[洛谷P2056][ZJOI2007]捉迷藏(2019-7-20考试)

题目大意：有一棵$n(n\leqslant10^6)$个点的树，上面所有点是黑点，有$m$次操作：

1. $C\;u$：把点$u$颜色翻转
2. $G$：问树上最远的两个黑点的距离，若没有黑点输出$0$

题解：有两个点集，已知它们的直径的端点，可以很快求出点集的并的直径。所有可以用线段树维护所有点的直径。

卡点：考试时没想到

C++ Code：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <vector>
const int maxn = 1e5 + 10, inf = 0x3f3f3f3f;

int head[maxn], cnt;
struct Edge {
	int to, nxt;
} e[maxn << 1];
inline void addedge(int a, int b) {
	e[++cnt] = (Edge) { b, head[a] }; head[a] = cnt;
	e[++cnt] = (Edge) { a, head[b] }; head[b] = cnt;
}

int n, m;
int Col[maxn];

int dep[maxn], idx;
int LG[maxn << 1], pos[maxn], st[22][maxn << 1], tot;
inline int _min(int a, int b) { return dep[a] < dep[b] ? a : b; }
inline int _max(int a, int b) { return dep[a] > dep[b] ? a : b; }
void dfs(int u, int fa = 0) {
	st[0][++tot] = u, pos[u] = tot;
	for (int i = head[u], v; i; i = e[i].nxt) {
		v = e[i].to;
		if (v != fa) {
			dep[v] = dep[u] + 1;
			dfs(v, u);
			st[0][++tot] = u;
		}
	}
}
inline int LCA(int x, int y) {
	static int l, r, len; l = pos[x], r = pos[y];
	if (l > r) std::swap(l, r);
	len = LG[r - l + 1];
	return _min(st[len][l], st[len][r - (1 << len) + 1]);
}
inline int dis(int x, int y) {
	return dep[x] + dep[y] - 2 * dep[LCA(x, y)];
}

struct Line {
	int l, r;
	Line() { }
	Line(int _l, int _r) : l(_l), r(_r) { }
	inline friend bool operator ! (const Line &rhs) {
		return !rhs.l && !rhs.r;
	}
	inline Line operator + (const Line rhs) const {
		if (!*this) return rhs;
		if (!rhs) return *this;
		const int x = rhs.l, y = rhs.r;
		int now = dis(l, r), t; Line ans(l, r);
		t = dis(l, x); if (now < t) now = t, ans = Line(l, x);
		t = dis(l, y); if (now < t) now = t, ans = Line(l, y);
		t = dis(r, x); if (now < t) now = t, ans = Line(r, x);
		t = dis(r, y); if (now < t) now = t, ans = Line(r, y);
		t = dis(x, y); if (now < t) now = t, ans = Line(x, y);
		return ans;
	}
} ;
namespace SgT {
	Line V[maxn << 2];
	void modify(int rt, int l, int r, int p, int v) {
		if (l == r) {
			V[rt] = Line(v, v);
			return ;
		}
		const int mid = l + r >> 1;
		if (p <= mid) modify(rt << 1, l, mid, p, v);
		else modify(rt << 1 | 1, mid + 1, r, p, v);
		V[rt] = V[rt << 1] + V[rt << 1 | 1];
	}
	int query() {
		return dis(V[1].l, V[1].r);
	}
}

int num;
int main() {
	std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
	std::cin >> n; num = n;
	for (int i = 1, a, b; i < n; ++i) {
		std::cin >> a >> b;
		addedge(a, b);
	}
	dfs(1); LG[0] = -1;
	for (int i = 1; i <= tot; ++i) LG[i] = LG[i >> 1] + 1;
	for (int i = 1; i <= LG[tot]; ++i)
		for (int j = 1, len = 1 << i; j + len - 1 <= tot; ++j)
			st[i][j] = _min(st[i - 1][j], st[i - 1][j + (len >> 1)]);
	for (int i = 1; i <= n; ++i) SgT::modify(1, 1, n, i, i);
	std::cin >> m;
	while (m --> 0) {
		char op;
		std::cin >> op;
		if (op == 'C') {
			static int x;
			std::cin >> x;
			Col[x] ^= 1;
			if (Col[x]) SgT::modify(1, 1, n, x, 0), --num;
			else SgT::modify(1, 1, n, x, x), ++num;
		} else {
			if (!num) std::cout << "-1\n";
			else std::cout << SgT::query() << '\n';
		}
	}
	return 0;
}