简化题面
维护四种操作和一颗树
(1)、断开(u)和(v)之间的边
(2)、在(u)和(v)之间连一条边
(3)、讲节点(c)更换为(i)种函数，且系数和常数项为(a)和(b)
(4)、查询(u)到(v)路径上所有点上函数带入等于(x)时的函数值之和
(nle 10^5)，(mle 2 imes 10^5)
解题思路
首先(1,2)号操作告诉我们这题肯定要用(mathcal{LCT})来做，但是考虑怎么维护函数和函数值，根据小绿本高等数学上有个叫泰勒展开的玩意，我们对于每个节点维护其函数的泰勒展开来代表其近似值，因为三个函数(x)的幂数都很小因此我们维护大概展开(10)次的函数即可，最后查询的时候直接代入即可，具体看代码。
(mathcal{Code})
// Author: Ame__
#include<bits/stdc++.h>
#define _ 0
//#define AME__
#define AME__DEBUG
#define bomb exit(0)
#define LOG(FMT...) fprintf(stderr , FMT)
#define TOWA(FMT...) fprintf(stdout , FMT)
using namespace std;
/*Grievous Lady*/
typedef int32_t i32;
typedef int64_t i64;
typedef double qwq
const int BUF_SIZE = 1 << 12;
char buf[BUF_SIZE] , *buf_s = buf , *buf_t = buf + 1;
#define PTR_NEXT()
{
buf_s ++;
if(buf_s == buf_t)
{
buf_s = buf;
buf_t = buf + fread(buf , 1 , BUF_SIZE , stdin);
}
}
#define mians(_s_)
{
while(!isgraph(*buf_s)) PTR_NEXT();
char register *_ptr_ = (_s_);
while(isgraph(*buf_s) || *buf_s == '-')
{
*(_ptr_ ++) = *buf_s;
PTR_NEXT();
}
(*_ptr_) = ' ';
}
template <typename _n_> void mian(_n_ & _x_){
char buf_s; while(buf_s != '-' && !isdigit(buf_s)) buf_s = getchar();
bool register _nega_ = false; if(buf_s == '-'){ _nega_ = true; buf_s = getchar(); }
_x_ = 0; while(isdigit(buf_s)){ _x_ = _x_ * 10 + buf_s - '0'; buf_s = getchar(); } if(_nega_) _x_ = -_x_;
}
const i32 kato = 2e5 + 10;
template <typename _n_> bool cmax(_n_ &a , const _n_ &b){ return a < b ? a = b , 1 : 0; }
template <typename _n_> bool cmin(_n_ &a , const _n_ &b){ return a > b ? a = b , 1 : 0; }
i32 n , m , op , x , y;
qwq a , b;
qwq qaq[2] , opt[20];
namespace towa{
struct Tylar{
qwq a[12];
void clear(){ memset(a , 0 , sizeof a); }
Tylar(){ clear(); }
Tylar(i32 opt , qwq x , qwq y){
y += x * 0.5;
if(opt == 1){
qwq res = 1 , f = sin(y) , g = cos(y);
for(int i = 1;i <= 11;i ++ , res = res * x , swap(f = -f , g)) this -> a[i] = res * f;
}else if(opt == 2){
qwq res = exp(y);
for(int i = 1;i <= 11;i ++ , res *= x) this -> a[i] = res;
}else *this = Tylar() , this -> a[1] = y , this -> a[2] = x;
}
friend Tylar operator + (const Tylar &a , const Tylar &b){
Tylar c;
for(i32 i = 1;i <= 11;i ++) c.a[i] = a.a[i] + b.a[i];
return c;
}
inline qwq get_ans(qwq x){
qwq res = 0; x -= 0.5;
for(i32 i = 11 ; i ; i --) res = this -> a[i] + res * x / i;
return res;
}
};
struct node{
node *ch[2] , *fa;
static queue<node*> q;
i32 rev;
Tylar val , tot;
node(node *fa = 0x0 , i32 rev = 0): fa(fa) , rev(rev){
ch[0] = ch[1] = 0x0;
}
inline bool ntr(){
return fa && (fa -> ch[0] == this || fa -> ch[1] == this);
}
inline bool isr(){
return this == fa -> ch[1];
}
inline void Modify_rev(){
rev ^= 1; swap(ch[0] , ch[1]);
}
inline void up(){
tot = this -> val;
if(ch[0]) this -> tot = this -> tot + ch[0] -> tot;
