题解 (by;zjvarphi)
一道并查集的题
对于它路径上点权，我们可以转化一下：对于一个点，它在哪些路径上是最小的点权
那么我们排个序，从大到小加入点，每回加入时，将这个点与它所相连的且权值比它大的点所在集合合并
那么这个新集合中，这个点的权值一定是最小的，所以求出这个集合的直径即可
对于这个新集合的直径，一定是由原来的集合的直径的端点组合而来的,或就直接是两个集合中直径大的那个
一共六种情况，枚举即可，复杂度可以做到 (O(nlogn))
Code
#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
using namespace std;
namespace IO{
char buf[1<<21],*p1=buf,*p2=buf;
#define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++
template<typename T>inline void read(T &x) {
ri f=1;x=0;register char ch=gc();
while(ch<'0'||ch>'9') {if (ch=='-') f=0;ch=gc();}
while(ch>='0'&&ch<='9') {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
x=f?x:-x;
}
}
using IO::read;
namespace nanfeng{
#define cmax(x,y) ((x)>(y)?(x):(y))
#define cmin(x,y) ((x)>(y)?(y):(x))
#define FI FILE *IN
#define FO FILE *OUT
typedef long long ll;
static const int N=1e5+7;
int first[N],fa[N],w[N],head[N],st[N<<1][19],lg[N<<1],dep[N],p[N],ol,n,T,t=1;
ll dis[N],wl[N],ans;
struct edge{int v,nxt,w;}e[N<<1];
struct node{int x1,x2;}pnt[N];
inline void add(int u,int v,int w) {
e[t].v=v,e[t].w=w,e[t].nxt=first[u],first[u]=t++;
e[t].v=u,e[t].w=w,e[t].nxt=first[v],first[v]=t++;
}
int find(int x) {return fa[x]==x?x:fa[x]=find(fa[x]);}
void dfs(int x,int f) {
head[st[p(ol)][0]=x]=ol;
for (ri i(first[x]),v;i;i=e[i].nxt) {
if ((v=e[i].v)==f) continue;
dep[v]=dep[x]+1,dis[v]=dis[x]+e[i].w;
dfs(v,x);
st[p(ol)][0]=x;
}
}
inline void init_rmq() {
dfs(1,0);
int k=lg[ol];
for (ri j(1);j<=k;p(j)) {
ri len=1<<j;
for (ri i(1);i+len-1<=ol;p(i)) {
ri x1=st[i][j-1],x2=st[i+(1<<j-1)][j-1];
st[i][j]=dep[x1]<dep[x2]?x1:x2;
}
}
}
inline int Getlca(int u,int v) {
if (head[u]>head[v]) swap(u,v);
ri k=lg[head[v]-head[u]+1];
ri x1=st[head[u]][k],x2=st[head[v]-(1<<k)+1][k];
return dep[x1]<dep[x2]?x1:x2;
}
inline int cmp(int x,int y) {return w[x]>w[y];}
inline int main() {
// FI=freopen("nanfeng.in","r",stdin);
// FO=freopen("nanfeng.out","w",stdout);
for (ri i(2);i<N<<1;p(i)) lg[i]=lg[i>>1]+1;
read(T);
for (ri z(1);z<=T;p(z)) {
memset(first,0,sizeof(first));
memset(wl,0,sizeof(wl));
t=1,ans=ol=0;
read(n);
for (ri i(1);i<=n;p(i)) read(w[i]),fa[i]=p[i]=i,pnt[i].x1=pnt[i].x2=i;
for (ri i(1),u,v,ew;i<n;p(i)) read(u),read(v),read(ew),add(u,v,ew);
init_rmq();
sort(p+1,p+n+1,cmp);
for (ri i(1);i<=n;p(i)) {
ri cur=p[i];
// printf("%d %d ans=%lld
",w[cur],cur,ans);
for (ri j(first[cur]),v;j;j=e[j].nxt) {
if (w[v=e[j].v]<w[cur]) continue;
ri r1=find(e[j].v),r2=find(cur);
ri r1x1=pnt[r1].x1,r1x2=pnt[r1].x2;
ri r2x1=pnt[r2].x1,r2x2=pnt[r2].x2;
ri l11=Getlca(r1x1,r2x1),l12=Getlca(r1x1,r2x2);
ri l21=Getlca(r1x2,r2x1),l22=Getlca(r1x2,r2x2);
ri nx1=r1x1,nx2=r1x2;
// if (cur==1) printf("in r2=%d nx1=%d nx2=%d %lld
",r2,r2x1,r2x2,wl[r2]);
register ll res=wl[r1];
if (wl[r1]<wl[r2]) res=wl[r2],nx1=r2x1,nx2=r2x2;
if (res<dis[r1x1]+dis[r2x1]-(dis[l11]<<1))
res=dis[r1x1]+dis[r2x1]-(dis[l11]<<1),nx1=r1x1,nx2=r2x1;
if (res<dis[r1x1]+dis[r2x2]-(dis[l12]<<1))
res=dis[r1x1]+dis[r2x2]-(dis[l12]<<1),nx1=r1x1,nx2=r2x2;
if (res<dis[r1x2]+dis[r2x1]-(dis[l21]<<1))
res=dis[r1x2]+dis[r2x1]-(dis[l21]<<1),nx1=r1x2,nx2=r2x1;
if (res<dis[r1x2]+dis[r2x2]-(dis[l22]<<1))
res=dis[r1x2]+dis[r2x2]-(dis[l22]<<1),nx1=r1x2,nx2=r2x2;
// res=max(res,di[r1x1]+dis[r2x1]-(dis[Getlca(r1x1,r2x1)]<<1));
// res=max(res,di[r1x1]+dis[r2x2]-(dis[Getlca(r1x1,r2x2)]<<1));
// res=max(res,di[r1x2]+dis[r2x1]-(dis[Getlca(r1x2,r2x1)]<<1));
// res=max(res,di[r1x2]+dis[r2x2]-(dis[Getlca(r1x2,r2x2)]<<1));
wl[fa[r1]=r2]=res;
ans=cmax(ans,res*w[cur]);
pnt[r2].x1=nx1,pnt[r2].x2=nx2;
// if (cur==1) printf("out r2=%d nx1=%d nx2=%d %lld
",r2,r2x1,r2x2,wl[r2]);
// printf("w[%d]=%d %d %d res=%lld ans=%lld
",cur,w[cur],nx1,nx2,res,ans);
}
// printf("%d %d ans=%lld
",w[cur],cur,ans);
}
printf("%lld
",ans);
}
return 0;
}
// #undef int
}
int main() {return nanfeng::main();}