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毒瘤出題人卡精度……

思路

看到森林裏加邊刪邊，容易想到LCT。

然而LCT上似乎很難實現往一條鏈裏代一個數進去求和，怎麽辦呢？

善良的出題人在下方給了提示：把奇怪的函數泰勒展開搞成多項式，就很好維護了。

註意到數都很小，精度問題不會太大（那你還被卡），可以直接在\(0\)處泰勒展開更為方便。

然後就做完啦~

代碼

要開O2才能過QwQ

#include<bits/stdc++.h>
namespace my_std{
    using namespace std;
    #define pii pair<int,int>
    #define fir first
    #define sec second
    #define MP make_pair
    #define rep(i,x,y) for (int i=(x);i<=(y);i++)
    #define drep(i,x,y) for (int i=(x);i>=(y);i--)
    #define go(x) for (int i=head[x];i;i=edge[i].nxt)
    #define sz 100101
    typedef long long ll;
    template<typename T>
    inline void read(T& t)
    {
        t=0;char f=0,ch=getchar();
        double d=0.1;
        while(ch>‘9‘||ch<‘0‘) f|=(ch==‘-‘),ch=getchar();
        while(ch<=‘9‘&&ch>=‘0‘) t=t*10+ch-48,ch=getchar();
        if(ch==‘.‘)
        {
            ch=getchar();
            while(ch<=‘9‘&&ch>=‘0‘) t+=d*(ch^48),d*=0.1,ch=getchar();
        }
        t=(f?-t:t);
    }
    template<typename T,typename... Args>
    inline void read(T& t,Args&... args){read(t); read(args...);}
    void file()
    {
        #ifndef ONLINE_JUDGE
        freopen("a.txt","r",stdin);
        #endif
    }
//  inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;
typedef long double db;

#define B 20
namespace LCT
{
    int fa[sz],ch[sz][2];
    db sum[sz][B+5],a[sz][B+5];
    bool tag[sz];
    #define ls ch[x][0]
    #define rs ch[x][1]
    #define I inline
    I bool get(int x){return ch[fa[x]][1]==x;}
    I bool nroot(int x){return ch[fa[x]][0]==x||ch[fa[x]][1]==x;}
    I void rev(int x){tag[x]^=1;swap(ls,rs);}
    I void pushup(int x){rep(i,0,B) sum[x][i]=sum[ls][i]+sum[rs][i]+a[x][i];}
    I void pushdown(int x){ if (!x||!tag[x]) return; rev(ls); rev(rs); tag[x]=0; }
    I void rotate(int x)
    {
        int y=fa[x],z=fa[y],k=get(x),w=ch[x][!k];
        if (nroot(y)) ch[z][get(y)]=x;ch[x][!k]=y;ch[y][k]=w;
        if (w) fa[w]=y;fa[y]=x;fa[x]=z;
        pushup(y);pushup(x);
    }
    void Pushdown(int x){if (nroot(x)) Pushdown(fa[x]);pushdown(x);}
    I void splay(int x)
    {
        Pushdown(x);
        while (nroot(x))
        {
            int y=fa[x];
            if (nroot(y)) rotate(get(x)==get(y)?y:x);
            rotate(x);
        }
    }
    void access(int x){for (int y=0;x;x=fa[y=x]) splay(x),rs=y,pushup(x);}
    void makeroot(int x){access(x);splay(x);rev(x);}
    int findroot(int x){access(x);splay(x);while (ls) x=ls;return x;}
    void split(int x,int y){makeroot(x);access(y);splay(y);}
    void link(int x,int y){makeroot(x);access(y);splay(y);fa[x]=y;}
    void cut(int x,int y){split(x,y);fa[x]=ch[y][0]=0;pushup(y);}
    #undef ls
    #undef rs
    #undef I
}

db fac[B+5];
void init(){fac[0]=1;rep(i,1,B) fac[i]=fac[i-1]*i;}

void calc(int x,int f,db a,db b)
{
    #define A LCT::a[x]
    if (f==1)
    {
        db Sin=sin(b),Cos=cos(b),t=1;
        rep(n,0,B)
        {
            db ret=t;
            if (n&1) ret*=Cos; else ret*=Sin;
            if (n%4>=2) ret=-ret;
            A[n]=ret;
            t*=a;
        }
    }
    if (f==2)
    {
        db t=exp(b);
        rep(i,0,B) A[i]=t,t*=a;
    }
    if (f==3) 
    {
        A[0]=b;A[1]=a;
        rep(i,2,B) A[i]=0;
    }
    LCT::pushup(x);
    #undef A
}

int n,m;

int main()
{
    file();
    init();
    string type;
    int f,x,y;db a,b;
    read(n,m);cin>>type;
    rep(i,1,n) read(f,a,b),calc(i,f,a,b);
    while (m--)
    {
        cin>>type;read(x,y);
        if (type[0]==‘a‘) LCT::link(x+1,y+1);
        else if (type[0]==‘d‘) LCT::cut(x+1,y+1);
        else if (type[0]==‘m‘) ++x,f=y,read(a,b),LCT::makeroot(x),calc(x,f,a,b);
        else
        {
            ++x,++y;read(a);
            LCT::makeroot(x);if (LCT::findroot(y)!=x) { puts("unreachable"); continue; }
            LCT::split(x,y);
            db ans=LCT::sum[y][0],t=a;
            rep(i,1,B) ans+=LCT::sum[y][i]*t/fac[i],t*=a;
            printf("%.10Lf\n",ans);
        }
    }
}
 

洛谷P4546 [THUWC2017]在美妙的數學王國中暢遊 [LCT，泰勒展開]