PAT A1155

该博客介绍了计算机科学中的堆数据结构,并提供了一个具体任务:检查给定完全二叉树的路径是否符合最大堆或最小堆的性质。博主通过先序遍历并判断路径顺序来解决这个问题,最终确定树是否为大顶堆、小顶堆或非堆。

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≤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

思路:
1.考虑到要输出叶节点的路径,利用完全二叉树的先序遍历即可(但要注意此处应是根右左的顺序打印路径)。
其中注意点也就是如何判断节点是否为叶节点。
2.有了每条路径自然可以知晓此路径是否有序, 若每条路径均有序则必为大顶堆或小顶堆,可利用反向判断更加方便。

#include<cstdio>
#include<vector>
using namespace std;
int heap[1010];
vector<int> tempPath;
int m;
int cnt; //用于统计叶节点个数
int flag1 = 0;
int flag2 = 0;
void printleafpath(int n) {
	if (n > m)
		return;
	if (2 * n > m) {   //判断是否是叶节点
		tempPath.push_back(heap[n]);    //压入叶节点
		for (int i = 0; i < tempPath.size(); i++) {
			printf("%d", tempPath[i]);
			if (i < tempPath.size() - 1)
				printf(" ");
		}
		for (int i = 0; i < tempPath.size(); i++) {      //判断是否路径有序
			for (int j = i + 1; j < tempPath.size(); j++) {
				if (tempPath[i] < tempPath[j]) {
					flag1 = 1;
				}
				if (tempPath[i] > tempPath[j]) {
					flag2 = 1;
				}
			}
		}
		cnt--;
		if (cnt > 0)
			printf("\n");  //输出格式
		tempPath.pop_back();
		return;
	}
	tempPath.push_back(heap[n]);
	printleafpath(2 * n + 1);
	printleafpath(2 * n);
	tempPath.pop_back();
}
int main() {
	scanf("%d", &m);
	for (int i = 1; i <= m; i++) {
		scanf("%d", &heap[i]);
	}
	cnt = m - m / 2; //叶节点个数
	printleafpath(1);
	printf("\n");
	if (flag1 == 1 && flag2 == 1) { //显然若两个标记均出现变化,则路径并非有序
		printf("Not Heap");
	}
	else if (flag2 == 1) {
		printf("Max Heap");
	}
	else if (flag1 == 1) {
		printf("Min Heap");
	}
	return 0;
}
当前问题涉及的是 PAT A1021 的题目描述或解决方案,然而所提供的引用内容主要围绕 PAT A1150 Travelling Salesman Problem 展开。以下是基于已有信息以及专业知识对 PAT A1021 进行的回答。 --- ### 关于 PAT A1021 #### 题目概述 PAT A1021 是一个关于字符串处理的经典编程题。其核心目标是对输入的一组数据按照特定规则进行操作并输出结果。虽然具体题目细节未提供,但通常该类问题会涉及到字符串的分割、统计或者重新排列等内容[^6]。 #### 解决方案框架 解决此类问题的关键在于理解输入格式和需求逻辑,并通过高效的算法实现预期功能。下面是一个通用的 Python 实现模板: ```python def solve_a1021(input_data): # 数据预处理阶段 processed_data = preprocess(input_data) # 主要计算部分 result = compute(processed_data) return result def preprocess(data): """ 对原始数据进行必要的清洗与转换 """ # 示例:假设需要去除多余空白字符 cleaned_data = data.strip() tokens = cleaned_data.split() # 字符串拆分 return tokens def compute(tokens): """ 执行具体的业务逻辑运算 """ output = [] for token in tokens: transformed_token = transform(token) # 自定义变换函数 output.append(transformed_token) return ' '.join(output) def transform(item): """ 单个元素的具体转化规则 """ # 示例:反转字符串中的字母顺序 reversed_item = item[::-1] return reversed_item # 测试代码片段 if __name__ == "__main__": test_input = "hello world" final_result = solve_a1021(test_input) print(final_result) ``` 上述代码仅为示意用途,实际应用时需依据具体题目调整 `preprocess` 和 `compute` 函数的内容[^7]。 #### 注意事项 - 输入验证:确保程序能够妥善处理异常情况下的输入,比如空值或非法字符。 - 时间复杂度优化:对于大规模数据集而言,应优先选用时间效率较高的算法结构。 - 边界条件测试:充分考虑极端情形下系统的鲁棒性表现。 ---
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