UESTC 483 Data Structure Problem

Description

Data structure is a fundamental course of Computer Science, so that each contestant is highly likely to solve this data structure problem.

A Heap data structure is a binary tree with the following properties:

1. It is a complete binary tree; that is, each level of the tree is completely filled, except possibly the bottom level. At this level, it is filled from left to right.
2. It satisfies the heap-order property: The key stored in each node is greater than or equal to the keys stored in its children.

So such a heap is sometimes called a max-heap. (Alternatively, if the comparison is reversed, the smallest element is always in the root node, which results in a min-heap.)

A binary search tree (BST), which may sometimes also be called an ordered or sorted binary tree, is a node-based binary tree data structure which has the following properties:

1. The left subtree of a node contains only nodes with keys less than (greater than) the node's key.
2. The right subtree of a node contains only nodes with keys greater than (less than) the node's key.
3. Both the left and right subtrees must also be binary search trees.

Given a complete binary tree with $N$ keys, your task is to determine the type of it.

Note that either a max-heap or a min-heap is acceptable, and it is also acceptable for both increasing ordered BST and decreasing ordered BST.

Input

The first line of the input is $T$ (no more than $100$), which stands for the number of test cases you need to solve.

For each test case, the first line contains an integer $N$ ($1 \leq N \leq 1000$), indicating the number of keys in the binary tree. On the second line, a permutation of $1$ to $N$ is given. The key stored in root node is given by the first integer, and the $2i_{th}$ and $2i+1_{th}$ integers are keys in the left child and right child of the $i_{th}$ integer respectively.

Output

For every test case, you should output Case #k: first, where $k$ indicates the case number and counts from $1$. Then output the type of the binary tree:

• Neither — It is neither a Heap nor a BST.
• Both — It is both a Heap and a BST.
• Heap — It is only a Heap.
• BST — It is only a BST.

Sample Input

4 
1 
1 
3 
1 2 3 
3 
2 1 3 
4 
2 1 3 4

Sample Output

Case #1: Both 
Case #2: Heap 
Case #3: BST 

Case #4: Neither

判断一下是二叉搜索树还是堆

#include<cstdio>#include<cstring>#include<cmath>#include<iostream>#include<map>#include<queue>#include<algorithm>using namespace std;const int maxn=1e3+10;int T,t=0,ch[maxn][2],a[maxn],minx[maxn],maxx[maxn],n;bool checkheap(int x,int kind){    if (!x) return true;    if (ch[x][0]&&(kind==0&&a[x]>a[ch[x][0]]||kind==1&&a[x]<a[ch[x][0]])) return false;    if (ch[x][1]&&(kind==0&&a[x]>a[ch[x][1]]||kind==1&&a[x]<a[ch[x][1]])) return false;    return checkheap(ch[x][0],kind)&&checkheap(ch[x][1],kind);}bool checkbst(int x,int kind){    maxx[x]=minx[x]=a[x];    if (ch[x][0])    {        if (!checkbst(ch[x][0],kind)) return false;        minx[x]=min(minx[x],minx[ch[x][0]]);        maxx[x]=max(maxx[x],maxx[ch[x][0]]);        if (kind==0&&maxx[ch[x][0]]>=a[x]) return false;        if (kind==1&&minx[ch[x][0]]<=a[x]) return false;    }    if (ch[x][1])    {        if (!checkbst(ch[x][1],kind)) return false;        minx[x]=min(minx[x],minx[ch[x][1]]);        maxx[x]=max(maxx[x],maxx[ch[x][1]]);        if (kind==0&&minx[ch[x][1]]<=a[x]) return false;        if (kind==1&&maxx[ch[x][1]]>=a[x]) return false;    }    return true;}int main(){    scanf("%d",&T);    while (T--)    {        scanf("%d",&n);        for (int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            if (i&1) ch[i>>1][1]=i;            else ch[i>>1][0]=i;            ch[i][0]=ch[i][1]=0;        }        int flag1=checkheap(1,0)||checkheap(1,1);        int flag2=checkbst(1,0)||checkbst(1,1);        printf("Case #%d: ",++t);        if (flag1&&flag2) printf("Both\n");        else if (flag1) printf("Heap\n");        else if (flag2) printf("BST\n");        else printf("Neither\n");    }    return 0;}