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;
}