題意略過，口胡正解

對於每個節點i，我們求出每個與i相連的節點的權值和s，那麼i的貢獻就是∑d[j]*(s-d[j])

求最大就排個序咯

#include<stdio.h>
#include<vector>
#include<algorithm>
#define M 10007
using namespace std;
int n,d[200010],mx=0,sum=0;
vector<int> G[200010];
inline bool c1(int a,int b){ return d[a]>d[b]; }
int main(){
	scanf("%d",&n);
	for
 
(int x,y,i=1;i<n;++i){
		scanf("%d%d",&x,&y);
		G[x].push_back(y);
		G[y].push_back(x);
	}
	for(int i=1;i<=n;++i) scanf("%d",d+i);
	for(int S,i=1;i<=n;++i){
		if(G[i].size()<2) continue;
		sort(G[i].begin(),G[i].end(),c1);
		mx=max(mx,d[G[i][0]]*d[G[i][1]]); S=0;
		for(int j=0,z=G[i].size();j<z;++j) S+=d[G[i][j]];
		for
 
(int j=0,z=G[i].size();j<z;++j) sum=((1ll*(S-d[G[i][j]]+M)*d[G[i][j]])%M+sum)%M;
	}
	printf("%d %d\n",mx,sum);
}