Side of the partition may be considered, for a certain color, we deal with this color all endpoint path length side partition left and right sides, then just go for like
For each set but this requires asking edge points again, this complexity is \ (O (nm + n \ log n) \) of
There is a more violent approach is to enumerate all the trees path, so that you can direct statistics, complexity is \ (O (n ^ 2) \) of
Combine these two violent moment, when a partition block size is less than \ (\ sqrt {n} \ ) , we will direct the second run of violence, or we ran side partition
The second run of violence when we need to quickly determine the current which corresponds to a query point, then a need for about two minutes, then a \ (\ log \) , so we set the threshold than \ (\ sqrt {n} \ ) slightly smaller so that the complexity of the rule inclined edge points, so that we can run past
Code
#include<bits/stdc++.h>
#define re register
#define LL long long
#pragma GCC optimize(3)
#define max(a,b) ((a)>(b)?(a):(b))
inline int read() {
char c=getchar();int x=0;while(c<'0'||c>'9') c=getchar();
while(c>='0'&&c<='9') x=(x<<3)+(x<<1)+c-48,c=getchar();return x;
}
const int maxn=1e5+5;const int M=4e5+3;
int rn,n,num,rt,Mnow,S,B,tot,Q,sz;
struct E{int v,nxt,w;}e[M<<1];
std::vector<int> son[M],w[M];
int vis[M],head[M],sum[M],col[maxn];
LL Ans[maxn],tax[2][maxn],b[maxn];
int st[2][M],top[2],t[2][maxn],c[maxn],qx[maxn],qy[maxn];
inline int find(LL x) {
int l=1,r=sz;
while(l<=r) {
int mid=l+r>>1;
if(b[mid]==x) return mid;
if(b[mid]<x) l=mid+1;else r=mid-1;
}
return 0;
}
inline void add(int x,int y,int w) {
e[++num].v=y;e[num].nxt=head[x];head[x]=num;e[num].w=w;
}
void dfs1(int x,int fa) {
for(re int i=head[x];i;i=e[i].nxt) {
if(e[i].v==fa) continue;
son[x].push_back(e[i].v);w[x].push_back(e[i].w);
dfs1(e[i].v,x);
}
}
void getrt(int x,int fa) {
sum[x]=1;
for(re int i=head[x];i;i=e[i].nxt) {
if(e[i].v==fa||vis[i>>1]) continue;
getrt(e[i].v,x);sum[x]+=sum[e[i].v];
int now=max(sum[x],S-sum[x]);
if(now<Mnow) Mnow=now,rt=i;
}
}
void dfs2(int x,LL dis,int fa,int o) {
if(x<=rn) tax[o][col[x]]+=dis,st[o][++top[o]]=col[x],t[o][col[x]]++;
for(re int i=head[x];i;i=e[i].nxt) {
if(e[i].v==fa||vis[i>>1]) continue;
dfs2(e[i].v,dis+e[i].w,x,o);
}
}
void calc(int x,int fa,int o,LL dis) {
if(x<=rn&&col[x]>=o) {
int to=c[find(1ll*o*rn+col[x])];
if(to) Ans[to]+=dis;
}
for(re int i=head[x];i;i=e[i].nxt) {
if(e[i].v==fa||vis[i>>1]) continue;
calc(e[i].v,x,o,dis+e[i].w);
}
}
void dfs(int x,int fa) {
if(x<=rn) calc(x,0,col[x],0);
for(re int i=head[x];i;i=e[i].nxt) {
if(e[i].v==fa||vis[i>>1]) continue;
dfs(e[i].v,x);
}
}
void solve(int x,int s) {
if(s<=B) {dfs(x,0);return;}
Mnow=M,S=s,getrt(x,0);
if(Mnow==M) return;vis[rt>>1]=1;
dfs2(e[rt].v,e[rt].w,0,0);dfs2(e[rt^1].v,0,0,1);
for(re int i=1;i<=Q;i++) {
Ans[i]+=1ll*t[0][qx[i]]*tax[1][qy[i]]+1ll*t[1][qx[i]]*tax[0][qy[i]]
+1ll*t[0][qy[i]]*tax[1][qx[i]]+1ll*t[1][qy[i]]*tax[0][qx[i]];
}
while(top[0]) tax[0][st[0][top[0]]]=t[0][st[0][top[0]]]=0,top[0]--;
while(top[1]) tax[1][st[1][top[1]]]=t[1][st[1][top[1]]]=0,top[1]--;
int k=e[rt^1].v,h=S-sum[e[rt].v];
solve(e[rt].v,sum[e[rt].v]);solve(k,h);
}
int main() {
rn=n=read();
for(re int i=1;i<=n;i++) col[i]=read();
for(re int x,y,z,i=1;i<n;i++)
x=read(),y=read(),z=read(),add(x,y,z),add(y,x,z);
dfs1(1,0);num=1;for(re int i=1;i<=n;i++) head[i]=0;int s[2];
for(re int i=1;i<=n;i++) {
int t=son[i].size();
if(t<=2) {
for(re int j=0;j<t;++j)
add(i,son[i][j],w[i][j]),add(son[i][j],i,w[i][j]);
continue;
}
s[0]=++n,s[1]=++n;
add(i,s[0],0),add(s[0],i,0);add(i,s[1],0),add(s[1],i,0);
for(re int j=0;j<t;++j)
son[s[j&1]].push_back(son[i][j]),w[s[j&1]].push_back(w[i][j]);
}
Q=read();B=0.8*std::sqrt(rn);
for(re int i=1;i<=Q;i++) {
qx[i]=read(),qy[i]=read();
if(qx[i]>qy[i]) std::swap(qx[i],qy[i]);
b[i]=1ll*qx[i]*rn+qy[i];
}
std::sort(b+1,b+Q+1);sz=std::unique(b+1,b+Q+1)-b-1;
for(re int i=Q;i;--i) c[find(1ll*qx[i]*rn+qy[i])]=i;
solve(1,n);
for(re int i=1;i<=Q;i++) {
int to=c[find(1ll*qx[i]*rn+qy[i])];
if(to!=i) {printf("%lld\n",Ans[to]);continue;}
Ans[i]/=(qx[i]==qy[i]?2ll:1);
printf("%lld\n",Ans[i]);
}
return 0;
}