description

有必要說明一下，下面的點分治在菊花圖上會tle。

solution

直接點分治，維護點權最小值和邊權和，按點權最小值排序，兩個指標維護一下最大值即可。

code

#include<set>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for
 
(int i=j;i>=k;i--)
#define fr(i,j) for(int i=begin[j];i;i=next[i])
using namespace std;
int const mn=1e5+9,mm=2*1e5+9,mo=1<<30;
int t,n,gra,begin[mn],to[mm],len[mm],next[mm],size[mn],vis[mn],cntb,w[mn];
LL ans;
void insert(int u,int v,int w){
    to[++gra]=v;
    len[gra]=w;
    next[gra]=begin
 
[u];
    begin[u]=gra;
}
void gsize(int p,int q){
    size[p]=1;
    fr(i,p)if((to[i]!=q)&&(!vis[to[i]])){
        gsize(to[i],p);
        size[p]+=size[to[i]];
    }
}
int groot(int p,int q,int lim){
    fr(i,p)if((to[i]!=q)&&(!vis[to[i]])){
        int tmp=groot(to[i],p,lim);
        if
 
(tmp)return tmp;
    }
    if(size[p]>lim)return p;
    return 0;
}
struct rec{
    int mi;
    LL sum;
    friend bool operator<(rec x,rec y){
        return x.mi<y.mi;
    }
};
rec a[mn],b[mn],c[mn];
void dfs(int p,int q,int mi,LL sum){
    b[++cntb].mi=mi;
    b[cntb].sum=sum;
    fr(i,p)if((to[i]!=q)&&(!vis[to[i]]))
        dfs(to[i],p,min(mi,w[to[i]]),sum+len[i]);
}
LL calc(rec *a,int cnta,rec *b,int cntb){
    LL tmp=0,mx=0;
    for(int i=cnta,j=cntb+1;i;i--){
        while((j>1)&&(b[j-1].mi>=a[i].mi))j--,mx=max(mx,b[j].sum);
        tmp=max(tmp,a[i].mi*(a[i].sum+mx));
    }
    return tmp;
}
void solve(int p){
    gsize(p,0);
    p=groot(p,0,size[p]/2);
    vis[p]=1;
    int cnta=0;
    fr(i,p)if(!vis[to[i]]){
        cntb=0;
        dfs(to[i],p,min(w[p],w[to[i]]),len[i]);
        sort(b+1,b+cntb+1);
        ans=max(ans,calc(a,cnta,b,cntb));
        ans=max(ans,calc(b,cntb,a,cnta));
        int cntc=0;
        for(int j=1,k=1;(j<=cnta)||(k<=cntb);){
            cntc++;
            if((k>cntb)||((j<=cnta)&&(a[j].mi<b[k].mi))){
                c[cntc].mi=a[j].mi;
                c[cntc].sum=a[j].sum;
                j++;
            }else{
                c[cntc].mi=b[k].mi;
                c[cntc].sum=b[k].sum;
                k++;
            }
        }
        cnta=cntc;
        fo(j,1,cntc){
            a[j].mi=c[j].mi;
            a[j].sum=c[j].sum;
        }
    }
    fr(i,p)if(!vis[to[i]])solve(to[i]);
}
int main(){
    freopen("d.in","r",stdin);
    freopen("d.out","w",stdout);
    scanf("%d",&t);
    fo(cas,1,t){
        scanf("%d",&n);
        fo(i,1,n)scanf("%d",&w[i]),vis[i]=0;
        gra=0;
        fo(i,1,n)begin[i]=0;
        fo(i,1,n-1){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            insert(u,v,w);
            insert(v,u,w);
        }
        ans=0;
        solve(1);
        printf("%lld\n",ans);
    }
    return 0;
}