PAT(Advanced)1155 Heap Paths (30 分)

1155 Heap Paths (30 分)

In computer science, a heap is a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either greater than or equal to (in a max heap) or less than or equal to (in a min heap) the key of C. A common implementation of a heap is the binary heap, in which the tree is a complete binary tree. (Quoted from Wikipedia at https://en.wikipedia.org/wiki/Heap_(data_structure))

One thing for sure is that all the keys along any path from the root to a leaf in a max/min heap must be in non-increasing/non-decreasing order.

Your job is to check every path in a given complete binary tree, in order to tell if it is a heap or not.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer $N$ ($1<N\le 1,000$), the number of keys in the tree. Then the next line contains $N$ distinct integer keys (all in the range of int), which gives the level order traversal sequence of a complete binary tree.

Output Specification:

For each given tree, first print all the paths from the root to the leaves. Each path occupies a line, with all the numbers separated by a space, and no extra space at the beginning or the end of the line. The paths must be printed in the following order: for each node in the tree, all the paths in its right subtree must be printed before those in its left subtree.

Finally print in a line Max Heap if it is a max heap, or Min Heap for a min heap, or Not Heap if it is not a heap at all.

Sample Input 1:

8
98 72 86 60 65 12 23 50

Sample Output 1:

98 86 23
98 86 12
98 72 65
98 72 60 50
Max Heap

Sample Input 2:

8
8 38 25 58 52 82 70 60

Sample Output 2:

8 25 70
8 25 82
8 38 52
8 38 58 60
Min Heap

Sample Input 3:

8
10 28 15 12 34 9 8 56

Sample Output 3:

10 15 8
10 15 9
10 28 34
10 28 12 56
Not Heap

题目解读：给出一个完全二叉树的层序遍历序列，从根遍历输出它的路径，右子树必须在左子树之前被输出，最后输出这个树是否是最大堆或最小堆。

解题思路： 由于是完全二叉树，所以可以用数组来存储heap，遍历只需要用dfs就可以了，因为数组存储heap方便计算，root应该是1而不是0。有一个小坑是要先输出右子树再输出左子树，更改一下遍历的顺序就行了。此外就是判断是否是最大堆最小堆，这里只需要检测不是的情况就可以了。AC之前错了一个判断条件是遍历完1~N的数组，i应该最后取得到N，直到N+1退出循环。

AC代码

#include <iostream>
#include <vector>
using namespace std;
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
int heap[1005];
int n;

void dfs(int v) {
	if (2 * v > n) {
		vector<int> num;
		for (int i = v; i >= 1; i = i / 2) {
			num.push_back(heap[i]);
		}
		for (int i = num.size()-1; i >= 0; i--) {
			printf("%d%c", num[i],i == 0 ? '\n':' ');
		}
		return;
	}
	if (2 * v + 1 <= n) {
		dfs(2 * v + 1);
	}
	if (2 * v <= n) {
		dfs(2 * v);
	}

}

int main(int argc, char** argv) {
	int isMAX = 1;
	int isMIN = 1;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &heap[i]);
	}
	dfs(1);
	for (int i = 2; i <= n; i++) {
		if (heap[i / 2] > heap[i]) isMIN = 0;
		if (heap[i / 2] < heap[i]) isMAX = 0;
	}
	if (isMAX) {
		printf("Max Heap\n");
	}
	if (isMIN) {
		printf("Min Heap\n");
	}
	else if (!isMIN && !isMAX) {
		printf("Not Heap\n");
	}
	return 0;
}