[洛谷P3950]部落冲突

题目大意：给你一棵树，有$3$个操作：

1. $Q;p;q：$询问$p,q$是否连通
2. $C;p;q：$把$p->q$这条边割断
3. $U;x：$恢复第$x$次操作二

题解：可以在割断时把这条边赋值上$1$，恢复时赋成$0$，只需要求$p->q$路径和是否为$0$即可，可以用$dfs$序+树状数组维护

卡点：$LCA$越界

C++ Code：

#include <cstdio>
#include <cctype>
#include <algorithm>
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 300010

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

int n, m;
#define M 20
int fa[maxn][M], sz[maxn], dep[maxn], dfn[maxn], idx;
void dfs(int u) {
	dfn[u] = ++idx;
	sz[u] = 1;
	for (int i = 1; i < M; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
	for (int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if (v != *fa[u]) {
			*fa[v] = u;
			dep[v] = dep[u] + 1;
			dfs(v);
			sz[u] += sz[v];
		}
	}
}

inline int LCA(int x, int y) {
	if (x == y) return x;
	if (dep[x] < dep[y]) std::swap(x, y);
	for (int i = dep[x] - dep[y]; i; i &= i - 1) x = fa[x][__builtin_ctz(i)];
	if (x == y) return x;
	for (int i = M - 1; ~i; i--) if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
	return *fa[x];
}

namespace BIT {
	int Tr[maxn], res;
	inline void modify(int p, int num) {for (; p <= n; p += p & -p) Tr[p] += num;}
	inline int query(int p) {for (res = 0; p; p &= p - 1) res += Tr[p]; return res;}
}

inline void modify(int x, int num) {
	BIT::modify(dfn[x], num);
	BIT::modify(dfn[x] + sz[x], -num);
}
inline void query(int x, int y) {
	int res = BIT::query(dfn[x]) + BIT::query(dfn[y]) - BIT::query(dfn[LCA(x, y)]) * 2;
	puts(res ? "No" : "Yes");
}

int war[maxn], Tim;
int main() {
	n = read(), m = read();
	for (int i = 1, a, b; i < n; i++) {
		a = read(), b = read();
		add(a, b);
		add(b, a);
	}
	dfs(1);
	while (m --> 0) {
		int x, y;
		char op = getchar();
		while (!isalpha(op)) op = getchar();
		switch (op) {
			case 'Q':
				x = read(), y = read();
				query(x, y);
				break;
			case 'C':
				x = read(), y = read();
				if (dep[x] < dep[y]) std::swap(x, y);
				war[++Tim] = x;
				modify(x, 1);
				break;
			case 'U':
				x = read();
				modify(war[x], -1);
				break;
		}
	}
	return 0;
}