack = new Stack;
	stack->capacity = CAPACITY;
	stack->size = 0;
	stack->H = new int[CAPACITY] {0};
	stack->H[0] = MINDATA;
	for (int i = 1; i <= N; i++)
	{
		cin >> k;
		insert(stack, k);
	}
	//sort(stack);
	for (int i = 0; i < M; i++)
	{
		//int k = 0;
		cin >> k;
		for (int j = k; j > 0; j /= 2)
		{
			cout << stack->H[j];
			if (j != 1)
				cout << " ";
		}
		if(i<M-1)
			cout << endl;
	}
	system("pause");
	return 0;
}