Submission #1259414

#	Time	Username	Problem	Language	Result	Execution time	Memory
1259414		sunflower	Circle selection (APIO18_circle_selection)	C++17	100 / 100	871 ms	14368 KiB

circle_selection

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define MASK(x) (1LL << (x))
#define BIT(x, i) (((x) >> (i)) & 1)
#define SZ(x) ((int) (x).size())
#define ALL(a) (a).begin(), (a).end()
#define FOR(i, a, b) for (int i = (a), _b = (b); i <= _b; ++i)
#define FORD(i, a, b) for (int i = (a), _b = (b); i >= _b; --i)
#define REP(i, n) for (int i = 0, _n = (n); i < _n; ++i)
#define debug(x) cout << #x << " = " << (x) << endl;

#define fi first
#define se second
#define left    __left
#define right   __right
#define prev    __prev
#define next    __next

template <class X, class Y>
    bool maximize(X &x, Y y) {
        if (x < y) return x = y, true;
        else return false;
    }

template <class X, class Y>
    bool minimize(X &x, Y y) {
        if (x > y) return x = y, true;
        else return false;
    }

template <class T>
    void remove_dup(T &v) {
        sort(ALL(v));
        v.resize(unique(ALL(v)) - v.begin());
    }
    int POS(int x, const vector <int> &v) {
        return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
    }

const int inf = (int) 1e9 + 7;

#define MAX_N 300'300
int n;

int del[MAX_N + 2];

struct Circle {
    int x, y, r, id;

    bool operator < (const Circle &other) const {
        if (r != other.r) return (r > other.r);
        return (id < other.id);
    }

    friend istream& operator >> (istream &in, Circle &c) {
        in >> c.x >> c.y >> c.r;
        return in;
    }
} cir[MAX_N + 2];

ll sqr(ll x) {
    return 1LL * x * x;
}

namespace subtask1 {
    bool check() {
        return (n <= 5000);
    }

    void solve() {
        sort(cir + 1, cir + 1 + n);

        memset(del, 0, sizeof(del));
        FOR(i, 1, n) {
            int curID = cir[i].id;
            if (del[curID]) continue;
            del[curID] = curID;
            FOR(j, i + 1, n) {
                if (del[cir[j].id]) continue;
                if (sqr(cir[i].x - cir[j].x) + sqr(cir[i].y - cir[j].y) <= sqr(cir[i].r + cir[j].r)) {
                    del[cir[j].id] = curID;
                }
            }
        }

        FOR(i, 1, n) cout << del[i] << " ";
    }
};

const int TREE_SIZE = MAX_N * 2;
struct Segtree {
    int seg[TREE_SIZE * 4 + 2], lazy[TREE_SIZE * 4 + 2];
    int n;

    void initTree(int _n) {
        n = _n;
        FOR(i, 0, 4 * _n + 7) {
            seg[i] = inf;
            lazy[i] = inf;
        }
    }

    void down(int id, int l, int r) {
        if (lazy[id] == inf || l >= r) return;

        int &v = lazy[id];

        FOR(j, 2 * id, 2 * id + 1) {
            minimize(seg[j], v);
            minimize(lazy[j], v);
        }

        v = inf;
    }

    void update(int id, int l, int r, int u, int v, int val) {
        if (l > v || u > r) return;
        if (u <= l && r <= v) {
            minimize(seg[id], val);
            minimize(lazy[id], val);
            return;
        }

        down(id, l, r);

        int g = (l + r) >> 1;
        update(id << 1, l, g, u, v, val);
        update(id << 1 | 1, g + 1, r, u, v, val);
        seg[id] = min(seg[id << 1], seg[id << 1 | 1]);
    }

    int get(int id, int l, int r, int u, int v) {
        if (l > v || u > r) return inf;
        if (u <= l && r <= v) return seg[id];

        down(id, l, r);

        int g = (l + r) >> 1;
        return min(get(id << 1, l, g, u, v), get(id << 1 | 1, g + 1, r, u, v));
    }

    void update(int u, int v, int val) {
        update(1, 1, n, u, v, val);
    }

    int get(int u, int v) {
        if (u > v) return inf;
        return get(1, 1, n, u, v);
    }
} st;

namespace subtask2 {
    bool check() {
        FOR(i, 1, n) if (cir[i].y != 0) return false;
        return true;
    }

    void solve() {
        sort(cir + 1, cir + 1 + n);

        vector <int> compressed;

