这题求最大异或值的时候直接按照 $x\times popcount(x)$ 作为关键字排序，暴力往后找 $50000$ 个即可通过。

这下充分发扬人类智慧了。

代码如下：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 5;
int n, d[maxn], cnt, ans = -1;
struct data {
    int to, weight;
};
vector<data> G[maxn];

bool cmp(int a, int b) {
	return a * __builtin_popcount(a) > b * __builtin_popcount(b);
}

inline void dfs(int u, int fa) {
    for (auto i : G[u]) {
        int v = i.to, w = i.weight;
        if (v == fa) continue;
        d[v] = d[u] ^ w;
        dfs(v, u);
    }
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> n;
    for (int i = 1; i < n; i++) {
        int u, v, w;
        cin >> u >> v >> w;
        G[u].push_back({v, w});
        G[v].push_back({u, w});
    }
    dfs(1, 1);
    sort(d + 1, d + n + 1, cmp);
    int m = min(n, 50000);
    for (int i = 1; i <= m; i++) {
    	for (int j = i + 1; j <= m; j++) {
    		ans = max(ans, d[i] ^ d[j]);
		}
	}
    cout << ans;
    return 0;
}