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1155 Heap Paths

Statement

Metadata

  • 作者: 陈越
  • 单位: 浙江大学
  • 代码长度限制: 16 KB
  • 时间限制: 400 ms
  • 内存限制: 64 MB

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

Solution

#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
int n, a[N], rt;
vector<vector<int>> G;
vector<int> vec;
void print(vector<int> &vec) {
    for (int i = 0, sze = vec.size(); i < sze; ++i) printf("%d%c", vec[i], " \n"[i == sze - 1]);
}

void dfs(int u) {
    if (u > n) {
        return;
    }
    vec.push_back(a[u]);
    if (u * 2 > n) {
        print(vec);
    }
    dfs(u << 1 | 1);
    dfs(u << 1);
    vec.pop_back();
}

int main() {
    while (scanf("%d", &n) != EOF) {
        G.clear();
        G.resize(n + 1);
        int l = 0, g = 0;
        for (int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
            if (i > 1) {
                if (a[i] > a[i / 2])
                    ++l;
                else
                    ++g;
            }
        }
        vec.clear();
        dfs(1);
        if (l && g)
            puts("Not Heap");
        else if (l)
            puts("Min Heap");
        else
            puts("Max Heap");
    }
    return 0;
}
Last update: May 4, 2022
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