[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;
}