PAT_A1147#Heaps

Source:

PAT A1147 Heaps (30 分)

Description:

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

Keys:

• 堆（Heap）
• 二叉树的建立和遍历

Attention:

• 有点水了哈0,0；

Code:

/*
Data: 2019-06-29 16:15:43
Problem: PAT_A1147#Heaps
AC: 19:20

题目大意：
判断给定的完全二叉树是否是堆
输入：
第一行给出测试数M(<=100)和结点数N[1,1e3]
接下来N行，逐层给出完全二叉树的各个键值
输出：
大根堆，小根堆，非堆；
接下来输出二叉树的后序遍历

基本思路：
静态树存储BST，遍历判断是否为堆
*/
#include<cstdio>
#include<vector>
using namespace std;
const int M=1e3+10;
int bst[M],Min,Max,n,m;
vector<int> path;

void Travel(int root)
{
    if(root > n)
        return;
    if(root!=1)
    {
        if(bst[root/2] > bst[root])
            Min=0;
        if(bst[root/2] < bst[root])
            Max=0;
    }
    Travel(root*2);
    Travel(root*2+1);
    path.push_back(bst[root]);
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("Test.txt", "r", stdin);
#endif // ONLINE_JUDGE

    scanf("%d%d", &m,&n);
    while(m--)
    {
        for(int i=1; i<=n; i++)
            scanf("%d", &bst[i]);
        Min=1;
        Max=1;
        path.clear();
        Travel(1);
        if(Max) printf("Max Heap\n");
        else if(Min)    printf("Min Heap\n");
        else    printf("Not Heap\n");
        for(int i=0; i<n; i++)
            printf("%d%c", path[i],i==n-1?'\n':' ');
    }

    return 0;
}