        FOR(i, 1, n) {
            compressed.push_back(cir[i].x - cir[i].r);
            compressed.push_back(cir[i].x + cir[i].r);
        }
        remove_dup(compressed);

        int lim = SZ(compressed) + 2;
        st.initTree(lim);

        memset(del, 0, sizeof(del));
        FOR(i, 1, n) {
            int curID = cir[i].id;
            int L = POS(cir[i].x - cir[i].r, compressed), R = POS(cir[i].x + cir[i].r, compressed);
            int get_val = st.get(L, R);

            if (get_val != inf) del[curID] = cir[get_val].id;
            else {
//                assert(get_val == inf);
                del[curID] = curID;
                st.update(L, R, i);
            }
        }

        FOR(i, 1, n) cout << del[i] << " ";

    }
};

const int dx[] = {-2, -1, 0, 1, 2};
const int dy[] = {-2, -1, 0, 1, 2};

pair <int, int> nen[MAX_N + 2];

struct Data {
    pair <int, int> dir;
    int id;

    bool operator < (const Data &other) const {
        if (dir != other.dir) return (dir < other.dir);
        return (id < other.id);
    }
};

namespace subtask3 {
    bool check() {
        FOR(i, 2, n) if (cir[i].r != cir[1].r) return false;
        return true;
    }

    // ko can xoa vi tong lan duyet ko qua lon
    /// sum visit all circle <= 50 * n;

    void solve() {
        /// compressed into board with each cell have size R x R;
        FOR(i, 1, n) nen[i] = {cir[i].x / cir[i].r, cir[i].y / cir[i].r};

        vector <Data> v;
        FOR(i, 1, n) v.push_back({nen[i], i});

        sort(ALL(v));

        FOR(i, 1, n) {
            if (del[i]) continue;

            /// kc toi da de 2 o giao nhau;
            FOR(nx, nen[i].fi -2, nen[i].fi + 2) FOR(ny, nen[i].se -2, nen[i].se + 2) {
                pair <int, int> curDir = {nx, ny};

                int l = 0, r = SZ(v) - 1, g, pos = -1;
                while (l <= r) {
                    g = (l + r) >> 1;
                    if (v[g].dir < curDir) l = g + 1;
                    else pos = g, r = g - 1;
                }

                if (pos == -1) continue;

                while (pos < SZ(v)) {
                    pair <int, int> tmp = v[pos].dir;
                    if (tmp != curDir) break;

                    int id = v[pos].id;
                    if (!del[id] && sqr(cir[i].x - cir[id].x) + sqr(cir[i].y - cir[id].y) <= sqr(cir[i].r + cir[id].r)) {
                        del[id] = i;
                    }

                    ++pos;
                }
            }
        }

        FOR(i, 1, n) cout << del[i] << " ";
    }
};

namespace subtask4 {

    vector <Data> v;
    void hashTable(int radius) {
        v.clear();
        FOR(i, 1, n) if (del[cir[i].id] == 0) {
            nen[i] = {cir[i].x / radius, cir[i].y / radius};
            v.push_back({nen[i], i});
        }

        sort(ALL(v));
    }

    void solve() {
        sort(cir + 1, cir + 1 + n);

        int radius = cir[1].r;
        hashTable(radius);

        FOR(i, 1, n) {
            if (del[cir[i].id] > 0) continue;
            if (cir[i].r * 2 < radius) {
                radius = cir[i].r;
                hashTable(radius);
            }

            FOR(nx, nen[i].fi - 2, nen[i].fi + 2)
                FOR(ny, nen[i].se - 2, nen[i].se + 2) {
                    pair <int, int> cur = {nx, ny};
                    Data tt = {cur, 0};
                    int pos = lower_bound(ALL(v), tt) - v.begin();

                    while (pos < SZ(v)) {
                        pair <int, int> tmp = v[pos].dir;
                        if (tmp != cur) break;

                        int j = v[pos].id;
                        if (!del[cir[j].id] && sqr(cir[i].x - cir[j].x) + sqr(cir[i].y - cir[j].y) <= sqr(cir[i].r + cir[j].r)) {
                            del[cir[j].id] = cir[i].id;
                        }

                        ++pos;
                    }
                }
        }

        FOR(i, 1, n) cout << del[i] << " ";
    }
};

int main() {
    ios_base::sync_with_stdio(false);cin.tie(nullptr);
    cin >> n;
    FOR(i, 1, n) cin >> cir[i], cir[i].id = i;

//    if (subtask1 :: check()) return subtask1 :: solve(), 0;
//    if (subtask2 :: check()) return subtask2 :: solve(), 0;
//    if (subtask3 :: check()) return subtask3 :: solve(), 0;
    subtask4 :: solve();

    return 0;
}

• Subtask 1: 7 / 7
• Subtask 2: 12 / 12
• Subtask 3: 15 / 15
• Subtask 4: 23 / 23
• Subtask 5: 30 / 30
• Subtask 6: 13 / 13