【题解】LOJ-2116-BZOJ-4012-「HNOI2015」开店

Posted on Edited on In Disqus:

题目

https://loj.ac/problem/2116

题意

询问一个点到所有点权在 \([l, r]\) 之间的点的距离和。（强制在线）

题解

方法1

  • \(\displaystyle \sum_{val_v \in [l, r]} lca(u, v)= \displaystyle \sum_{val_v \in [l, r]} dep_v+dep_u-2dep_{lca(u,v)}\)
  • 前两项可以用前缀和相减 \(O(1)\) 得到。
  • 如果不考虑 \(val_v\) 的限制，计算 \(u\) 与所有点的 \(lca\) 到根的距离和，这部分可以用树链剖分 + 线段树完成。
  • 每个点为从自己到根的所有边贡献 1，最后查询 \(u\) 到根所有边权与贡献数之积。
  • 由于有区间限制，使用函数式线段树，按 \(val\) 从小到大计算贡献，保留每个 \(val\) 对应的根，最后对两棵历史版本的线段树上的查询答案相减即可。
  • \(val\) 需要离散化，空间玄学（理论上要开 \(2n\log^2n\)，但开不下）。

代码

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#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define FOR(i, x, y) for (decay<decltype(y)>::type i = (x), _##i = (y); i < _##i; ++i)
#define FORD(i, x, y) for (decay<decltype(x)>::type i = (x), _##i = (y); i > _##i; --i)
#ifdef zerol
#define dbg(args...) do { cout << "\033[32;1m" << #args << " -> "; err(args); } while (0)
#else
#define dbg(...)
#endif
void err() { cout << "\033[39;0m" << endl; }
template<template<typename...> class T, typename t, typename... Args>
void err(T<t> a, Args... args) { for (auto x: a) cout << x << ' '; err(args...); }
template<typename T, typename... Args>
void err(T a, Args... args) { cout << a << ' '; err(args...); }
// -----------------------------------------------------------------------------
const int maxn = 15E4 + 100;
int v[maxn];
struct E {
    int to, d;
};
vector<E> G[maxn];
int n, Q;
LL ans, A;
int sz[maxn], son[maxn], fa[maxn], idx[maxn], top[maxn], dis[maxn], clk, len[maxn];

void predfs(int u, int d = 0) {
    dis[u] = d; sz[u] = 1;
    int& maxs = son[u] = -1;
    for (E& e: G[u]) {
        int v = e.to;
        if (v == fa[u]) continue;
        fa[v] = u;
        predfs(v, d + e.d);
        sz[u] += sz[v];
        if (maxs == -1 || sz[v] > sz[maxs]) maxs = v;
    }
}

void dfs(int u, int tp) {
    top[u] = tp;
    idx[u] = ++clk;
    if (~son[u]) dfs(son[u], tp);
    for (E& e: G[u]) {
        int v = e.to;
        if (v != son[u] && v != fa[u]) dfs(v, v);
        if (v != fa[u]) len[idx[v]] = e.d;
    }
}

template<typename T>
void mod(int u, T&& fn) {
    while (top[u] != 1) {
        fn(idx[top[u]], idx[u]);
        u = fa[top[u]];
    }
    if (u != 1) fn(2, idx[u]);
}
LL dsum[maxn], cnt[maxn];
struct P {
    int u, v;
};
vector<P> a;

namespace tree {
#define mid ((l + r) >> 1)
#define lson ql, qr, l, mid
#define rson ql, qr, mid + 1, r
    struct P {
        LL add, sum;
        int ls, rs;
    } tr[maxn * 45 * 2];
    int sz = 1;
    int N(LL add, int l, int r, int ls, int rs) {
        tr[sz] = {add, tr[ls].sum + tr[rs].sum + add * (len[r] - len[l - 1]), ls, rs};
        return sz++;
    }
    int update(int o, int ql, int qr, int l, int r, LL add) {
        if (ql > r || l > qr) return o;
        const P& t = tr[o];
        if (ql <= l && r <= qr) return N(add + t.add, l, r, t.ls, t.rs);
        return N(t.add, l, r, update(t.ls, lson, add), update(t.rs, rson, add));
    }
    LL query(int o, int ql, int qr, int l, int r, LL add = 0) {
        if (ql > r || l > qr) return 0;
        const P& t = tr[o];
        if (ql <= l && r <= qr) return add * (len[r] - len[l - 1]) + t.sum;
        return query(t.ls, lson, add + t.add) + query(t.rs, rson, add + t.add);
    }
}

int rt[maxn];
int main() {
    cin >> n >> Q >> A;
    vector<int> vv;
    FOR (i, 0, n) {
        scanf("%d", &v[i]);
        vv.push_back(v[i]);
    }
    sort(vv.begin(), vv.end());
    vv.erase(unique(vv.begin(), vv.end()), vv.end());
    FOR (i, 0, n) a.push_back({i + 1, int(lower_bound(vv.begin(), vv.end(), v[i]) - vv.begin() + 1)});
    FOR (_, 1, n) {
        int u, v, d; scanf("%d%d%d", &u, &v, &d);
        G[u].push_back({v, d});
        G[v].push_back({u, d});
    }
    predfs(1); dfs(1, 1);
    FOR (i, 1, n + 1) len[i] += len[i - 1];
    for (P& x: a) dsum[x.v] += dis[x.u], cnt[x.v]++;
    FOR (i, 1, vv.size() + 1) dsum[i] += dsum[i - 1], cnt[i] += cnt[i - 1];
    sort(a.begin(), a.end(), [](const P& a, const P& b)->bool { return a.v < b.v; });
    int pt = 0, o = 0;
    FOR (val, 1, (int)vv.size() + 1) {
        while (pt < n && a[pt].v == val) mod(a[pt++].u, [&](int ql, int qr) {
                o = tree::update(o, ql, qr, 1, n, 1);
            });
        rt[val] = o;
    }
    while (Q--) {
        int u; LL a, b; scanf("%d%lld%lld", &u, &a, &b);
        a = (a + ans) % A; b = (b + ans) % A;
        LL l = min(a, b), r = max(a, b);
        l = lower_bound(vv.begin(), vv.end(), l) - vv.begin() + 1;
        r = upper_bound(vv.begin(), vv.end(), r) - vv.begin();
        if (l > r) { cout << (ans = 0) << endl; continue; }
        ans = dsum[r] - dsum[l - 1] + 1LL * (cnt[r] - cnt[l - 1]) * dis[u];
        mod(u, [&](int ql, int qr) {
            ans -= (tree::query(rt[r], ql, qr, 1, n) -
                    tree::query(rt[l - 1], ql, qr, 1, n)) * 2;
        });
        cout << ans << endl;
    }
}

