PAT 甲级 1155 Heap Paths

Posted 2021-02-06 zlrrrr

https://pintia.cn/problem-sets/994805342720868352/problems/1071785408849047552

 

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

 

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;
int N;
int a[maxn];
vector<int> v;

void dfs(int st) {
    if(st * 2 > N && st * 2 + 1 > N) {
        if(st <= N) {
            for(int i = 0; i < v.size(); i ++) {
                printf("%d", v[i]);
                printf("%s", i != v.size() - 1 ? " " : "
");
            }
        }
    } else {
        v.push_back(a[st * 2 + 1]);
        dfs(st * 2 + 1);
        v.pop_back();
        v.push_back(a[st * 2]);
        dfs(st * 2);
        v.pop_back();
    }
}

int main() {
    scanf("%d", &N);
    for(int i = 1; i <= N; i ++)
        scanf("%d", &a[i]);
    v.push_back(a[1]);
    dfs(1);

    int MaxHeap = 1, MinHeap = 1;
    for(int i = 2; i <= N; i ++) {
        if(a[i / 2] > a[i]) MinHeap = 0;
        if(a[i / 2] < a[i]) MaxHeap = 0;
    }

    if(MaxHeap == 1)
        printf("Max Heap
");
    else if(MinHeap == 1)
        printf("Min Heap
");
    else printf("Not Heap
");
    return 0;
}

　　dfs 搜索路径 然后根据最大堆最小堆的性质判断 刚刚好上午写好了堆排序 (FH 的堆排序)

FHFHFH  33 天倒计时