Easy Tree Query（二叉搜索树 规律）

Easy Tree Query

时间限制: 3 Sec  内存限制: 128 MB
提交: 130  解决: 15
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题目描述

You are given a binary search tree with depth k, whose nodes are valued from 1 to (2k − 1) and then Q queries.
For each query, you are given p nodes. Find the root of a smallest subtree which contains all p nodes and print its value.

输入

The first line of input contains an integer T (T ≤ 100), the number of test cases. The first line of each test case contains two integers k (1 ≤ k ≤ 60) and Q (1 ≤ Q ≤ 10 4 ). In next Q lines, each line contains an integer p and then p integers — the values of nodes in this query. It is guaranteed that the total number of nodes given in each test case will not exceed 105.

输出

For each query, print a line contains the answer of that query.

样例输入

1
4 1
3 10 15 13

样例输出

12

提示

也算是个规律题，不过需要把这个树先写出来，然后找出最大值和最小值，从根节点开始遍历就好了，先画个图。。。

开始都没注意到二叉搜索树，然后wyp也是浪了一波，以为题目出错了，然后开始敲，样例都不过，谁给他的勇气！！！狂的一PI

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const LL INF = 0x7f3f3f3f3f3f3f3f;

 
int main(){
    int T,len,Q,n;
    LL tmp,mmax,mmin,root;
    scanf("%d", &T );
    while(T--) {
        scanf("%d%d", &n ,&Q);
        for(int i=0;i<Q;i++){
            root=(1LL)<<(n-1);
            scanf("%d",&len);
            mmax=-1,mmin=INF;
            for(int j=0;j<len;j++){
                scanf("%lld",&tmp);
                mmax=max(mmax,tmp);
                mmin=min(mmin,tmp);
            }
            int x=n-1;
            while(1){
                if(root<=mmax&&root>=mmin)   break;
                if(root<mmin)  root=root+((1LL)<<(x-1));
                else   root=root-((1LL)<<(x-1));
                x--;
            }
            printf("%lld\n",root);
        }
    }
    return 0;
}