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.
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 ",root); } } return 0; }