Find MaxXorSum NBUT - 1597 （01字典树求异或最大值）

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Given n non-negative integers, you need to find two integers a and b that a xor b is maximum. xor is exclusive-or.

Input

Input starts with an integer T(T <= 10) denoting the number of tests. 
For each test case, the first line contains an integer n(n <= 100000), the next line contains a1, a2, a3, ......, an(0 <= ai <= 1000000000);

Output

For each test case, print you answer.

Sample Input

2
4
1 2 3 4
3
7 12 5

Sample Output

7
11

Hint

无

题目大意：给出一组数，求任意两个数的异或最大值。

解题思路：把每一个数转化成01串构建字典树。最后再对于数组中没个数在字典树中查找最大的异或值是多少即可。

题目链接：http://fastvj.rainng.com/problem/NBUT-1597

/*
@Author: Top_Spirit
@Language: C++
*/
//#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std ;
typedef long long ll ;
const int Maxn = 1e5 + 10 ;
const int Maxn1 = 1e6 + 10 ;
const int INF = 0x3f3f3f3f ;

ll a[Maxn] ;
ll son[Maxn << 4][2] ,cnt ;

void init(){
    memset(son, 0, sizeof(son)) ;
    cnt = 0 ;
}

void Insert(ll k){
    ll p = 0 ;
    for (int i = 31; i >= 0; i--){
        ll id = (k >> i) & 1 ;
        if (!son[p][id]) son[p][id] = ++cnt ;
        p = son[p][id] ;
    }
}

ll Find(ll k){
    ll p = 0, ans = 0 ;
    for (int i = 31; i >= 0; i--){
        int id = !((k >> i) & 1) ;
        ans <<= 1 ;
        if (son[p][id]) p = son[p][id], ans++ ;
        else p = son[p][!id] ;
    }
    return ans ;
}

int main (){
    ios_base::sync_with_stdio(false) ;
    cin.tie(0) ;
    cout.tie(0) ;
    int T, n ;
    cin >> T ;
    while (T--){
        init() ;
        cin >> n ;
        for (ll i = 1; i <= n; i++){
            cin >> a[i] ;
            Insert(a[i]) ;
        }
        ll ans = -1 ;
        for (int i = 1; i <= n; i++){
            ans = max(ans, Find(a[i])) ;
        }
        cout << ans << endl ;
    }
    return 0 ;
}