CF1446C Xor Tree

合集 - 题解(19)

1.[博客迁移20190713]题解 P4169 【[Violet]天使玩偶/SJY摆棋子】2024-05-272.模拟题2024-06-263.SP666 VOCV - Con-Junctions 题解2024-08-254.Codeforces Round #900 (Div. 3)2024-08-255.Codeforces Round 962 (Div. 3)2024-08-276.题解 CF993E 【Nikita and Order Statistics】2024-09-167.题解 P6891 【[JOISC2020]ビルの飾り付け 4】2024-09-188.P2163 [SHOI2007] 园丁的烦恼2024-09-189.[POI2011] LIZ-Lollipop2024-09-1910.Codeforces Round 973 (Div. 2)2024-09-2111.Codeforces Round 974 (Div. 3)2024-09-2212.NEERC2013题解2024-09-23

13.CF1446C Xor Tree2024-09-25

14.Codeforces Round 709 (Div. 1)2024-09-2715.Codeforces Round 975 (Div. 2)2024-09-2816.NEERC2014题解2024-09-2917.题解 P2726 【[SHOI2005]树的双中心】2024-10-0118.CF2019 F. Max Plus Min Plus Size2024-10-0319.Codeforces Round 757 (Div. 2)03-05

收起

很有意思的题目，我们考虑能连边的两个数一定是在 01-Trie 上距离最近的两个点。于是我们先把所有数插入到 01-Trie 上去，然后 $d{p}_u$ 考虑以 $u$ 为根的子树中最多能留几个数，他的两个儿子内部的点只能在内部转移，你只能取一个儿子和另一个儿子的一个，也就是说我们的转移为 $d{p}_u=max(d{p}_{lson},d{p}_{rson})+1$

const int M = 30;
const int N = 2e5 + 5;
int n, a[N]; 

struct Trie {
	int t[N * M][2], ed[N * M], dp[N * M], tot;
	inline void clear(void) {
		for (int i = 0; i <= tot; i++) t[i][0] = t[i][1] = ed[i] = dp[i] = 0;
		tot = 0;
	}
	
	Trie(void) { clear(); }
	inline void insert(int val) {
		int pos = 0;
		for (int i = M; i >= 0; i--) {
			int b = val >> i & 1;
			if (!t[pos][b]) t[pos][b] = ++tot;
			pos = t[pos][b];
		}
		ed[pos]++;
	}
	
	inline void remove(int val) {
		int pos = 0;
		for (int i = M; i >= 0; i--) {
			int b = val >> i & 1;
			if (!t[pos][b]) break;
			pos = t[pos][b];
		}
		if (ed[pos]) ed[pos]--;
	}
	
	inline int query_min(int val) {
		int pos = 0, ret = 0;
		for (int i = M; i >= 0; i--) {
			int b = val >> i & 1;
			if (t[pos][b]) pos = t[pos][b];
			else if (t[pos][1 - b]) ret = ret | (1 << i), pos = t[pos][1 - b];
			else break;
		} return ret;
	}
	
	inline int query_max(int val) {
		int pos = 0, ret = 0;
		for (int i = M; i >= 0; i--) {
			int b = val >> i & 1; b = 1 - b;
			if (t[pos][b]) pos = t[pos][b];
			else if (t[pos][1 - b]) ret = ret | (1 << i), pos = t[pos][1 - b];
			else break;
		} return ret;
	}
	
	inline void dfs(int pos) {
		int l = t[pos][0], r = t[pos][1];
		if (l) dfs(l), chkmax(dp[pos], dp[l]);
		if (r) dfs(r), chkmax(dp[pos], dp[r]);
		if (l && r) dp[pos]++;
		if (ed[pos]) dp[pos] = 1;
	}
} T;

signed main(void) {
	read(n);
	for (int i = 1; i <= n; i++) read(a[i]), T.insert(a[i]);
	T.dfs(0); writeln(n - T.dp[1]);
	//fwrite(pf, 1, o1 - pf, stdout);
	return 0;
}