Solution -「WC 2014」「洛谷 P3920」紫荆花之恋

\(\mathscr{Description}\)

  Link.

  维护一棵树，初始时树空。接下来 \(n\) 次操作，每次操作加入一片叶子 \(u\)，\(u\) 到其邻接点的边权为 \(w\)，\(u\) 的半径为 \(r_u\)。每次加入结点后，求出 \(\sum_{u<v}[r_u+r_v\ge\operatorname{dist}(u,v)]\) 的值。强制在线。

  \(n\le10^5\)。

\(\mathscr{Solution}\)

  初学 OI 的时候，第一次听说所谓“超级难写的毒瘤题”就是《紫荆花之恋》，后来每次向尝试也不知道为什么都咕掉了。这几天抽空写了一发却发现……这种码量顶多叫难写，写出来也没啥新奇了。还是挺感慨。（

  显然答案可以随着树的变化维护增量。考虑在一棵静态的树上，对于固定的 \(x\)，如何快速求出 \(\sum_{u\neq x}[r_u+r_x\ge\operatorname{dist}(u,x)]\)——要说“维护”这一值，那么可以点分树套平衡树维护。注意本题得用小常数的平衡树（例如替罪羊树）。

  怎么动态维护点分树树形？也可以像替罪羊一样，发现子树不平衡就点分治重构。总复杂度套用替罪羊的证明可知为 \(\mathcal O(n\log^2 n)\)。

\(\mathscr{Code}\)

  巨大常数，代码就自己写嘛 qwq。

/*+Rainybunny+*/

#include <bits/stdc++.h>

#define rep(i, l, r) for (int i = l, rep##i = r; i <= rep##i; ++i)
#define per(i, r, l) for (int i = r, per##i = l; i >= per##i; --i)

typedef long long LL;

inline char fgc() {
    static char buf[1 << 17], *p = buf, *q = buf;
    return p == q && (q = buf + fread(p = buf, 1, 1 << 17, stdin), p == q) ?
      EOF : *p++;
}

template <typename Tp = int>
inline Tp rint() {
    Tp x = 0, s = fgc(), f = 1;
    for (; s < '0' || '9' < s; s = fgc()) f = s == '-' ? -f : f;
    for (; '0' <= s && s <= '9'; s = fgc()) x = x * 10 + (s ^ '0');
    return x * f;
}

template <typename Tp>
inline void wint(Tp x) {
    if (x < 0) putchar('-'), x = -x;
    if (9 < x) wint(x / 10);
    putchar(x % 10 ^ '0');
}

template <typename Tp>
inline void chkmin(Tp& u, const Tp& v) { v < u && (u = v, 0); }
template <typename Tp>
inline void chkmax(Tp& u, const Tp& v) { u < v && (u = v, 0); }
template <typename Tp>
inline Tp imin(const Tp& u, const Tp& v) { return u < v ? u : v; }
template <typename Tp>
inline Tp imax(const Tp& u, const Tp& v) { return u < v ? v : u; }

const int MAXN = 1e5;
const double LUCK = 0.712;
// const double LUCK = 1.;
int n;

class ScapegoatTree {
private:
    static const int MAXND = 8e6;
    int node, ch[MAXND][2], siz[MAXND];
    LL val[MAXND];
    int cnt[MAXND], sum[MAXND], rcyc[MAXND];

    inline int newnd() {
        int u = rcyc[0] ? rcyc[rcyc[0]--] : ++node;
        ch[u][0] = ch[u][1] = val[u] = 0, cnt[u] = sum[u] = siz[u] = 1;
        return u;
    }

    inline void pushup(const int u) {
        siz[u] = siz[ch[u][0]] + siz[ch[u][1]] + 1;
        sum[u] = sum[ch[u][0]] + sum[ch[u][1]] + cnt[u];
    }

    inline bool valid(const int u) {
        return imax(siz[ch[u][0]], siz[ch[u][1]]) <= siz[u] * LUCK + 5;
    }

    inline void collect(const int u, int*& idx) {
        if (!u) return ;
        collect(ch[u][0], idx), *idx++ = u, collect(ch[u][1], idx);
    }

    inline int rebuild(const int* buc, const int l, const int r) {
        if (l > r) return 0;
        int mid = l + r >> 1, u = buc[mid];
        ch[u][0] = rebuild(buc, l, mid - 1);
        ch[u][1] = rebuild(buc, mid + 1, r);
        return pushup(u), u;
    }

    inline void balance(int& u) {
        if (valid(u)) return ;
        static int buc[MAXN + 5], *idx;
        collect(u, idx = buc), u = rebuild(buc, 0, idx - buc - 1);
    }

public:
    inline void recycle(const int u) {
        if (!u) return ;
        rcyc[++rcyc[0]] = u;
        recycle(ch[u][0]), recycle(ch[u][1]);
    }

    inline void insert(int& u, const LL x) {
        if (!u) return u = newnd(), val[u] = x, void();
        balance(u);
        if (val[u] == x) return ++cnt[u], ++sum[u], void();
        else if (val[u] < x) insert(ch[u][1], x);
        else insert(ch[u][0], x);
        pushup(u);
    }

