Luogu3379 【模板】最近公共祖先（LCA）

题面

题解

这里讲一种硬核做法。

首先\(\mathrm{dfs}\)整棵树，求出这棵树的欧拉序，然后\(\mathrm{LCA}\)问题就变成了\(\pm 1\mathrm{RMQ}\)问题。

考虑\(\mathrm{O}(n)\)解决\(\pm 1\mathrm{RMQ}\)问题。

将原序列分块，每一块长度为\(\dfrac {\log_2 n}2\)，块外用\(\mathrm{ST}\)表预处理，复杂度\(\mathrm{O}(n)\)，考虑块内如何\(\mathrm{O}(1)\)回答。

因为相邻两项之差最多为\(1\)，所以块内本质不同的状态只有\(2 ^ {\frac {\log n} 2} = \sqrt n\)种。

那么可以设\(f[S][l][r]\)表示状态为\(S\)时，区间\([l, r]\)的最小值。

于是块内就能\(\mathrm{O}(1)\)解决了，这一部分预处理的复杂度为\(\mathrm{O}(\sqrt n \log^2n)\)。

因为以上操作复杂度均没有超过\(\mathrm{O}(n)\)，所以预处理的复杂度为\(\mathrm{O}(n)\)，总复杂度为\(\mathrm{O}(n) - \mathrm{O}(1)\)。

代码

#include <cstdio>
#include <cmath>
#include <algorithm>
#define file(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)

inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if (ch == '-') w = -1, ch = getchar();
	while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

const int maxn(500010), LEN(1050), BLK(10), N(200010);
struct edge { int next, to; } e[maxn << 1];
int head[maxn], e_num, n, m, S, dep[maxn], f[1 << BLK][BLK][BLK];
int A[maxn << 1], ST[BLK << 1][N], Log[maxn << 1], pos[maxn], cnt;
int Len, Blk, minv[N], set[N], B[maxn << 1];
inline int min(const int &x, const int &y) { return B[x] < B[y] ? x : y; }
inline int Pos(const int &x) { return (x - 1) / Len + 1; }
inline int Posy(const int &x) { return (x - 1) % Len; }
inline void add_edge(int from, int to)
{
	e[++e_num] = (edge) {head[from], to};
	head[from] = e_num;
}

void dfs(int x, int fa)
{
	A[pos[x] = ++cnt] = x, B[cnt] = dep[x];
	for (int i = head[x]; i; i = e[i].next)
	{
		int to = e[i].to; if (to == fa) continue;
		dep[to] = dep[x] + 1, dfs(to, x), A[++cnt] = x, B[cnt] = dep[x];
	}
}

void Init()
{
	Len = std::max(1, (int) (log(cnt * 1.) / log(2.) * .5));
	Blk = cnt / Len + (cnt % Len > 0); int SIZ = 1 << (Len - 1);
	for (int i = 2; i <= cnt; i++) Log[i] = Log[i >> 1] + 1;
	for (int i = 0; i < SIZ; i++) for (int l = 0; l < Len; l++)
		for (int r = (f[i][l][l] = l) + 1, now = 0, _min = 0; r < Len; r++)
		{
			f[i][l][r] = f[i][l][r - 1];
			if (i & (1 << (r - 1))) ++now;
			else { --now; if(now < _min) _min = now, f[i][l][r] = r; }
		}
	for (int i = 1; i <= cnt; i++)
		if (!Posy(i)) minv[Pos(i)] = i, set[Pos(i)] = 0;
		else
		{
			if (B[i] < B[minv[Pos(i)]]) minv[Pos(i)] = i;
			if (B[i] > B[i - 1]) set[Pos(i)] |= 1 << (Posy(i) - 1);
		}
	for (int i = 1; i <= Blk; i++) ST[0][i] = minv[i];
	for (int i = 1; i <= Log[Blk]; i++)
		for (int j = 1; j <= Blk - (1 << i) + 1; j++)
			ST[i][j] = min(ST[i - 1][j], ST[i - 1][j + (1 << (i - 1))]);
}

int Query(int l, int r)
{
	l = pos[l], r = pos[r]; if(l > r) std::swap(l, r);
	int idl = Pos(l), idr = Pos(r);
	if (idl == idr) return (idl - 1) * Len + f[set[idl]][Posy(l)][Posy(r)] + 1;
	else
	{
		int a1 = (idl - 1) * Len + f[set[idl]][Posy(l)][Len - 1] + 1;
		int a2 = (idr - 1) * Len + f[set[idr]][0][Posy(r)] + 1;
		int ans = min(a1, a2), _l = Log[idr - idl - 1];
		if (idr - idl - 1)
			return min(ans, min(ST[_l][idl + 1], ST[_l][idr - (1 << _l)]));
		return ans;
	}
}

int main()
{
	n = read(), m = read(), S = read();
	for (int i = 1, a, b; i < n; i++)
		a = read(), b = read(), add_edge(a, b), add_edge(b, a);
	dep[S] = 1, dfs(S, 0); Init();
	for (int a, b; m--; ) a = read(), b = read(), printf("%d\n", A[Query(a, b)]);
	return 0;
}