洛谷P4546 [THUWC2017]在美妙的数学王国中畅游 [LCT，泰勒展开]

传送门

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毒瘤出题人卡精度……

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思路

看到森林里加边删边，容易想到LCT。

然而LCT上似乎很难实现往一条链里代一个数进去求和，怎么办呢？

善良的出题人在下方给了提示：把奇怪的函数泰勒展开搞成多项式，就很好维护了。

注意到数都很小，精度问题不会太大（那你还被卡），可以直接在\(0\)处泰勒展开更为方便。

然后就做完啦~

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代码

要开O2才能过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);
		}
	}
}