汇总贴

2020年3月PAT甲级满分必备刷题技巧

题目

来源：https://pintia.cn/problem-sets/994805342720868352/problems/994805342821531648

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

题目大意

给定堆的层序，要求判断堆的种类（大顶堆、小顶堆、不是堆）和输出后序

题目分析

用数组存就行了，不建树，DFS生成后序遍历序列，堆的类型就直接用定义排除掉明显不对的可能。

满分代码

#include<iostream>
#include<vector>
using namespace std;
const int maxn=1005;
int n,m;
vector<int> v(maxn);
void dfs(int i){
    if(i>m)return;
    dfs(2*i);
    dfs(2*i+1);
    printf("%d%s",v[i],i==1?"\n":" ");
}
int main(){
    cin>>n>>m;
    for(int i=0;i<n;i++){
        int minheap=1,maxheap=1;
        for(int j=1;j<=m;j++){
            scanf("%d",&v[j]);
            if(j>1 && v[j/2]>v[j])minheap=0;
            if(j>1 && v[j/2]<v[j])maxheap=0;
        }
        if(minheap==1)printf("Min Heap\n");
        else printf("%s\n",maxheap==1?"Max Heap":"Not Heap");
        dfs(1);
    }
    return 0;
}