解题：HNOI 2015 开店

题面

根据树上距离的计算方法，可以先把答案化成$\sum dep_i+n*dep_u-\sum 2*dep[LCA(i,u)]$的形式，然后维护$\sum 2*dep[LCA(i,u)]$

把妖怪们按年龄排序，轻重剖分后插入每个点到根的路径，记录经过次数，询问也是往根跳然后每次统计边权*次数。

可持久化线段树+差分

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int N=,M=1e7+,inf=1e9;
 int root[N],son[M][],num[M];
 int p[N],noww[N],goal[N],val[N];
 int siz[N],far[N],imp[N],top[N],dfn[N];
 long long dis[N],sum1[N],sum2[N],sum[M],ans;
 int n,m,nm,rt,t1,t2,t3,rot,cnt,tot;
 struct a
 {
     int age,idx;
 }mon[N];
 bool operator < (a x,a y)
 {
     return x.age==y.age?x.idx<y.idx:x.age<y.age;
 }
 void Link(int f,int t,int v)
 {
     noww[++cnt]=p[f],p[f]=cnt;
     goal[cnt]=t,val[cnt]=v;
     noww[++cnt]=p[t],p[t]=cnt;
     goal[cnt]=f,val[cnt]=v;
 }
 void DFS(int nde,int fth)
 {
     int tmp=;
     siz[nde]=,far[nde]=fth;
     for(int i=p[nde],g;i;i=noww[i])
         if((g=goal[i])!=fth)
         {
             dis[g]=dis[nde]+val[i];
             DFS(g,nde),siz[nde]+=siz[g];
             if(siz[g]>tmp) tmp=siz[g],imp[nde]=g;
         }
 }
 void Mark(int nde,int tpp)
 {
     top[nde]=tpp,dfn[nde]=++tot;
     sum1[tot]=dis[nde]-dis[far[nde]];
     if(imp[nde])
     {
         Mark(imp[nde],tpp);
         for(int i=p[nde],g;i;i=noww[i])
             if((g=goal[i])!=far[nde]&&g!=imp[nde]) Mark(g,g);
     }
 }
 
 int Insert(int pre,int l,int r,int ll,int rr)
 {
     int nde=++tot;
     son[nde][]=son[pre][];
     son[nde][]=son[pre][];
     num[nde]=num[pre],sum[nde]=sum[pre];
     if(l==ll&&r==rr) num[nde]++;
     else
     {
         int mid=(l+r)>>;
         sum[nde]+=sum1[rr]-sum1[ll-];
         if(mid>=rr) son[nde][]=Insert(son[pre][],l,mid,ll,rr);
         else if(mid<ll) son[nde][]=Insert(son[pre][],mid+,r,ll,rr);
         else son[nde][]=Insert(son[pre][],l,mid,ll,mid),
              son[nde][]=Insert(son[pre][],mid+,r,mid+,rr);
     }
     return nde;
 }
 long long Query(int nde,int l,int r,int ll,int rr)
 {
     long long ret=(sum1[rr]-sum1[ll-])*num[nde];
     if(l==ll&&r==rr)
         return ret+sum[nde];
     else
     {
         int mid=(l+r)>>;
         if(mid>=rr) return ret+Query(son[nde][],l,mid,ll,rr);
         else if(mid<ll) return ret+Query(son[nde][],mid+,r,ll,rr);
         else return ret+Query(son[nde][],l,mid,ll,mid)+Query(son[nde][],mid+,r,mid+,rr);
     }
 }
 
 int Change(int nde)
 {
     while(top[nde]!=rt)
         rot=Insert(rot,,n,dfn[top[nde]],dfn[nde]),nde=far[top[nde]];
     return rot=Insert(rot,,n,,dfn[nde]);
 }
 long long Ask(int trt,int nde)
 {
     long long ret=;
     while(top[nde]!=rt)
         ret+=Query(trt,,n,dfn[top[nde]],dfn[nde]),nde=far[top[nde]];
     return ret+Query(trt,,n,,dfn[nde]);
 }
 
 int main()
 {
     scanf("%d%d%d",&n,&m,&nm),rt=;
     for(int i=;i<=n;i++)
         scanf("%d",&t1),mon[i]=(a){t1,i};
     sort(mon+,mon++n);
     for(int i=;i<n;i++)
         scanf("%d%d%d",&t1,&t2,&t3),Link(t1,t2,t3);
     DFS(,),Mark(,);
     for(int i=;i<=n;i++)
         sum1[i]+=sum1[i-],sum2[i]=sum2[i-]+dis[mon[i].idx];
     for(int i=;i<=n;i++)
         root[i]=Change(mon[i].idx);
     for(int i=;i<=m;i++)
     {
         scanf("%d%d%d",&t1,&t2,&t3);
         t2=(ans+t2)%nm,t3=(ans+t3)%nm;
         if(t2>t3) swap(t2,t3);
         int ll=lower_bound(mon+,mon++n,(a){t2,})-mon;
         int rr=upper_bound(mon+,mon++n,(a){t3,inf})-mon-;
     //    printf("%d==%d==%d==%d==%d==",ll,rr,dis[t1]*(rr-ll+1),sum2[rr]-sum2[ll-1],2*(Ask(root[rr],t1)-Ask(root[ll-1],t1)));
         printf("%lld\n",ans=dis[t1]*(rr-ll+)+sum2[rr]-sum2[ll-]-*(Ask(root[rr],t1)-Ask(root[ll-],t1)));
     }
     return ;
 }