Codeforces 1110F Nearest Leaf 线段树（看题解）

Nearest Leaf

没想到在dfs的过程中用线段树维护所有点到当前点的距离。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 5e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}


int n, q, idx, L[N], R[N], who[N];
bool leaf[N];
vector<PII> G[N];
vector<pair<PLL, int>> qus[N];
LL dis[N], ans[N];

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
template<class T> struct SGT {
    T mn[N << 2], lazy[N << 2];
    inline void pull(int rt) {
        mn[rt] = min(mn[rt << 1], mn[rt << 1 | 1]);
    }
    inline void push(int rt) {
        if(lazy[rt]) {
            mn[rt << 1] += lazy[rt]; mn[rt << 1 | 1] += lazy[rt];
            lazy[rt << 1] += lazy[rt]; lazy[rt << 1 | 1] += lazy[rt];
            lazy[rt] = 0;
        }
    }
    void build(int l, int r, int rt) {
        mn[rt] = INF; lazy[rt] = 0;
        if(l == r) {
            mn[rt] = leaf[who[l]] ? dis[who[l]] : INF;
            return;
        }
        int mid = l + r >> 1;
        build(lson); build(rson);
        pull(rt);
    }
    void update(int L, int R, int val, int l, int r, int rt) {
        if(L > R) return;
        if(R < l || r < L) return;
        if(L <= l && r <= R) {
            mn[rt] += val;
            lazy[rt] += val;
            return;
        }
        push(rt);
        int mid = l + r >> 1;
        update(L, R, val, lson);
        update(L, R, val, rson);
        pull(rt);
    }
    T query(int L, int R, int l, int r, int rt) {
        if(L > R) return INF;
        if(R < l || r < L) return INF;
        if(L <= l && r <= R) return mn[rt];
        push(rt);
        int mid = l + r >> 1;
        return min(query(L, R, lson), query(L, R, rson));
    }
};
SGT<LL> Tree;

void dfs(int u) {
    L[u] = ++idx;
    who[idx] = u;
    if(u != 1 && !SZ(G[u])) leaf[u] = true;
    for(auto& e : G[u]) dis[e.fi] = dis[u] + e.se, dfs(e.fi);
    R[u] = idx;
}
void solve(int u) {
    for(auto& q : qus[u]) {
        ans[q.se] = Tree.query(q.fi.fi, q.fi.se, 1, n, 1);
    }
    for(auto& e : G[u]) {
        int v = e.fi, w = e.se;
        Tree.update(L[v], R[v], -w, 1, n, 1);
        Tree.update(1, L[v] - 1, w, 1, n, 1);
        Tree.update(R[v] + 1, n, w, 1, n, 1);

        solve(e.fi);

        Tree.update(L[v], R[v], w, 1, n, 1);
        Tree.update(1, L[v] - 1, -w, 1, n, 1);
        Tree.update(R[v] + 1, n, -w, 1, n, 1);
    }
}
int main() {
    scanf("%d%d", &n, &q);
    for(int i = 2; i <= n; i++) {
        int p, w; scanf("%d%d", &p, &w);
        G[p].push_back(mk(i, w));
    }
    for(int i = 1; i <= n; i++) sort(ALL(G[i]));
    dfs(1);
    Tree.build(1, n, 1);
    for(int i = 1; i <= q; i++) {
        int v, l, r; scanf("%d%d%d", &v, &l, &r);
        qus[v].push_back(mk(mk(l, r), i));
    }
    solve(1);
    for(int i = 1; i <= q; i++) printf("%lld\n", ans[i]);
    return 0;
}

/*
*/