BZOJ 2631 tree / Luogu P1501 [国家集训队]Tree II (LCT，多重标记)

题意

一棵树,有删边加边,有一条链加/乘一个数,有询问一条链的和

分析

LCT,像线段树一样维护两个标记(再加上翻转标记就是三个),维护size,就行了

CODE

#include<bits/stdc++.h>
using namespace std;
inline void read(int &num) {
	char ch; int flg = 1; while(!isdigit(ch=getchar()))if(ch=='-')flg = -flg;
	for(num=0; isdigit(ch); num=num*10+ch-'0', ch=getchar()); num*=flg;
}
const int MAXN = 100005;
const int mod = 51061;
namespace LCT {
	#define ls ch[x][0]
	#define rs ch[x][1]
	int ch[MAXN][2], fa[MAXN], w[MAXN], sz[MAXN], mul[MAXN], add[MAXN], sum[MAXN];
	bool rev[MAXN];
	inline bool isr(int x) { return ch[fa[x]][0] != x && ch[fa[x]][1] != x; }
	inline bool get(int x) { return ch[fa[x]][1] == x; }
	inline void upd(int x) {
		sum[x] = (sum[ls] + sum[rs] + w[x]) % mod;
		sz[x] = sz[ls] + sz[rs] + 1;
	}
	inline void pusha(int x, int val) {
		if(!x) return;
		sum[x] = (sum[x] + 1ll * val * sz[x]) % mod;
		w[x] = (w[x] + val) % mod;
		add[x] = (add[x] + val) % mod;
	}
	inline void pushm(int x, int val) {
		if(!x) return;
		sum[x] = 1ll * sum[x] * val % mod;
		w[x] = 1ll * w[x] * val % mod;
		add[x] = 1ll * add[x] * val % mod;
		mul[x] = 1ll * mul[x] * val % mod;
	}
	inline void mt(int x) {
		if(rev[x]) rev[ls]^=1, rev[rs]^=1, rev[x]^=1, swap(ls, rs);
		if(mul[x]!=1) pushm(ls, mul[x]), pushm(rs, mul[x]), mul[x] = 1;
		if(add[x]) pusha(ls, add[x]), pusha(rs, add[x]), add[x] = 0;
	}
	void mtpath(int x) { if(!isr(x)) mtpath(fa[x]); mt(x); }
	inline void rot(int x) {
		int y = fa[x], z = fa[y]; bool l = get(x), r = l^1;
		if(!isr(y)) ch[z][get(y)] = x;
		fa[ch[x][r]] = y; fa[y] = x; fa[x] = z;
		ch[y][l] = ch[x][r]; ch[x][r] = y;
		upd(y), upd(x);
	}
	inline void splay(int x) {
		mtpath(x);
		for(; !isr(x); rot(x))
			if(!isr(fa[x])) rot(get(fa[x])==get(x)?fa[x]:x);
	}
	inline int access(int x) { int y = 0;
		for(; x; x=fa[y=x]) splay(x), ch[x][1] = y, upd(x);
		return y;
	}
	inline void bert(int x) { access(x), splay(x), rev[x]^=1; }
	inline int sert(int x) {
		access(x), splay(x);
		for(; ch[x][0]; x=ch[x][0]);
		return x;
	}
	inline void Link(int x, int y) { bert(x); if(sert(y) != x) fa[x] = y; }
	inline void Cut(int x, int y) { bert(x); if(sert(y) == x && fa[x] == y && !ch[x][1]) fa[x] = ch[y][0] = 0, upd(y); }
	inline int split(int x, int y) { bert(x), access(y), splay(y); return y; }
	inline void Add(int x, int y, int val) { pusha(split(x, y), val); }
	inline void Mul(int x, int y, int val) { pushm(split(x, y), val); }
	inline int Query(int x, int y) { return sum[split(x, y)]; }
	#undef ls
	#undef rs
}
using namespace LCT;
int n, q;
int main() {
	//freopen("sample.txt", "r", stdin);
	read(n), read(q);
	for(int i = 1; i <= n; ++i)
		w[i] = sum[i] = mul[i] = sz[i] = 1;
	for(int i = 1, x, y; i < n; ++i)
		read(x), read(y), Link(x, y);
	char ch[2];
	int x, y, u, v;
	while(q--) {
		scanf("%s", ch);
		switch(ch[0]) {
			case '+':
				read(u), read(v), read(x), Add(u, v, x%mod);
			break;
			case '-':
				read(u), read(v), read(x), read(y), Cut(u, v), Link(x, y);
			break;
			case '*':
				read(u), read(v), read(x), Mul(u, v, x%mod);
			break;
			case '/':
				read(u), read(v), printf("%d\n", Query(u, v));
			break;
		}
	}
}