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 = 150005;

int n,Q,A;

int x[maxn];

vector <pair<int,int> > g[maxn];
int euler[maxn<<1],nE,app[maxn][2];
int RMQ[maxn<<1][19],f[maxn],dep[maxn],climb[maxn];

template<typename T>
void in(T &x){
    x = 0; char ch = getchar();
    while(ch > '9' || ch < '0') ch = getchar();
    while(ch >= '0' && ch <= '9') x = x*10+ch-'0',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][0] = app[now][1] = nE;
    for(int i=0;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+1);
    euler[++nE] = now; app[now][1] = nE;
    }
}

void BuildRMQ(){
    for(int i=1;i<=nE;i++) RMQ[i][0] = euler[i];
    for(int k=1;(1<<k)<=nE;k++){
    for(int i=1;i<=nE;i++){
        if(i+(1<<k-1) > nE) {RMQ[i][k] = RMQ[i][k-1];continue;}
        if(climb[RMQ[i][k-1]] < climb[RMQ[i+(1<<k-1)][k-1]]){
        RMQ[i][k] = RMQ[i][k-1];
        }else RMQ[i][k] = RMQ[i+(1<<k-1)][k-1];
    }
    }
}

int QueryLCA(int u,int v){
    int fst = min(app[u][0],app[v][0]),sec = max(app[u][1],app[v][1]);
    int len = sec-fst+1,k = 0;
    while((1<<k+1) <= len) k++;
    if(climb[RMQ[fst][k]] < climb[RMQ[sec-(1<<k)+1][k]]) return RMQ[fst][k];
    else return RMQ[sec-(1<<k)+1][k];
}

int dist(int u,int v){ return dep[u]+dep[v]-2*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] = 0;seg[now] = 0;
    for(int i=0;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]-1);
    for(int i=0;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 != 0) v2[root].push_back(make_pair(x[now],dist(now,oldroot)));
    for(int i=0;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,0);
    int p = GetG(now,0,sz[now]);

    LenQuery(p,0,0,p,last);    sort(v[p].begin(),v[p].end()); sum[p].push_back(0);
    sort(v2[p].begin(),v2[p].end()); sum2[p].push_back(0);
    for(int i=0;i<v[p].size();i++) sum[p].push_back(sum[p][i]+1ll*v[p][i].second);
    for(int i=0;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=0;i<g[p].size();i++){        
        int nxt = g[p][i].first;
        if(arr[nxt]) continue;
        divide(nxt,p,ht+1);
    }
    }

public:
    void BuildTree(){divide(1,0,1);}
    void printans(int now,int l,int r,long long &lastans){
    int p = now;
    while(p != 0) sta.push(p),p = fa[p];
    long long ans = 0;
    while(!sta.empty()){
        int tp = sta.top();sta.pop();
        int pp = lower_bound(v[tp].begin(),v[tp].end(),make_pair(l,0))-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()-1;
        int len = ed-pp+1;
        ans += sum[tp][ed+1] - sum[tp][pp] + 1ll*len*dist(tp,now);
        if(fa[tp]!=0)
        ans -= (sum2[tp][ed+1] - sum2[tp][pp] + 1ll*len*dist(fa[tp],now));
    }
    while(!sta.empty())sta.pop();
    printf("%lld\n",ans);lastans = ans;
    }
}T1;

void init(){
    dfs(1,0,0,1);
    BuildRMQ();
    T1.BuildTree();
}

void work(){
    long long lastans = 0;
    for(int i=1;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=1;i<=n;i++) in(x[i]);
    for(int i=1;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 0;
}