方法2

  • 动态点分治，对每个重心维护一个按照点权排序 vector，从而可以二分查询区间和。
  • 代码比方法1短（似乎也更加好写），时间复杂度一致，空间复杂度少一个 \(\log\) 。
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#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define FOR(i, x, y) for (decay<decltype(y)>::type i = (x), _##i = (y); i < _##i; ++i)
#define FORD(i, x, y) for (decay<decltype(x)>::type i = (x), _##i = (y); i > _##i; --i)
#ifdef zerol
#define dbg(args...) do { cout << "\033[32;1m" << #args << " -> "; err(args); } while (0)
#else
#define dbg(...)
#endif
void err() { cout << "\033[39;0m" << endl; }
template<template<typename...> class T, typename t, typename... Args>
void err(T<t> a, Args... args) { for (auto x: a) cout << x << ' '; err(args...); }
template<typename T, typename... Args>
void err(T a, Args... args) { cout << a << ' '; err(args...); }
// -----------------------------------------------------------------------------
const int maxn = 15E4 + 100, INF = 1E9;
struct E {
    int to, d;
};
vector<E> G[maxn];
int n, Q, w[maxn];
LL A, ans;

bool vis[maxn];
int sz[maxn];
int get_sz(int u, int fa) {
    int& s = sz[u] = 1;
    for (E& e: G[u]) {
        int v = e.to;
        if (vis[v] || v == fa) continue;
        s += get_sz(v, u);
    }
    return s;
}

void get_rt(int u, int fa, int s, int& m, int& rt) {
    int t = s - sz[u];
    for (E& e: G[u]) {
        int v = e.to;
        if (vis[v] || v == fa) continue;
        get_rt(v, u, s, m, rt);
        t = max(t, sz[v]);
    }
    if (t < m) { m = t; rt = u; }
}

int dep[maxn], md[maxn];
void get_dep(int u, int fa, int d) {
    dep[u] = d; md[u] = 0;
    for (E& e: G[u]) {
        int v = e.to;
        if (vis[v] || v == fa) continue;
        get_dep(v, u, d + e.d);
        md[u] = max(md[u], md[v] + 1);
    }
}

struct P {
    int w;
    LL s;
};
using VP = vector<P>;
struct R {
    VP *rt, *rt2;
    int dep;
};
VP pool[maxn << 1], *pit = pool;
vector<R> tr[maxn];

void go(int u, int fa, VP* rt, VP* rt2) {
    tr[u].push_back({rt, rt2, dep[u]});
    for (E& e: G[u]) {
        int v = e.to;
        if (v == fa || vis[v]) continue;
        go(v, u, rt, rt2);
    }
}

void dfs(int u) {
    int tmp = INF; get_rt(u, -1, get_sz(u, -1), tmp, u);
    vis[u] = true;
    get_dep(u, -1, 0);
    VP* rt = pit++; tr[u].push_back({rt, nullptr, 0});
    for (E& e: G[u]) {
        int v = e.to;
        if (vis[v]) continue;
        go(v, u, rt, pit++);
        dfs(v);
    }
}

bool cmp(const P& a, const P& b) { return a.w < b.w; }

LL query(VP& p, int d, int l, int r) {
    l = lower_bound(p.begin(), p.end(), P{l, -1}, cmp) - p.begin();
    r = upper_bound(p.begin(), p.end(), P{r, -1}, cmp) - p.begin() - 1;
    return p[r].s - p[l - 1].s + 1LL * (r - l + 1) * d;
}

int main() {
    cin >> n >> Q >> A;
    FOR (i, 1, n + 1) scanf("%d", &w[i]);
    FOR (_, 1, n) {
        int u, v, d; scanf("%d%d%d", &u, &v, &d);
        G[u].push_back({v, d}); G[v].push_back({u, d});
    }
    dfs(1);
    FOR (i, 1, n + 1)
        for (R& x: tr[i]) {
            x.rt->push_back({w[i], x.dep});
            if (x.rt2) x.rt2->push_back({w[i], x.dep});
        }
    FOR (it, pool, pit) {
        it->push_back({-INF, 0});
        sort(it->begin(), it->end(), cmp);
        FOR (i, 1, it->size())
            (*it)[i].s += (*it)[i - 1].s;
    }
    while (Q--) {
        int u; LL a, b; scanf("%d%lld%lld", &u, &a, &b);
        a = (a + ans) % A; b = (b + ans) % A;
        int l = min(a, b), r = max(a, b);
        ans = 0;
        for (R& x: tr[u]) {
            ans += query(*(x.rt), x.dep, l, r);
            if (x.rt2) ans -= query(*(x.rt2), x.dep, l, r);
        }
        printf("%lld\n", ans);
    }
}