Submission #1256479

#	Time	Username	Problem	Language	Result	Execution time	Memory
1256479		biank	Obstacles for a Llama (IOI25_obstacles)	C++20	83 / 100	167 ms	35272 KiB

obstacles

#include "obstacles.h"
#include <bits/stdc++.h>

using namespace std;

#define forsn(i, s, n) for (int i = int(s); i < int(n); i++)
#define forn(i, n) forsn(i, 0, n)
#define dforsn(i, s, n) for (int i = int(n) - 1; i >= int(s); i--)
#define dforn(i, n) dforsn(i, 0, n)

#define sz(x) int(x.size())
#define all(x) begin(x), end(x)

using ii = pair<int, int>;
using vi = vector<int>;
using ll = long long;
using vll = vector<ll>;
using pll = pair<ll, ll>;

#define fst first
#define snd second

#define pb push_back
#define eb emplace_back

struct DSU {
    vi par, mini, tim;
    vector<ii> edges;
    DSU(vi h) : par(sz(h)), mini(h), tim(sz(h), 0) {
        iota(all(par), 0);
    }
    int find(int x) {
        if (par[x] == x) return x;
        return par[x] = find(par[x]);
    }
    bool unite(int x, int y, int t) {
        x = find(x), y = find(y);
        if (x == y) return false;
        int p = sz(par);
        par[x] = p, par[y] = p;
        par.pb(p);
        mini.pb(min(mini[x], mini[y]));
        tim.pb(t);
        edges.eb(p, x);
        edges.eb(p, y);
        return true;
    }
};

const int INF = 1e9;

struct SegTree {
    int n;
    vi st;
    SegTree(vi &v) : n(sz(v)), st(2 * n) {
        forn(i, n) st[i + n] = v[i];
        dforsn(i, 1, n) st[i] = min(st[2 * i], st[2 * i + 1]);
    }
    int query(int l, int r) {
        int ret = INF;
        for (l += n, r += n; l < r; l /= 2, r /= 2) {
            if (l & 1) ret = min(ret, st[l++]);
            if (r & 1) ret = min(st[--r], ret);
        }
        return ret;
    }
};

DSU ret(vi{});

void initialize(vi T, vi H) {
    const int n = sz(T), m = sz(H);
    vector<bool> active(m, false);
    DSU dsu(H);
    vi order(m);
    iota(all(order), 0);
    sort(all(order), [&](const int &lhs, const int &rhs) {
        return H[lhs] < H[rhs];
    });
    auto activate = [&](int i, int t) {
        if (i > 0 && active[i - 1]) dsu.unite(i - 1, i, t);
        if (i + 1 < m && active[i + 1]) dsu.unite(i, i + 1, t);
        active[i] = true;
    };
    {
        int maxi = -INF, j = 0;
        forn(i, n) if (T[i] > maxi) {
            maxi = T[i];
            while (j < m && H[order[j]] < T[i]) {
                activate(order[j++], i);
            }
        }
    }
    const int N = sz(dsu.par);
    vi placeholder(N, 0);
    SegTree st(T);
    ret = DSU(placeholder);
    for (auto [p, u] : dsu.edges) {
        int j = dsu.tim[p], i = dsu.tim[u];
        if (st.query(i, j) > dsu.mini[u]) ret.unite(u, p, -1);
    }
}

bool can_reach(int L, int R, int S, int D) {
  return ret.find(S) == ret.find(D);
}

• Subtask 0: 0 / 0
• Subtask 1: 10 / 10
• Subtask 2: 14 / 14
• Subtask 3: 13 / 13
• Subtask 4: 21 / 21
• Subtask 5: 25 / 25
• Subtask 6: 0 / 17