LOJ2116 [HNOI2015] 开店 【点分治】

题目分析：

　　观察题目发现度数不小于三，考虑从边分治入手，用点分治代替。将树划分成重心链接的结构，称为点分树。令当前询问的点为$ u $。那么我们考虑点分树的根到$ u $的一条路径。考虑根结点，排除掉与$ u $有关的点所在的子树，剩下的点到$ u $的路径长度等价于它们到根的路径加上根到$ u $的路径。所以我们要做的是消去$ u $所在子树的影响，然后递归问题。这个做法在点分治意义下是常用做法，前缀和维护即可。由于每个点都在点分路径上的前缀和上出现了一次，而点分树高为$ O(\log n) $，所以空间为$ O(n*\log n) $。时间是$ O(n*\log^2n) $

代码：

 #include<bits/stdc++.h>
 using namespace std;
 
 const int maxn = ;
 
 int n,Q,A;
 
 int x[maxn];
 
 vector <pair<int,int> > g[maxn];
 int euler[maxn<<],nE,app[maxn][];
 int RMQ[maxn<<][],f[maxn],dep[maxn],climb[maxn];
 
 template<typename T>
 void in(T &x){
     x = ; char ch = getchar();
     while(ch > '' || ch < '') ch = getchar();
     while(ch >= '' && ch <= '') x = x*+ch-'',ch=getchar();
 }
 
 void dfs(int now,int last,int dp,int cl){
     dep[now] = dp; f[now] = last; climb[now] = cl;
     euler[++nE] = now; app[now][] = app[now][] = nE;
     for(int i=;i<g[now].size();i++){
     int pp = g[now][i].first,len = g[now][i].second;
     if(pp == last) continue;
     dfs(pp,now,dp+len,cl+);
     euler[++nE] = now; app[now][] = nE;
     }
 }
 
 void BuildRMQ(){
     for(int i=;i<=nE;i++) RMQ[i][] = euler[i];
     for(int k=;(<<k)<=nE;k++){
     for(int i=;i<=nE;i++){
         if(i+(<<k-) > nE) {RMQ[i][k] = RMQ[i][k-];continue;}
         if(climb[RMQ[i][k-]] < climb[RMQ[i+(<<k-)][k-]]){
         RMQ[i][k] = RMQ[i][k-];
         }else RMQ[i][k] = RMQ[i+(<<k-)][k-];
     }
     }
 }
 
 int QueryLCA(int u,int v){
     int fst = min(app[u][],app[v][]),sec = max(app[u][],app[v][]);
     int len = sec-fst+,k = ;
     while((<<k+) <= len) k++;
     if(climb[RMQ[fst][k]] < climb[RMQ[sec-(<<k)+][k]]) return RMQ[fst][k];
     else return RMQ[sec-(<<k)+][k];
 }
 
 int dist(int u,int v){ return dep[u]+dep[v]-*dep[QueryLCA(u,v)]; }
 
 class Pdivide{
 private:
     int sz[maxn],arr[maxn],seg[maxn],fa[maxn];
     vector <pair<int,int> > v[maxn],v2[maxn];
     vector <long long> sum[maxn],sum2[maxn];
     stack <int> sta;
     void GetSize(int now,int last){
     sz[now] = ;seg[now] = ;
     for(int i=;i<g[now].size();i++){
         int pp = g[now][i].first;
         if(arr[pp] || pp == last) continue;
         GetSize(pp,now);
         sz[now] += sz[pp];
         seg[now] = max(seg[now],sz[pp]);
     }
     sz[now]++;
     }
     int GetG(int now,int last,int tot){
     int res = now; seg[now] = max(seg[now],tot-sz[now]-);
     for(int i=;i<g[now].size();i++){
         int pp = g[now][i].first;
         if(arr[pp] || pp == last) continue;
         int p = GetG(pp,now,tot);
         if(seg[p] < seg[res]) res = p;
     }
     return res;
     }
     void LenQuery(int now,int last,int dst,int root,int oldroot){
     v[root].push_back(make_pair(x[now],dst));
     if(oldroot != ) v2[root].push_back(make_pair(x[now],dist(now,oldroot)));
     for(int i=;i<g[now].size();i++){
         int pp = g[now][i].first, len = g[now][i].second;
         if(arr[pp] || pp == last) continue;
         LenQuery(pp,now,dst+len,root,oldroot);
     }
     }
     void divide(int now,int last,int ht){
     GetSize(now,);
     int p = GetG(now,,sz[now]);
 
     LenQuery(p,,,p,last);    sort(v[p].begin(),v[p].end()); sum[p].push_back();
     sort(v2[p].begin(),v2[p].end()); sum2[p].push_back();
     for(int i=;i<v[p].size();i++) sum[p].push_back(sum[p][i]+1ll*v[p][i].second);
     for(int i=;i<v2[p].size();i++) sum2[p].push_back(sum2[p][i]+1ll*v2[p][i].second);
 
     fa[p] = last;arr[p] = ht;
     for(int i=;i<g[p].size();i++){
         int nxt = g[p][i].first;
         if(arr[nxt]) continue;
         divide(nxt,p,ht+);
     }
     }
 
 public:
     void BuildTree(){divide(,,);}
     void printans(int now,int l,int r,long long &lastans){
     int p = now;
     while(p != ) sta.push(p),p = fa[p];
     long long ans = ;
     while(!sta.empty()){
         int tp = sta.top();sta.pop();
         int pp = lower_bound(v[tp].begin(),v[tp].end(),make_pair(l,))-v[tp].begin();
         if(pp == v[tp].size()||v[tp][pp].first > r) break;
         int ed = upper_bound(v[tp].begin(),v[tp].end(),make_pair(r,(int)1E9))-v[tp].begin()-;
         int len = ed-pp+;
         ans += sum[tp][ed+] - sum[tp][pp] + 1ll*len*dist(tp,now);
         if(fa[tp]!=)
         ans -= (sum2[tp][ed+] - sum2[tp][pp] + 1ll*len*dist(fa[tp],now));
     }
     while(!sta.empty())sta.pop();
     printf("%lld\n",ans);lastans = ans;
     }
 }T1;
 
 void init(){
     dfs(,,,);
     BuildRMQ();
     T1.BuildTree();
 }
 
 void work(){
     long long lastans = ;
     for(int i=;i<=Q;i++){
     int a,u,v; in(a),in(u),in(v);
     u = (1ll*u+lastans)%A;
     v = (1ll*v+lastans)%A;
     if(u > v) swap(u,v);
      T1.printans(a,u,v,lastans);
     }
 }
 
 int main(){
     in(n),in(Q),in(A);
     for(int i=;i<=n;i++) in(x[i]);
     for(int i=;i<n;i++){
     int u,v,w; in(u),in(v),in(w);
     g[u].push_back(make_pair(v,w));
     g[v].push_back(make_pair(u,w));
     }
     init();
     work();
     return ;
 }