二叉搜索树上的LCA（最近公共祖先）

1017: Easy Tree Query

时间限制: 3 Sec   内存限制: 128 MB
提交: 184   解决: 30
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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

#include
#include
#include

比赛的时候想复杂了。通过将二叉搜索树上的值转化为二进制，然后找到了规律，每次都将当前的两个数转化为二进制，然后找到第一个不同的二进制位，先判断置为0可否，再判断了置为1可否。超时。

超时代码

#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
ll arr[maxn];int n=4,m;
ll getone(ll pre, ll old){
    if(pre == old) return pre;
    if(pre >old) swap(pre,old);
    ll ppre = pre;
    ll pold = old;
    vectorv1;
    vectorv2;
    while(pre){
        v1.push_back(pre%2);
        pre>>=1;
    }
    while(old){
        v2.push_back(old%2);
        old>>=1;
    }
    while(v1.size() < n) v1.push_back(0);
    while(v2.size() < n) v2.push_back(0);
 
//    reverse(v1.begin(),v1.end());
//    reverse(v2.begin(),v2.end());
    ll ans = 0;
    int k = n-1;
//
//        for(int i=0;i=0;i--){
        if(v1[i] != v2[i]){
            if(ans >= ppre && ans<= pold){
                return ans;
            }else{
                   ll ans1 = ans + (1<