[51nod 1766] 樹上最遠點對（線段樹+樹的直徑）

題意

• 多次詢問一個點在$[a, b]$, 另一個點在$[c, d]$內的樹上最遠距離。

有一個結論對於兩個聯通塊$S, T$，設$d(S)$表示聯通塊直徑的兩個端點，那麼$d(S∪T) ∈d(S)∪d(T)$，這個東西仔細想下不是很難證，然後我們用線段樹維護就好了。

但是每次$pushup$的時候都要求$lca$，複雜度是$O(nlog^2n)$的，我們可以用$ST$表求$lca$來優化，我們維護一下原樹的歐拉序，也就是一個點進棧的時候把它加入序列，出棧的時候把它的父親加入序列，那麼兩個點的$lca$就是兩個點在序列中第一次出現位置中深度最小的點。

這樣子求$lca$的複雜度就降到了$O(nlogn + q)$了，總複雜度就是$O(nlogn)$的。

#include<bits/stdc++.h>
#include<bits/extc++.h>

#define file(s) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)
#define go(x, i) for(register int i = head[x]; i; i = nxt[i])
#define For(i, a, b) for(register int i = (a), i##_end_ = (b); i <= i##_end_; ++ i)
#define FOR(i, a, b) for(register int i = (a), i##_end_ = (b); i >= i##_end_; -- i)
#define debug(x) cout << #x << " = " << x << endl
#define mem(a, b) memset(a, b, sizeof(a))
#define cpy(a, b) memcpy(a, b, sizeof(a))
#define min(a, b) (a < b ? a : b)
#define max(a, b) (b < a ? a : b)
#define inf (0x3f3f3f3f)
#define INF (1e18)
#define pb push_back
#define mp make_pair
#define x first
#define y second

typedef unsigned long long ull;
typedef unsigned int uint;
typedef long long ll;
typedef std::pair<ll, int> PLI;
typedef std::pair<int, int> PII;
typedef long double ldb;
typedef double db;

namespace IO {
#define getc() ((S_ == T_) && (T_ = (S_ = Ch_) + fread(Ch_, 1, Buffsize, stdin), S_ == T_) ? 0 : *S_ ++)
#define putc(x) *nowps ++ = (x)
	const uint Buffsize = 1 << 15, Output = 1 << 23;
	static char Ch_[Buffsize], *S_ = Ch_, *T_ = Ch_;
	static char Out[Output], *nowps = Out;
	inline void flush(){fwrite(Out, 1, nowps - Out, stdout); nowps = Out;}
	template<class T>inline void read(T &_) {
		_ = 0; static char __; T ___ = 1;
		for(__ = getc(); !isdigit(__); __ = getc()) if(__ == '-') ___ = -1;
		for(; isdigit(__); __ = getc()) _ = (_ << 3) + (_ << 1) + (__ ^ 48);
		_ *= ___;
	}
	template<class T>inline void write(T _, char __ = '\n') {
		if(!_) putc('0');
		if(_ < 0) putc('-'), _ = -_;
		static uint sta[111], tp;
		for(tp = 0; _; _ /= 10) sta[++ tp] = _ % 10;
		for(; tp; putc(sta[tp --] ^ 48)); putc(__);
	}
	template<class T>inline bool chkmax(T &_, T __) {return _ < __ ? _ = __, 1 : 0;}
	template<class T>inline bool chkmin(T &_, T __) {return _ > __ ? _ = __, 1 : 0;}
}

using namespace std;
using namespace IO;

const int N = 1e5 + 10;
const int M = 2e5 + 10;

int head[N], to[M], v[M], nxt[M], e;
int dep[N], lg[M], pos[N], st[M][21];
int n, q, cnt;

void add(int x, int y, int z) {
	to[++ e] = y; nxt[e] = head[x]; head[x] = e; v[e] = z;
}

void dfs(int x, int fa) {
	st[pos[x] = ++ cnt][0] = x;
	go(x, i) if(to[i] ^ fa) {
		dep[to[i]] = dep[x] + v[i];
		dfs(to[i], x), st[++ cnt][0] = x;
	}
}

int _min(int x, int y) {
	return dep[x] < dep[y] ? x : y;
}

int lca(int x, int y) {
	x = pos[x], y = pos[y]; if(x > y) swap(x, y);
	int len = y - x + 1, lenn = lg[len];
	return _min(st[x][lenn], st[y - (1 << lenn) + 1][lenn]);
}

int get_dep(int x, int y) {
	return dep[x] + dep[y] - dep[lca(x, y)] * 2;
}

void Init() {
	int x, y, z, now = 0;
	read(n);
	For(i, 2, n) {
		read(x), read(y), read(z);
		add(x, y, z), add(y, x, z);
	}
	dfs(1, 0);
	For(i, 1, n << 1) {
		lg[i] = now; 
		if((1 << (now + 1)) == i) ++ now;
	}
	For(j, 1, 20) For(i, 1, cnt)
		if(i + (1 << (j - 1)) <= cnt)
			st[i][j] = _min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
}

struct Segment_Tree {
#define ls (bh << 1)
#define rs (ls | 1)
#define mid ((l + r) >> 1)
#define lson ls, l, mid
#define rson rs, mid + 1, r
	PII S[N << 2];

	PII merge(PII x, PII y) {
		PII res; int ans = 0;
		if(chkmax(ans, get_dep(x.x, x.y))) res = x;
		if(chkmax(ans, get_dep(y.x, y.y))) res = y;
		if(chkmax(ans, get_dep(x.x, y.y))) res = mp(x.x, y.y);
		if(chkmax(ans, get_dep(x.x, y.x))) res = mp(x.x, y.x);
		if(chkmax(ans, get_dep(x.y, y.x))) res = mp(x.y, y.x);
		if(chkmax(ans, get_dep(x.y, y.y))) res = mp(x.y, y.y);
		return res;
	}

	void build(int bh, int l, int r) {
		if(l == r) S[bh] = mp(l, r);
		else {
			build(lson), build(rson);
			S[bh] = merge(S[ls], S[rs]);
		}
	}

	PII query(int bh, int l, int r, int x, int y) {
		if(x <= l && r <= y) return S[bh];
		if(y <= mid) return query(lson, x, y);
		if(x > mid) return query(rson, x, y);
		return merge(query(lson, x, y), query(rson, x, y));
	}

}T;

int main() {
#ifdef ylsakioi
	file("1766");
#endif
	Init();
	T.build(1, 1, n);
	for(read(q); q -- ; ) {
		int l1, r1, l2, r2, ans = 0;
		read(l1), read(r1), read(l2), read(r2);
		PII L = T.query(1, 1, n, l1, r1), R = T.query(1, 1, n, l2, r2);
		chkmax(ans, get_dep(L.x, R.x)), chkmax(ans, get_dep(L.x, R.y));
		chkmax(ans, get_dep(L.y, R.x)), chkmax(ans, get_dep(L.y, R.y));
		write(ans);
	}
	return flush(), 0;
}