    inline int rank(const int rt, const LL x) { // count elements <= x.
        int u = rt, ret = 0;
        while (u) {
            if (val[u] == x) return ret + sum[ch[u][0]] + cnt[u];
            else if (val[u] < x) ret += sum[ch[u][0]] + cnt[u], u = ch[u][1];
            else u = ch[u][0];
        }
        return ret;
    }
} sct;

namespace LCA {

const int MAXLG = 16;
int dep[MAXN + 5], fa[MAXN + 5][MAXLG + 2];
LL dis[MAXN + 5];

inline void append(const int u, const int f, const int w) {
    fa[u][0] = f, dep[u] = dep[f] + 1, dis[u] = dis[f] + w;
    for (int i = 1; fa[u][i - 1]; fa[u][i] = fa[fa[u][i - 1]][i - 1], ++i);
}

inline int lca(int u, int v) {
    if (dep[u] < dep[v]) u ^= v ^= u ^= v;
    per (i, MAXLG, 0) if (dep[fa[u][i]] >= dep[v]) u = fa[u][i];
    if (u == v) return u;
    per (i, MAXLG, 0) if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
    return fa[u][0];
}

inline LL dist(const int u, const int v) {
    return dis[u] + dis[v] - 2 * dis[lca(u, v)];
}

} // namespace LCA.

namespace DivideTree {

int vfa[MAXN + 5], rad[MAXN + 5], ecnt, head[MAXN + 5];
struct Edge { int to, nxt; } graph[MAXN * 2 + 5];

inline void link(const int u, const int v) {
    graph[++ecnt] = { v, head[u] }, head[u] = ecnt;
    graph[++ecnt] = { u, head[v] }, head[v] = ecnt;
}

std::vector<int> inc[MAXN + 5];
bool foc[MAXN + 5];
int siz[MAXN + 5], wgt[MAXN + 5], srt[MAXN + 5][2];

inline void collect(const int u, const int fa, std::vector<int>& rec) {
    rec.push_back(u);
    for (int i = head[u], v; i; i = graph[i].nxt) {
        if (foc[v = graph[i].to] && v != fa) {
            collect(v, u, rec);
        }
    }
}

inline void findG(const int u, const int fa, const int all, int& rt) {
    siz[u] = 1, wgt[u] = 0;
    for (int i = head[u], v; i; i = graph[i].nxt) {
        if (foc[v = graph[i].to] && v != fa) {
            findG(v, u, all, rt), siz[u] += siz[v];
            chkmax(wgt[u], siz[v]);
        }
    }
    chkmax(wgt[u], all - siz[u]);
    if (!rt || wgt[rt] > wgt[u]) rt = u;
}

inline void divide(const int u) {
    foc[u] = false;
    sct.recycle(srt[u][0]), srt[u][0] = 0;
    sct.recycle(srt[u][1]), srt[u][1] = 0;
    for (int i = head[u], v; i; i = graph[i].nxt) if (foc[v = graph[i].to]) {
        int rt = 0;
        std::vector<int> tmp; collect(v, 0, tmp);
        findG(v, 0, tmp.size(), rt), inc[rt].swap(tmp);
        vfa[rt] = u, divide(rt);
    }
}

void rebuild(const int); // pre-declare for function `update`.

inline void update(int u, const int til, const bool op) {
    int pia = 0; LL tmpd = 0;
    for (int las = 0, v = u; v != til; v = vfa[las = v]) {
        if (op) inc[v].push_back(u);
        sct.insert(srt[v][0], tmpd - rad[u]);
        if (vfa[v]) {
            sct.insert(srt[v][1], (tmpd = LCA::dist(u, vfa[v])) - rad[u]);
        }
        if (las && inc[las].size() > inc[v].size() * LUCK + 5) pia = v;
    }
    if (pia) rebuild(pia);
}

inline void rebuild(const int u) {
    int vf = vfa[u], rt = 0;
    for (int v: inc[u]) foc[v] = true;
    findG(u, 0, inc[u].size(), rt), inc[rt].swap(inc[u]);
    vfa[rt] = vf, divide(rt);
    for (int v: inc[rt]) update(v, vfa[rt], 0);
}

inline void append(int u, const int rfa, const int r) {
    link(u, rfa);
    vfa[u] = rfa, rad[u] = r, update(u, 0, 1);
}

inline int contri(const int u) {
    int ret = sct.rank(srt[u][0], rad[u]) - 1;
    for (int las = u, v = vfa[u]; v; v = vfa[las = v]) {
        LL d = LCA::dist(u, v);
        ret += sct.rank(srt[v][0], rad[u] - d)
          - sct.rank(srt[las][1], rad[u] - d);
    }
    return ret;
}

} // namespace DivideTree.

int main() {
    rint(), n = rint();
    LL ans = 0;
    rep (i, 1, n) {
        int a = rint() ^ (ans % 1'000'000'000), c = rint(), r = rint();
        LCA::append(i, a, c);
        DivideTree::append(i, a, r);
        ans += DivideTree::contri(i);
        wint(ans), putchar('\n');
    }
    return 0;
}