[洛谷P4312][COCI 2009] OTOCI / 极地旅行社

题目大意：有$n(nleqslant3 imes10^4)$个点，每个点有点权，$m(mleqslant3 imes10^5)$个操作，操作分三种：

1. $bridge;x;y：$询问节点$x$与节点$y$是否连通，若不连通则连一条边
2. $penguins;x;y：$把节点$x$点权改为$y$
3. $excursion;x;y：$询问$x->y$路径上点权和

题解：$LCT$直接维护即可

卡点：无

C++ Code：

#include <algorithm>
#include <cstdio>
#define maxn 30010
#define lc(rt) son[rt][0]
#define rc(rt) son[rt][1]

int son[maxn][2], fa[maxn], tg[maxn];
int w[maxn], S[maxn];
inline bool get(int rt, int tg = 1) { return son[fa[rt]][tg] == rt; }
inline bool is_root(int rt) { return !(get(rt, 0) || get(rt)); }
inline void update(int rt) {
	S[rt] = S[lc(rt)] + S[rc(rt)] + w[rt];
}
inline void pushdown(int rt) {
	if (tg[rt]) {
		tg[rt] ^= 1, tg[lc(rt)] ^= 1, tg[rc(rt)] ^= 1;
		std::swap(lc(rt), rc(rt));
	}
}
inline void rotate(int x) {
	int y = fa[x], z = fa[y], b = get(x);
	if (!is_root(y)) son[z][get(y)] = x;
	fa[son[y][b] = son[x][!b]] = y, son[x][!b] = y;
	fa[y] = x, fa[x] = z;
	update(y), update(x);
}
inline void splay(int x) {
	static int S[maxn], top;
	S[top = 1] = x;
	for (int y = x; !is_root(y); S[++top] = y = fa[y]) ;
	for (; top; top--) pushdown(S[top]);
	for (; !is_root(x); rotate(x)) if (!is_root(fa[x]))
		get(x) ^ get(fa[x]) ? rotate(x) : rotate(fa[x]);
	update(x);
}
inline void access(int x) { for (int t = 0; x; rc(x) = t, t = x, x = fa[x]) splay(x); }
inline void make_root(int rt) { access(rt), splay(rt), tg[rt] ^= 1; }
inline void link(int x, int y) { make_root(x), fa[x] = y; }
inline void split(int x, int y) { make_root(x), access(y), splay(y); }
inline void cut(int x, int y) { split(x, y), lc(y) = fa[x] = 0; }
inline int findroot(int x) {
	access(x), splay(x);
	while (lc(x)) x = lc(x), pushdown(x);
	splay(x);
	return x;
}
inline bool connected(int x, int y) {
	split(x, y);
	return x == findroot(y);
}

int n, m;
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", w + i);
		S[i] = w[i];
	}
	scanf("%d", &m);
	while (m --> 0) {
		char op[15];
		int x, y;
		scanf("%s%d%d", op, &x, &y);
		switch (*op) {
			case 'b':
				if (connected(x, y)) puts("no");
				else {
					puts("yes");
					link(x, y);
				}
				break;
			case 'p':
				make_root(x);
				w[x] = y;
				update(x);
				break;
			case 'e':
				if (connected(x, y)) {
					split(x, y);
					printf("%d
", S[y]);
				} else puts("impossible");
		}
	}
	return 0;
}