A very interesting question.

You can number each leaf first. Just follow the order in which DFS arrives. (assume $1 to $k $)

Then the number of all leaves under its jurisdiction is calculated for each point. Because it's a DFS sequence, it must be an interval. Set the interval of $u $[l_, r_] $.

Interval plus operation, considering the difference, then the operation of each point becomes $l $plus a number, $R + 1 $minus a number. (also consider $k+1 $)

Then the question becomes $0 for all numbers.

Feel that $(l u u, R UU + 1, C Uu) $is regarded as a weighted edge, so if and only if the selected edge constitutes a connected graph, the requirements are met.

Then it becomes the minimum spanning tree.

Time complexity $O(n\log n) $.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=200020;
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define ROF(i,a,b) for(int i=(a);i>=(b);i--)
#define MEM(x,v) memset(x,v,sizeof(x))
inline int read(){
    int x=0,f=0;char ch=getchar();
    while(ch<'0' || ch>'9') f|=ch=='-',ch=getchar();
    while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
    return f?-x:x;
}
struct edge{
    int u,v,w,id;
    bool operator<(const edge &e)const{
        return w<e.w;
    }
}e[maxn];
int n,c[maxn],el,head[maxn],to[maxn*2],nxt[maxn*2],el2;
int lft[maxn],rig[maxn],dfn[maxn],dfs_clock,ccc,fa[maxn],at[maxn],sss[maxn],al;
bool good[maxn];
ll ans;
inline void add(int u,int v){
    to[++el]=v;nxt[el]=head[u];head[u]=el;
}
int getfa(int x){
    return x==fa[x]?x:fa[x]=getfa(fa[x]);
}
void dfs(int u,int f){
    dfn[u]=++dfs_clock;
    lft[u]=n+1;rig[u]=0;
    for(int i=head[u];i;i=nxt[i]){
        int v=to[i];
        if(v==f) continue;
        dfs(v,u);
        lft[u]=min(lft[u],lft[v]);
        rig[u]=max(rig[u],rig[v]);
    }
    if(!rig[u]) lft[u]=rig[u]=++ccc;
}
int main(){
    n=read();
    FOR(i,1,n) c[i]=read();
    FOR(i,1,n-1){
        int u=read(),v=read();
        add(u,v);add(v,u);
    }
    dfs(1,0);
    FOR(i,1,n) e[++el2]=(edge){lft[i],rig[i]+1,c[i],i};
    sort(e+1,e+el2+1);
    FOR(i,1,ccc) fa[i]=i;
    FOR(i,1,el2){
        int j=i;
        while(j<=el2 && e[j].w==e[i].w) j++;
        j--;
        FOR(k,i,j){
            int u=e[k].u,v=e[k].v;
            u=getfa(u);v=getfa(v);
            if(u!=v) good[e[k].id]=true,al++;
        }
        FOR(k,i,j){
            int u=e[k].u,v=e[k].v;
            u=getfa(u);v=getfa(v);
            if(u!=v) fa[u]=v,ans+=e[k].w;
        }
        i=j;
    }
    cout<<ans<<" "<<al<<endl;
    FOR(i,1,n) if(good[i]) printf("%d ",i);
}