P1501 [国家集训队]Tree II

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一棵n个点的树，每个点的初始权值为1。对于这棵树有q个操作，每个操作为以下四种操作之一：

• + u v c：将u到v的路径上的点的权值都加上自然数c；
• - u1 v1 u2 v2：将树中原有的边(u1,v1)删除，加入一条新边(u2,v2)，保证操作完之后仍然是一棵树；
• * u v c：将u到v的路径上的点的权值都乘上自然数c；
• / u v：询问u到v的路径上的点的权值和，求出答案对于51061的余数。

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第一行两个整数n，q

接下来n-1行每行两个正整数u，v，描述这棵树

接下来q行，每行描述一个操作

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对于每个/对应的答案输出一行

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3 2
1 2
2 3
* 1 3 4
/ 1 1

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4

(color{#0066ff}{数据范围与提示})

10%的数据保证，(1leq n，qleq 2000)

另外15%的数据保证，(1leq n，qleq 5*10^4)，没有-操作，并且初始树为一条链

另外35%的数据保证，(1leq n，qleq 5*10^4)，没有-操作

100%的数据保证，(1leq n，qleq 10^5，0leq cleq 10^4)

(color{#0066ff}{ 题解 })

显然有link，cut的操作

所以用LCT来维护这个树

对于那些加，乘，类比线段树打标记来维护

#include<bits/stdc++.h>
using namespace std;
#define LL long long
LL in() {
	char ch; int x = 0, f = 1;
	while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
	for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
	return x * f;
}
const int mod = 51061;
const int maxn = 1e5 + 5;
struct LCT {
protected: 
	struct node {
		node *ch[2], *fa;
		LL val, num, add, mul, siz, rev;
		node(LL val = 1, LL num = 1, LL add = 0, LL mul = 1, LL siz = 1, int rev = 0)
			: val(val), num(num), add(add), mul(mul), siz(siz), rev(rev) { ch[0] = ch[1] = fa = NULL; }
		void g(LL a, LL m) {
			(val *= m) %= mod, (num *= m) %= mod, (add *= m) %= mod, (mul *= m) %= mod;
			(val += a) %= mod, (num += siz * a % mod) %= mod, (add += a) %= mod;
		}
		bool ntr() { return fa && (fa->ch[0] == this || fa->ch[1] == this); }
		bool isr() { return this == fa->ch[1]; }
		void trn() { std::swap(ch[0], ch[1]), rev ^= 1; }
		void dwn() {
			if(rev) {
				if(ch[0]) ch[0]->trn();
				if(ch[1]) ch[1]->trn();
				rev = 0;
			}
			if(ch[0]) ch[0]->g(add, mul);
			if(ch[1]) ch[1]->g(add, mul);
			add = 0, mul = 1;
		}
		void upd() {
			siz = 1, num = val;
			if(ch[0]) siz += ch[0]->siz, num += ch[0]->num;
			if(ch[1]) siz += ch[1]->siz, num += ch[1]->num;
		}
	}s[maxn], *t[maxn];
	int top;
	void rot(node *x) {
		node *y = x->fa, *z = y->fa;
		int k = x->isr(); node *w = x->ch[!k];
		if(y->ntr()) z->ch[y->isr()] = x;
		x->ch[!k] = y, y->ch[k] = w;
		y->fa = x, x->fa = z;
		if(w) w->fa = y;
		y->upd(), x->upd();
	}
	void splay(node *o) {
		t[top = 1] = o;
		while(t[top]->ntr()) t[top + 1] = t[top]->fa, top++;
		while(top) t[top--]->dwn();
		while(o->ntr()) {
			if(o->fa->ntr()) rot(o->isr() ^ o->fa->isr()? o : o->fa);
			rot(o);
		}
	}
	void access(node *x) {
		for(node *y = NULL; x; x = (y = x)->fa)
			splay(x), x->ch[1] = y, x->upd();
	}
	void makeroot(node *x) { access(x), splay(x), x->trn(); }
	node *findroot(node *x) {
		access(x), splay(x);
		while(x->dwn(), x->ch[0]) x = x->ch[0];
		return splay(x), x;
	}
	void link(node *x, node *y) {
		if(findroot(x) == findroot(y)) return;
		makeroot(x), x->fa = y;
	}
	void cut(node *x, node *y) {
		makeroot(x), access(y), splay(y);
		if(y->ch[0] == x) y->ch[0] = x->fa = NULL;
	}
	int query(node *x, node *y) {
		makeroot(x), access(y), splay(y);
		return y->num % mod;
	}
	void addpath(node *x, node *y, LL c) {
		makeroot(x), access(y), splay(y);
		y->g(c, 1);
	}
	void mulpath(node *x, node *y, LL c) {
		makeroot(x), access(y), splay(y);
		y->g(0, c);
	}
public:
	void link(int x, int y) { link(s + x, s + y); }
	void cut(int x, int y) { cut(s + x, s + y); }
	void add(int x, int y, LL c) { addpath(s + x, s + y, c); }
	void mul(int x, int y, LL c) { mulpath(s + x, s + y, c); }
	int query(int x, int y) { return query(s + x, s + y); }
}v;
char getch() {
	char ch = getchar();
	while(ch != '+' && ch != '-' && ch != '*' && ch != '/') ch = getchar();
	return ch;
}
int main() {
	int n = in(), q = in();
	for(int i = 1; i < n; i++) v.link(in(), in());
	int a, b, c, d;
	while(q --> 0) {
		char ch = getch();
		if(ch == '+') a = in(), b = in(), c = in(), v.add(a, b, c);
		if(ch == '-') a = in(), b = in(), c = in(), d = in(), v.cut(a, b), v.link(c, d);
		if(ch == '*') a = in(), b = in(), c = in(), v.mul(a, b, c);
		if(ch == '/') a = in(), b = in(), printf("%d
", v.query(a, b));
	}
	return 0;
}