[THUWC2017]在美妙的数学王国中畅游 LCT+泰勒展开+求导

p.s. 复合函数求导时千万不能先带值，再求导.

一定要先将符合函数按照求导的规则展开，再带值.

设 $f(x)=g(h(x))$，则对 $f(x)$ 求导： $f'(x)=h'(x)g'(h(x))$

此题中，我们用 LCT 维护 $x^{i}$ 前的系数和，每次询问时将一条链的系数和提出，将 $x$ 带入其前 15 项即可.

Code:

#include<bits/stdc++.h>
using namespace std;   
#define maxn 500000 
#define M 17  
#define setIO(s) freopen(s".in","r",stdin)  //,freopen(s".out","w",stdout)    
namespace tree{
    #define ls ch[x][0]
    #define rs ch[x][1]     
    #define lson ch[x][0]
    #define rson ch[x][0] 
    int ch[maxn][2],f[maxn],op[maxn],rev[maxn];       
    int sta[maxn];            
    double s[maxn][30],a[maxn],b[maxn]; 
    int get(int x){ return ch[f[x]][1]==x; }
    int isrt(int x){ return !(ch[f[x]][0]==x||ch[f[x]][1]==x); }
    void rever(int x){
        if(!x) return;
        rev[x]^=1;                        
        swap(ch[x][0],ch[x][1]); 
    }        
    void pd(int x){
        if(!rev[x]||!x) return;  
        if(rev[x]) rever(ch[x][0]),rever(ch[x][1]),rev[x]=0; 
    }                    
    void up(int x){  
        for(int i=0;i<M;++i)s[x][i]=s[ch[x][0]][i]+s[ch[x][1]][i];  
        if(op[x]==1){ 
        	double val=1.00000,Sin=sin(b[x]),Cos=cos(b[x]);  
            for(int i=0;i<M;i+=4){           
            	s[x][i]+=val*Sin,val*=a[x];
            	s[x][i+1]+=val*Cos,val*=a[x];
            	s[x][i+2]-=val*Sin,val*=a[x];
            	s[x][i+3]-=val*Cos,val*=a[x];  
            }
        }
        if(op[x]==2){
        	double EXP=exp(b[x]),val=1.000000; 
        	for(int i=0;i<M;++i){
        		s[x][i]+=EXP*val,val*=a[x]; 
        	}
        }
        if(op[x]==3){                   
        	s[x][0]+=b[x],s[x][1]+=a[x]; 
        }
    }
    void rotate(int x){     
        int old=f[x],oldf=f[old],which=get(x);
        if(!isrt(old))ch[oldf][ch[oldf][1]==old]=x;            
        ch[old][which]=ch[x][which^1],f[ch[old][which]]=old;
        ch[x][which^1]=old,f[old]=x,f[x]=oldf; 
        up(old),up(x); 
    }
    void splay(int x){
        int v=0,u=x; 
        sta[++v]=u; 
        while(!isrt(u))  sta[++v]=f[u],u=f[u];     
        while(v) pd(sta[v--]);  
        u=f[u]; 
        for(int fa;(fa=f[x])!=u;rotate(x)) 
            if(f[fa]!=u) rotate(get(fa)==get(x)?fa:x);                       
    } 
    void Access(int x){ 
        for(int y=0;x;y=x,x=f[x]) splay(x),ch[x][1]=y,up(x);  
    }
    void makert(int x){
        Access(x),splay(x),rever(x); 
    }
    void split(int x,int y){
        makert(x),Access(y),splay(y); 
    }
    void del(int x,int y){
        split(x,y); f[x]=ch[y][0]=0; up(y); 
    }
    void link(int x,int y){
        makert(x),f[x]=y;                             
    }
    int fd(int x){
        Access(x);
        splay(x);                     
        while(ch[x][0]) x=ch[x][0];  
        splay(x); return x; 
    } 
};     
double jc[maxn];
void init(){
    jc[0]=1.000;
    for(int i=1;i<M;++i) jc[i]=jc[i-1]*i; 
}
int main(){
    //setIO("input");         
    init(); 
    char str[20];
    int n,m; 
    scanf("%d%d%s",&n,&m,str); 
    for(int i=1;i<=n;++i) scanf("%d%lf%lf",&tree::op[i],&tree::a[i],&tree::b[i]); 
    while(m--){ 
        scanf("%s",str);   
        if(str[0]=='a') {
        	int x,y;  
        	scanf("%d%d",&x,&y);
        	++x,++y; 
        	tree::link(x,y);  
        }
        if(str[0]=='d') {
        	int x,y;  
        	scanf("%d%d",&x,&y);
        	++x,++y; 
        	tree::del(x,y);    
        }
        if(str[0]=='m'){
        	int x,y;
        	double w,k; 
            scanf("%d%d%lf%lf",&x,&y,&w,&k);
            ++x;                                                
            tree::Access(x),tree::splay(x);
            tree::op[x]=y,tree::a[x]=w,tree::b[x]=k;  
            tree::up(x); 
        }
        if(str[0]=='t'){
        	int u,v;
        	double w;  
            scanf("%d%d%lf",&u,&v,&w); 
            ++u,++v; 
            if(tree::fd(u)!=tree::fd(v)){
                printf("unreachable\n");
            }else{           
                tree::split(u,v);   
                double ans=0.0,val=1.00000;  
                for(int i=0;i<M;++i){
                    ans+=(double)tree::s[v][i]*val/jc[i]; 
                    val*=w;                                             
                }
                printf("%.8e\n",ans); 
            } 
        }
    } 
    return 0; 
}