【笨方法学PAT】1147 Heaps （30 分）

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本文探讨了堆数据结构的定义与应用，通过实例解析如何判断一棵完全二叉树是否满足最大堆或最小堆的性质。此外，还介绍了一个字符串处理问题，利用STL中的unordered_map统计并找出给定句子中出现频率最高的单词。

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一、题目

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))

Your job is to tell if a given complete binary tree is a heap.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 100), the number of trees to be tested; and N (1 < N ≤ 1,000), the number of keys in each tree, respectively. Then M lines follow, each 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, 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. Then in the next line print the tree's postorder traversal sequence. All the numbers are separated by a space, and there must no extra space at the beginning or the end of the line.

Sample Input:

3 8
98 72 86 60 65 12 23 50
8 38 25 58 52 82 70 60
10 28 15 12 34 9 8 56

Sample Output:

Max Heap
50 60 65 72 12 23 86 98
Min Heap
60 58 52 38 82 70 25 8
Not Heap
56 12 34 28 9 8 15 10

二、题目大意

题目大意是找出句中出现次数最多的单词，并输出。

三、考点

STL

四、注意

1、使tolower变小写，注意分词的方式;

2、使用map<string,int>计数。

五、代码

#include<iostream>
#include<string>
#include<unordered_map>
using namespace std;
int main() {
	//read
	string s;
	getline(cin, s);

	//solve
	unordered_map<string, int> un_map;
	for (int i = 0; i < s.length(); ++i) {
		if (isalnum(s[i])) {
			string t = "";
			while (i < s.length() && isalnum(s[i])) {
				s[i] = tolower(s[i]);
				t += s[i];
				i++;
			}
			if (un_map.find(t) == un_map.end())
				un_map[t] = 1;
			else
				un_map[t]++;
		}
	}

	//find the max
	int max = 0;
	string max_s;
	for (auto it = un_map.begin(); it != un_map.end(); ++it) {
		if (it->second > max) {
			max = it->second;
			max_s = it->first;
		}
		else if (it->second == max) {
			if (it->first < max_s) {
				max_s = it->first;
			}
		}
	}

	//output
	cout << max_s << " " << max << endl;

	system("pause");
	return 0;
}