[Trie、贪心] Acwing143. 最大异或对

题意：输入n个数，选出异或最大的一对

暴力\(O(n^{2})\)：枚举每一对
优化：把每个数转化成01串插入trie，枚举每个数，在trie中从高位到低位寻找每一位尽量与它不同的字符串，这样时间复杂度就优化到了\(O(n)\)

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;

int n, t;
int son[maxn][2], idx, ans;
vector<vector<int> > a;


void insert(vector<int> v) {
	int p = 0;
	for (auto u:v) {
		if (!son[p][u]) son[p][u] = ++idx;
		p = son[p][u];
	}
}

int solve(vector<int> v) {
	int p = 0, r = 31, ans = 0;
	for (auto u:v) {
		r--;
		if (son[p][!u]) {
			p = son[p][!u];
			u = !u;
			ans += 1 << r;
		}
		else p = son[p][u];
	}
	return ans;
}

int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
		cin >> t;
		vector<int> v;
		for (int i = 1; i <= 31; i++) {
			v.push_back(t & 1);
			t >>= 1;
		}
		reverse(v.begin(), v.end());
		insert(v);
		a.push_back(v);
	}
	for (auto v:a) ans = max(ans, solve(v)); 
	cout << ans;
  return 0;
}