[SDOI2008] 洞穴勘测

Portal

要求维护一个可以支持断开树边的并查集。

\[n <= 10000, m <= 200009 \]

---

LCT的最普通的应用。 直接上模板就好了。

---

#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
#define drep(i, a, b) for(int i = (a), i##_end_ = (b); i >= i##_end_; --i)
#define clar(a, b) memset((a), (b), sizeof(a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define pass {if (true){;}}
typedef long long LL;
typedef long double LD;
int read() {
    char ch = getchar();
    int x = 0, flag = 1;
    for (;!isdigit(ch); ch = getchar()) if (ch == '-') flag *= -1;
    for (;isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    return x * flag;
}
void write(int x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + 48);
}

template <int N> struct LCT {
	struct node {
		int fa, ch[2], revTag, size, wide;
	}t[N];	
	int _top, stk[N];

#define fa(x) (t[(x)].fa)
#define lc(x) (t[(x)].ch[0])
#define rc(x) (t[(x)].ch[1])

	inline bool isroot(int rt) { return lc(fa(rt)) != rt && rc(fa(rt)) != rt; }
	void pushup(int rt) { 
		t[rt].size = t[lc(rt)].size + t[rc(rt)].size + 1; 
	}
	void setRev(int rt) { t[rt].revTag ^= 1, swap(lc(rt), rc(rt)); }
	void pushdown(int rt) {
		if (t[rt].revTag) {
			setRev(lc(rt)), setRev(rc(rt));
			t[rt].revTag = 0;
		}
	}   

	void rotate(int rt) {
		int y = fa(rt), z = fa(y), dir = (rc(y) == rt);
		if (!isroot(y)) t[z].ch[t[z].ch[1] == y] = rt; t[rt].fa = z;
		t[y].ch[dir] = t[rt].ch[dir ^ 1]; t[t[rt].ch[dir ^ 1]].fa = y;
		t[rt].ch[dir ^ 1] = y; t[y].fa = rt;
		pushup(y); pushup(rt);
	}

	void splay(int rt) {
		stk[_top = 1] = rt;
/**/	for (int u = rt; !isroot(u); u = fa(u)) stk[++_top] = fa(u);
		while (_top) pushdown(stk[_top--]);
		while (!isroot(rt)) {
			int y = fa(rt), z = fa(y);
/**//**/	if (!isroot(y))
				(t[z].ch[1] == y) ^ (t[y].ch[1] == rt) ? rotate(rt) : rotate(y);
			rotate(rt);
		}
		pushup(rt);
	}

	void access(int u) { for (int y = 0; u; u = t[y = u].fa) splay(u), t[u].ch[1] = y, pushup(u); }
	void makeRoot(int u) { access(u); splay(u); setRev(u); }
	int findRoot(int u) {
		access(u); splay(u);
		while (t[u].ch[0]) pushdown(u), u = t[u].ch[0];
		return u;
	}
	void spilt(int u, int v) { makeRoot(u); access(v); /*?*/ splay(v); }
	void link(int u, int v) { makeRoot(u); if (findRoot(u) != v) t[u].fa = v; }
	void cut(int u, int v) {
		spilt(u, v);
		if (findRoot(v) == u && t[u].fa == v && t[v].ch[0] == u) {
			t[u].fa = t[v].ch[0] = 0;
			pushup(v);
		}
	}
#undef fa                   
#undef lc
#undef rc
};

const int Maxn = 10009;
LCT <Maxn> s;
int n, m;

void init() {
	n = read(), m = read();
}

char Opt[19];
void solve() {
	while (m--) {
		scanf("%s", Opt);
		int u = read(), v = read();
		if (Opt[0] == 'C') s.link(u, v);
		if (Opt[0] == 'D') s.cut(u, v);
		if (Opt[0] == 'Q') puts(s.findRoot(u) == s.findRoot(v) ? "Yes" : "No");
	}
}

int main() {
	freopen("LG2147.in", "r", stdin);
	freopen("LG2147.out", "w", stdout);

	init();
	solve();

#ifdef Qrsikno
	debug("\nRunning time: %.3lf(s)\n", clock() * 1.0 / CLOCKS_PER_SEC);
#endif
	return 0;
}