BZOJ4012 [HNOI2015]开店 【动态点分治 + splay】

题目链接

BZOJ4012

题解

Mychael并没有A掉,而是T掉了

讲讲主要思路
在点分树上每个点开两棵\(splay\)
平衡树\(A\)维护子树中各年龄到根的距离
平衡树\(B\)维护子树中各年龄到点分树父亲的距离

然后询问就可以在点分树上用两棵平衡树相减计算了
大常数\(O(nlog^2n)\)被卡死

// luogu-judger-enable-o2
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (register int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (register int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,LL>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,LL>
#define LL long long int
#define ls ch[u][0]
#define rs ch[u][1]
#define isr(u) (fa[u] && ch[fa[u]][1] == u)
#define res register
using namespace std;
const int maxn = 200005,maxm = 10000005,INF = 1000000000;
inline int read(){
    res int out = 0,flag = 1; res char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
inline void write(LL x){
    if (x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int Siz[maxm],siz[maxm],w[maxm],ch[maxm][2],fa[maxm],cnt;
LL Sum[maxm],sum[maxm];
struct Splay_Tree{
    int rt;
    void upd(int u){
        Siz[u] = Siz[ls] + Siz[rs] + siz[u];
        Sum[u] = Sum[ls] + Sum[rs] + sum[u];
    }
    void spin(int u){
        int s = isr(u),f = fa[u];
        fa[u] = fa[f]; if (fa[f]) ch[fa[f]][isr(f)] = u;
        ch[f][s] = ch[u][s ^ 1]; if (ch[u][s ^ 1]) fa[ch[u][s ^ 1]] = f;
        fa[f] = u; ch[u][s ^ 1] = f;
        upd(f); upd(u);
    }
    void splay(int u,int f = 0){
        for (; fa[u] != f; spin(u))
            if (fa[fa[u]] != f) spin((isr(u) ^ isr(fa[u])) ? u : fa[u]);
        if (!f) rt = u;
    }
    void insert(int& u,int f,int v,int Val){
        if (!u){
            w[u = ++cnt] = v; Sum[u] = sum[u] = Val;
            fa[u] = f; siz[u] = Siz[u] = 1; splay(u);
        }
        else if (w[u] > v) insert(ls,u,v,Val);
        else if (w[u] < v) insert(rs,u,v,Val);
        else {Sum[u] += Val; sum[u] += Val; Siz[u]++; siz[u]++; splay(u);}
    }
    void ins(int pos,int v){insert(rt,0,pos,v);}
    int pre(int u,int v){
        if (!u) return 0;
        if (w[u] >= v) return pre(ls,v);
        else {
            int t = pre(rs,v);
            return t ? t : u;
        }
    }
    int post(int u,int v){
        if (!u) return 0;
        if (w[u] <= v) return post(rs,v);
        else {
            int t = post(ls,v);
            return t ? t : u;
        }
    }
    cp query(int l,int r){
        int L = pre(rt,l),R = post(rt,r);
        splay(L); splay(R,L);
        if (!ch[R][0]) return mp(0,0);
        return mp(Siz[ch[R][0]],Sum[ch[R][0]]);
    }
    void init(){rt = 0; ins(-INF,0); ins(INF,0);}
}A[maxn],B[maxn];
LL ans;
int n,m,Limit,val[maxn],L,R;
int h[maxn],ne = 1;
struct EDGE{int to,nxt,w;}ed[maxn << 1];
inline void build(int u,int v,int w){
    ed[++ne] = (EDGE){v,h[u],w}; h[u] = ne;
    ed[++ne] = (EDGE){u,h[v],w}; h[v] = ne;
}
LL Dis[maxn][23];
int F[maxn],Fa[maxn],size[maxn],vis[maxn],N,rt;
void getrt(int u){
    F[u] = 0; size[u] = 1;
    Redge(u) if (!vis[to = ed[k].to] && to != Fa[u]){
        Fa[to] = u; getrt(to);
        size[u] += size[to];
        F[u] = max(F[u],size[to]);
    }
    F[u] = max(F[u],N - size[u]);
    if (F[u] < F[rt]) rt = u;
}
int c[maxn],d[maxn],ci;
void dfs1(int u){
    size[u] = 1; c[++ci] = u;
    Redge(u) if (!vis[to = ed[k].to] && to != Fa[u]){
        Fa[to] = u; d[to] = d[u] + ed[k].w;
        dfs1(to);
        size[u] += size[to];
    }
}
int pre[maxn],dep[maxn];
void solve(int u,int D){
    vis[u] = true; size[u] = 1; ci = 0; d[u] = 0; dep[u] = D;
    A[u].init(); A[u].ins(val[u],0); B[u].init();
    Redge(u) if (!vis[to = ed[k].to]){
        Fa[to] = u; d[to] = d[u] + ed[k].w;
        dfs1(to);
    }
    int v = pre[u];
    REP(i,ci){
        A[u].ins(val[c[i]],d[c[i]]);
        Dis[c[i]][D] = d[c[i]];
    }
    if (v){
        B[u].ins(val[u],Dis[u][D - 1]);
        REP(i,ci) B[u].ins(val[c[i]],Dis[c[i]][D - 1]);
    }
    Redge(u) if (!vis[to = ed[k].to]){
        N = size[to]; F[rt = 0] = INF;
        getrt(to); pre[rt] = u;
        solve(rt,D + 1);
    }
}

void work(int x){
    ans = A[x].query(L,R).second;
    LL dd; cp t1,t2;
    for (res int u = x,i = dep[x] - 1; pre[u]; u = pre[u],i--){
        dd = Dis[x][i];
        t1 = A[pre[u]].query(L,R);
        t2 = B[u].query(L,R);
        ans += t1.second - t2.second + dd * (t1.first - t2.first);
    }
    printf("%lld\n",ans);
}
int main(){
    n = read(); m = read(); Limit = read();
    LL a,b,w;
    for (res int i = 1; i <= n; i++) val[i] = read();
    for (res int i = 1; i < n; i++){
        a = read(); b = read(); w = read();
        build(a,b,w);
    }
    N = n; F[rt = 0] = INF;
    getrt(1);
    solve(rt,0);
    int u;
    while (m--){
        u = read(); a = read(); b = read();
        L = (a + ans) % Limit;
        R = (b + ans) % Limit;
        if (L > R) swap(L,R);
        work(u);
    }
    return 0;
}

posted @ 2018-05-18 22:04  Mychael  阅读(152)  评论(0编辑  收藏  举报