【Codeforces 566 C】Logistical Questions（求导+点分治）

传送门

由于$x^{\frac 3 2 }$是单调函数
在一条链上可以二分看往哪边更优
求导加起来即可得到变化量

在树上就点分就可以了

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define fi first
#define se second
#define bg begin
cs int RLEN=(1<<20)+1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob)?EOF:*ib++;
}
inline int read(){
	char ch=gc();
	int res=0;bool f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
template<class tp>inline void chemx(tp&a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp&a,tp b){a>b?a=b:0;}
cs int N=200005;
int n,maxn,siz[N],rt,mas;
int w[N],vis[N];
int adj[N],nxt[N<<1],to[N<<1],val[N<<1],cnt;
inline void addedge(int u,int v,int w){
	nxt[++cnt]=adj[u],adj[u]=cnt,to[cnt]=v,val[cnt]=w;
}
void findrt(int u,int fa){
	siz[u]=1;int son=0;
	for(int e=adj[u];e;e=nxt[e]){
		int v=to[e];
		if(v==fa||vis[v])continue;
		findrt(v,u),siz[u]+=siz[v];
		chemx(son,siz[v]);
	}
	chemx(son,maxn-siz[u]);
	if(son<mas)mas=son,rt=u;
}
double f[N],ans;
int ansrt;
void getans(int u,int fa,int dis){
	ans+=sqrt(dis)*dis*w[u];
	for(int e=adj[u];e;e=nxt[e]){
		int v=to[e];
		if(v==fa)continue;
		getans(v,u,dis+val[e]);
	}
}
void calc(int u,int fa,int dis){
	f[u]=1.5*sqrt(dis)*w[u];
	for(int e=adj[u];e;e=nxt[e]){
		int v=to[e];
		if(v==fa)continue;
		calc(v,u,dis+val[e]);
		f[u]+=f[v];
	}
}
double rans=1e30;
int rrt;
void solve(int u){
	vis[u]=1;
	ans=0;
	getans(u,0,0);
	if(ans<rans)rans=ans,rrt=u;
	calc(u,0,0);
	for(int e=adj[u];e;e=nxt[e]){
		int v=to[e];
		if(vis[v])continue;
		if(f[u]-2*f[v]<=0){
			maxn=mas=siz[v];
			findrt(v,0);
			solve(rt);break;
		}
	}
}
int main(){
	#ifdef Stargazer
	freopen("lx.cpp","r",stdin);
	#endif
	mas=maxn=n=read();
	for(int i=1;i<=n;i++)w[i]=read();
	for(int i=1;i<n;i++){
		int u=read(),v=read(),w=read();
		addedge(u,v,w),addedge(v,u,w);
	}
	findrt(1,0);
	solve(rt);
	printf("%d %.8lf",rrt,rans);
}