[Codeforces 893F. Subtree Minimum Query]线段树合并

分类：Data Structure SegMent Tree Merge

1. 题目链接

[Codeforces 893F. Subtree Minimum Query]

2. 题意描述

一个n个节点的有根树，每个节点有一个边权ai，每条边的边长为1。然后是m个询问。对于第i次询问，求在点xi为根节点的子树中且到点xi距离小于等于ki的点权最小值。询问要求强制在线。
数据范围：1  ≤ n ≤ 105,1 ≤ m ≤ 106,1 ≤ ai ≤ 109

3. 解题思路

qwb说这是线段树合并的经典套路题。
我第一次写线段树合并。合并的地方写残了。无限RE...
思路就是对每个节点建一个线段树，这个线段树维护了该节点为根节点的子树的最小点权。自底向上，将子节点的线段树合并到父节点上去。太精妙了。

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 = 100005;int n, m, r;ll a[MAXN];struct Edge {    int v, next;} edge[MAXN << 1];int head[MAXN], etot, dep[MAXN], null;void ini(int n) {    etot = 0;    for (int i = 0; i <= n; ++i) head[i] = -1;}void ins(int u, int v) {    edge[etot] = Edge{v, head[u]};    head[u] = etot ++;}#define lch     nd[rt].ch[0]#define rch     nd[rt].ch[1]struct TNode {    ll val;    int ch[2];    void ini() { ch[0] = ch[1] = 0; val = infl; }} nd[MAXN * 50];int root[MAXN], rsz;void pushUp(int rt) {    nd[rt].val = min(nd[lch].val, nd[rch].val);}void update(int p, ll v, int l, int r, int& rt) {    nd[rt = ++ rsz].ini();    if (l == r) {        nd[rt].val = 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 (!v) return u;    if (!u) return v;    int ret = ++ rsz; nd[ret].ini();    nd[ret].ch[0] = merge(nd[u].ch[0], nd[v].ch[0]);    nd[ret].ch[1] = merge(nd[u].ch[1], nd[v].ch[1]);    nd[ret].val = min(nd[u].val, nd[v].val);    return ret;}void dfs(int u, int fa, int d) {    dep[u] = d;    update(dep[u], a[u], 1, n, root[u]);    for (int i = head[u]; ~i; i = edge[i].next) {        int v = edge[i].v;        if (v == fa) continue;        dfs(v, u, d + 1);        root[u] = merge(root[u], root[v]);    }}ll query(int L, int R, int l, int r, int rt) {    if (L <= l && r <= R) return nd[rt].val;    int md = (l + r) >> 1; ll ret = infl;    if (L <= md) umin(ret, query(L, R, l, md, nd[rt].ch[0]));    if (R > md) umin(ret, query(L, R, md + 1, r, nd[rt].ch[1]));    return ret;}int main() {#ifdef ___LOCAL_WONZY___    freopen("input.txt", "r", stdin);#endif // ___LOCAL_WONZY___    int u, v, p, q, x, k;    ll last = 0;    scanf("%d %d", &n, &r);    for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);    ini(n);    for (int i = 2; i <= n; ++i) {        scanf("%d %d", &u, &v);        ins(u, v), ins(v, u);;    }    rsz = 0;    nd[null = 0].ini();    dfs(r, r, 1);    scanf("%d", &m);    for (int i = 1; i <= m; ++i) {        scanf("%d %d", &p, &q);        x = (p + last) % n + 1, k = (q + last) % n;        last = query(dep[x], dep[x] + k, 1, n, root[x]);        printf("%lld\n", last);    }#ifdef ___LOCAL_WONZY___    cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC * 1000 << "ms." << endl;#endif // ___LOCAL_WONZY___    return 0;}