1147 Heaps (30 分)

本文探讨了计算机科学中堆数据结构的概念,包括最大堆和最小堆的特点,并通过一个具体的编程问题来检测给定的完全二叉树是否满足堆属性。输入包含多个测试用例,每个用例给出一棵完全二叉树的层级遍历序列,任务是判断其是否为最大堆、最小堆或根本不是堆,并输出树的后序遍历。

1147 Heaps (30 分)

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

后序遍历数组存储形式二叉树参考:https://blog.youkuaiyun.com/qq_36926514/article/details/80273551

要注意输出格式!

#include <iostream>
#include <cstdlib>
#include <cstdio>
using namespace std;

int m,n;

void postOrder(int a[],int i)
{
    if(i>n)
        return ;
    postOrder(a,i*2);
    postOrder(a,i*2+1);
    printf("%d",a[i]);
    if(i==1) printf("\n");
    else printf(" ");


}
int main()
{

    scanf("%d%d",&m,&n);

    for(int i=0; i<m; i++)
    {
        int a[1010];
        int flag[1010];
        for(int x=0; x<1010; x++)
            flag[x]=0;
        for(int j=1; j<=n; j++)
        {
            scanf("%d",&a[j]);
        }
        if(n%2!=0)
        {
            for(int k=1; k<=n/2; k++)
            {
                if(a[k]>=a[2*k]&&a[k]>=a[2*k+1])
                    flag[k]=1;
                else if(a[k]<=a[2*k]&&a[k]<=a[2*k+1])
                    flag[k]=2;
            }
        }
        else
        {
            for(int k=1; k<n/2; k++)
            {
                if(a[k]>=a[2*k]&&a[k]>=a[2*k+1])
                    flag[k]=1;
                else if(a[k]<=a[2*k]&&a[k]<=a[2*k+1])
                    flag[k]=2;
            }
            if(a[n/2]>=a[n]) flag[n/2]=1;
            else if(a[n/2]<=a[n]) flag[n/2]=2;
        }

        int flag1=0;
        for(int k=1; k<n/2; k++)
        {
            if(flag[k]!=flag[k+1])
            {
                printf("Not Heap\n");
                flag1=1;
                break;
            }

        }
        if(flag1==0)
        {
            if(flag[1]==1)
                printf("Max Heap\n");
            else if(flag[1]==2)
                printf("Min Heap\n");
        }

        postOrder(a,1);


    }
    return 0;
}

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值