[CF888G]Xor-MST

题目大意：给一个$n$个点的完全图，第$i$个点有点权$v_i$，一条边$x-y$的边权为$v_xoplus v_y$，求最小生成树

题解：明显$Kruskal$和$Prim$都会$TLE$，有一种别的生成树的算法为$Sollin$。它对棵树找到离它最近的不连通的一棵树，对每棵树找好后若可以连这一条边就连接这条边。可以证明每次连通块个数至少减少一半，每次找最近的树可以枚举每一条边，复杂度$O(m)$，所以总复杂度是$O(mlog_2 n)$的。

在这一题中，可以用$Trie$来优化找最近的树的过程，可以优化为$O(log_2 v)$，可以通过本题

卡点：发现两个点点权相同就不会产生贡献，于是就可以排序去重，记录最开始连通块个数时记录的是没有去重的点数，于是就一直连不完，一直$TLE$

C++ Code：

#include <algorithm>
#include <cctype>
#include <cstdio>
namespace __IO {
	namespace R {
		int x, ch;
		inline int read() {
			ch = getchar();
			while (isspace(ch)) ch = getchar();
			for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);
			return x;
		}
	}
}
using __IO::R::read;

#define maxn 200010
const int inf = 0x3f3f3f3f;

namespace Trie {
#define M 29
#define N (maxn * (M + 2))
	int V[N], nxt[2][N], root, idx;

	void modify(int x, int num = 1) {
		int p = root;
		for (register int i = M; ~i; i--) {
			int tmp = x >> i & 1;
			if (!nxt[tmp][p]) nxt[tmp][p] = ++idx;
			p = nxt[tmp][p];
			V[p] += num;
		}
	}

	int query(int x) {
		int p = root, res = 0;
		for (register int i = M; ~i; i--) {
			int tmp = x >> i & 1;
			if (V[nxt[tmp][p]]) p = nxt[tmp][p];
			else p = nxt[!tmp][p], res |= 1 << i;
		}
		return res;
	}
#undef N
#undef M
}

int n, num;
std::pair<int, int> M[maxn];
int s[maxn], rnk[maxn];
long long ans;

int f[maxn];
int find(int x) {return x == f[x] ? x : (f[x] = find(f[x]));}

inline bool cmp(int a, int b) {return f[a] < f[b];}
int main() {
	n = read();
	for (int i = 1; i <= n; i++) s[i] = read();
	n = (std::sort(s + 1, s + n + 1), std::unique(s + 1, s + n + 1) - s - 1);
	for (int i = 1; i <= n; i++) {
		rnk[i] = f[i] = i;
		Trie::modify(s[i]);
	}
	num = n - 1;
	while (num) {
		for (int i = 1; i <= n; i++) find(i);
		std::sort(rnk + 1, rnk + n + 1, cmp);
		for (int i = 1; i <= n; i++) M[i] = std::make_pair(i, inf);
		for (int l = 1, r, father; l <= n; l = r + 1) {
			father = f[rnk[r = l]];
			while (r < n && father == f[rnk[r + 1]]) r++;
			for (int i = l; i <= r; i++) Trie::modify(s[rnk[i]], -1);
			for (int i = l; i <= r; i++) {
				int val = Trie::query(s[rnk[i]]), pos = std::lower_bound(s + 1, s + n + 1, s[rnk[i]] ^ val) - s;
				if (val < M[father].second) M[father] = std::make_pair(pos, val);
			}
			for (int i = l; i <= r; i++) Trie::modify(s[rnk[i]]);
		}
		for (int i = 1; i <= n; i++) {
			int x = find(i), y = find(M[i].first);
			if (x != y) {
				f[x] = y;
				ans += M[i].second;
				num--;
			}
		}
	}
	printf("%lld
", ans);
	return 0;
}