bzoj 2049 [Sdoi2008]Cave 洞穴勘测 LCT

题面

题目传送门

解法

LCT模板题

有些博客讲得挺好的，转一下

博客

这个挺详细的

时间复杂度：\(O(m\ log\ n)\)，尽管常数巨大

代码

#include <bits/stdc++.h>
#define N 20010
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
	x = 0; int f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Node {
	int fa, rev, pathfa, child[2];
} t[N];
int n, m;
int son(int x, int y) {return t[x].child[1] == y;}
void rev(int x) {t[x].rev ^= 1;}
void Clear() {
	for (int i = 1; i <= n; i++)
		t[i] = (Node) {0, false, 0, 0, 0};
}
void pushdown(int k) {
	int x = t[k].rev;
	if (x) {
		swap(t[k].child[0], t[k].child[1]);
		rev(t[k].child[0]), rev(t[k].child[1]);
		t[k].rev = false; 
	}
}
void Connect(int x, int y, int k) {t[x].child[k] = y, t[y].fa = x;}
void Rotate(int x) {
	int y = t[x].fa, z = t[y].fa; pushdown(y);
	pushdown(x); int d = son(y, x);
	swap(t[x].pathfa, t[y].pathfa);
	Connect(z, x, son(z, y));
	Connect(y, t[x].child[!d], d);
	Connect(x, y, !d);
}
void Splay(int x) {
	while (t[x].fa) {
		int y = t[x].fa, z = t[y].fa;
		if (z) {
			pushdown(z), pushdown(y);
			(son(z, y) ^ son(y, x)) ? Rotate(x) : Rotate(y);
		}
		Rotate(x);
	}
}
void expose(int x) {
	Splay(x); pushdown(x);
	int y = t[x].child[1];
	t[y].fa = 0, t[y].pathfa = x;
	t[x].child[1] = 0;
}
void access(int x) {
	for (expose(x); t[x].pathfa; Splay(x)) {
		expose(t[x].pathfa);
		Connect(t[x].pathfa, x, 1);
		t[x].pathfa = 0;
	}
}
void evert(int x) {access(x), Splay(x), rev(x);}
void link(int x, int y) {evert(y), t[y].pathfa = x;}
void cut(int x, int y) {
	evert(x), access(y);
	Splay(y); pushdown(y);
	t[t[y].child[0]].fa = 0;
	t[y].child[0] = 0;
}
int Find(int x) {
	access(x);
	for (pushdown(x); t[x].child[0]; x = t[x].child[0], pushdown(x));
	return x;
}
int main() {
	read(n), read(m); Clear();
	while (m--) {
		char st[10];
		scanf(" %s", st);
		int x, y; read(x), read(y);
		if (st[0] == 'C') link(x, y);
		if (st[0] == 'D') cut(x, y);
		if (st[0] == 'Q') {
			int tx = Find(x), ty = Find(y);
			if (tx == ty) cout << "Yes\n";
				else cout << "No\n";
		}
	}
	return 0;
}