if(ch[1]) this -> tot = this -> tot + ch[1] -> tot;
}
inline void down(){
if(rev){
if(ch[0]) ch[0] -> Modify_rev(); if(ch[1]) ch[1] -> Modify_rev();
rev = 0;
}
}
void *operator new(size_t){
static node *S = 0x0 , *T = 0x0; node *tmp;
return q.empty() ? (S == T && (T = (S = new node[1024]) + 1024 , S == T) ? 0x0 : S ++) : (tmp = q.front() , q.pop() , tmp);
}
void operator delete(void *qaq){ q.push(static_cast<node*>(qaq)); }
}pool[kato];
queue<node*> node::q;
inline void split(node *x){
node *y = x -> fa , *z = y -> fa;
y -> down() , x -> down();
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 -> up() , x -> up();
}
inline void splay(node *o){
static node *tree[kato]; int top = 0;
tree[top = 1] = o;
while(tree[top] -> ntr()) tree[top + 1] = tree[top] -> fa , top ++;
while(top) tree[top --] -> down();
while(o -> ntr()){
if(o -> fa -> ntr()) split(o -> isr() ^ o -> fa -> isr() ? o : o -> fa);
split(o);
}
}
inline void access(node *x){
for(node *y = 0x0 ; x ; x = (y = x) -> fa){
splay(x) , x -> ch[1] = y , x -> up();
}
}
inline void Move_to_root(node *x){
access(x) , splay(x) , x -> Modify_rev();
}
inline node *find_root(node *x){
access(x) , splay(x);
while(x -> down() , x -> ch[0]) x = x -> ch[0];
return splay(x) , x;
}
inline void link(node *x , node *y){
if(find_root(x) == find_root(y)) return;
Move_to_root(x); x -> fa = y;
}
inline void cut(node *x , node *y){
Move_to_root(x); access(y); splay(y);
y -> ch[0] = x -> fa = 0x0 , y -> up();
}
inline void change(int id , int tp , qwq a , qwq b){
node *o = pool + id;
splay(o);
o -> val = Tylar(tp , a , b);
o -> up();
}
inline Tylar query(node *x , node *y){
Move_to_root(x) , access(y) , splay(y);
return y -> tot;
}
inline void link(int x , int y){
link(pool + x , pool + y);
}
inline void cut(int x , int y){
cut(pool + x , pool + y);
}
inline void get_ans(int x , int y , qwq z){
if(find_root(pool + x) == find_root(pool + y)) TOWA("%.10f
" , query(pool + x , pool + y).get_ans(z));
else TOWA("unreachable
");
}
}
inline int Ame_(){
#ifdef AME__
freopen("b1.in" , "r" , stdin); freopen("1.out" , "w" , stdout); int nol_cl = clock();
#endif
mian(n) , mian(m); scanf("%s" , qaq);
for(int i = 0;i < n;i ++) mian(op) , scanf("%lf%lf" , &a , &b) , towa::change(i , op , a , b);
for(; m --> 0 ;){
scanf("%s" , opt);
if(opt[0] == 'a') mian(x) , mian(y) , towa::link(x , y);
if(opt[0] == 'm') mian(x) , mian(y) , scanf("%lf%lf" , &a , &b) , towa::change(x , y , a , b);
if(opt[0] == 'd') mian(x) , mian(y) , towa::cut(x , y);
if(opt[0] == 't') mian(x) , mian(y) , scanf("%lf" , &a) , towa::get_ans(x , y , a);
}
#ifdef AME__TIME
LOG("Time: %dms
", int((clock() - nol_cl) / (qwq)CLOCKS_PER_SEC * 1000));
#endif
return ~~(0^_^0); /*さようならプログラム*/
}
int Ame__ = Ame_();
int main(){;}