\(\color{#0066ff}{ 題目描述 }\)

一棵n個點的樹，每個點的初始權值為1。對於這棵樹有q個操作，每個操作為以下四種操作之一：

• + u v c：將u到v的路徑上的點的權值都加上自然數c；
• - u1 v1 u2 v2：將樹中原有的邊(u1,v1)刪除，加入一條新邊(u2,v2)，保證操作完之後仍然是一棵樹；
• * u v c：將u到v的路徑上的點的權值都乘上自然數c；
• / u v：詢問u到v的路徑上的點的權值和，求出答案對於51061的余數。

\(\color{#0066ff}{輸入格式}\)

第一行兩個整數n，q

接下來n-1行每行兩個正整數u，v，描述這棵樹

接下來q行，每行描述一個操作

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對於每個/對應的答案輸出一行

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

\(\color{#0066ff}{輸出樣例}\)

4

\(\color{#0066ff}{數據範圍與提示}\)

10%的數據保證，\(1\leq n，q\leq 2000\)

另外15%的數據保證，\(1\leq n，q\leq 5*10^4\)，沒有-操作，並且初始樹為一條鏈

另外35%的數據保證，\(1\leq n，q\leq 5*10^4\)，沒有-操作

100%的數據保證，\(1\leq n，q\leq 10^5，0\leq c\leq 10^4\)

\(\color{#0066ff}{ 題解 }\)

顯然有link，cut的操作

所以用LCT來維護這個樹

對於那些加，乘，類比線段樹打標記來維護

#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\n", v.query(a, b));
    }
    return 0;
}
 

P1501 [國家集訓隊]Tree II