[牛客网#35D 树的距离]离散化+线段树合并

分类：Data Structure SegMent Tree Merge

1. 题目链接

[牛客网#35D 树的距离]

2. 题意描述

wyf非常喜欢树。一棵有根数树上有n个节点，1号点是他的根，每条边都有一个距离，而wyf是个爱问奇怪问题的熊孩子，他想知道对于某个点x,以x为根的子树上，所有与x距离大于等于k的点与x的距离之和。
数据范围：1≤n,q≤2∗105,1≤d,k≤106

3. 解题思路

同《[Codeforces 893F. Subtree Minimum Query]线段树合并》这题的思路。
通过这两个题目总算对线段树合并比较理解了。
对每个节点建一个线段树，维护每个距离出现的次数以及区间距离之和。

4. 实现代码

#include <bits/stdc++.h>using namespace std;typedef long long ll;typedef long double lb;typedef unsigned int uint;typedef unsigned long long ull;typedef pair<int, int> pii;typedef pair<ll, ll> pll;typedef pair<ull, ull> puu;typedef pair<lb, lb> pbb;typedef vector<int> vi;const int inf = 0x3f3f3f3f;const ll infl = 0x3f3f3f3f3f3f3f3fLL;template<typename T> inline void umax(T &a, T b) { a = max(a, b); }template<typename T> inline void umin(T &a, T b) { a = min(a, b); }template<typename T> inline T randIntv(const T& a, const T& b) { return (T)rand() % (b - a + 1) + a; }void debug() { cout << endl; }template<typename T, typename ...R> void debug (T f, R ...r) { cout << "[" << f << "]"; debug (r...); }const int MAXN = 200005;int n, q;struct Edge {    int v, next;    ll w;} edge[MAXN << 1];int head[MAXN], etot;void ini(int n) {    etot = 0;    for (int i = 0; i <= n; ++i) head[i] = -1;}void ins(int u, int v, ll w) {    edge[etot].v = v;    edge[etot].w = w;    edge[etot].next = head[u];    head[u] = etot ++;}struct TNode {    int cnt;    ll sum;    int ch[2];    void ini() { cnt = sum = ch[0] = ch[1] = 0; }} nd[MAXN * 50];int root[MAXN], nsz, null = 0;ll dep[MAXN];ll f[MAXN], fsz;#define lch     nd[rt].ch[0]#define rch     nd[rt].ch[1]void pushUp(int rt) {    nd[rt].cnt = nd[lch].cnt + nd[rch].cnt;    nd[rt].sum = nd[lch].sum + nd[rch].sum;}void update(int p, ll v, int l, int r, int& rt) {    nd[rt = ++ nsz].ini();    if (l == r) {        nd[rt].cnt = 1;        nd[rt].sum = v;        return;    }    int md = (l + r) >> 1;    if (p <= md) update(p, v, l, md, lch);    else update(p, v, md + 1, r, rch);    pushUp(rt);}int merge(int u, int v) {    if (!u || !v) return u ^ v;    int rt = ++ nsz; nd[rt].ini();    nd[rt].ch[0] = merge(nd[u].ch[0], nd[v].ch[0]);    nd[rt].ch[1] = merge(nd[u].ch[1], nd[v].ch[1]);    nd[rt].cnt = nd[u].cnt + nd[v].cnt;    nd[rt].sum = nd[u].sum + nd[v].sum;    return rt;}void dfs1(int u, int fa, ll d) {    dep[u] = d;    for (int i = head[u]; ~i; i = edge[i].next) {        int v = edge[i].v; ll w = edge[i].w;        if (v == fa) continue;        dfs1(v, u, d + w);    }}void dfs2(int u, int fa) {    int pos = lower_bound(f + 1, f + fsz + 1, dep[u]) - f;    update(pos, dep[u], 1, fsz, root[u]);    for (int i = head[u]; ~i; i = edge[i].next) {        int v = edge[i].v;        if (v == fa) continue;        dfs2(v, u);        root[u] = merge(root[u], root[v]);    }}pair<int, ll> query(int L, int R, int l, int r, int rt) {    // if (L <= l && r <= R) debug("query", rt, nd[rt].cnt, nd[rt].sum);    if (L <= l && r <= R) return make_pair(nd[rt].cnt, nd[rt].sum);    int md = (l + r) >> 1;    pair<int, ll> lop(0, 0), rop(0, 0);    if (L <= md) lop = query(L, R, l, md, lch);    if (R > md) rop = query(L, R, md + 1, r, rch);    return make_pair(lop.first + rop.first, lop.second + rop.second);}int main() {#ifdef ___LOCAL_WONZY___    freopen("input.txt", "r", stdin);#endif // ___LOCAL_WONZY___    int u, x, cnt; ll w, k, sum, ans;    while (~scanf("%d", &n)) {        ini(n);        for (int i = 2; i <= n; ++i) {            scanf("%d %lld", &u, &w);            ins(i, u, w), ins(u, i, w);;        }        nsz = null = 0;        dfs1(1, 1, 1);        for (int i = 1; i <= n; ++i) f[i] = dep[i];        sort(f + 1, f + n + 1);        fsz = unique(f + 1, f + n + 1) - (f + 1);        dfs2(1, 1);        scanf("%d", &q);        for (int i = 1; i <= q; ++i) {            scanf("%d %lld", &x, &k);            int pos = lower_bound(f + 1, f + fsz + 1, dep[x] + k) - f;            tie(cnt, sum) = query(pos, fsz, 1, fsz, root[x]);            ans = sum - cnt * dep[x];            // debug(cnt, sum, ans);            printf("%lld\n", ans);        }    }#ifdef ___LOCAL_WONZY___    cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC * 1000 << "ms." << endl;#endif // ___LOCAL_WONZY___    return 0;}