제출 #742951

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
742951		GrindMachine	Road Construction (JOI21_road_construction)	C++17	100 / 100	9197 ms	203080 KiB

road_construction

// Om Namah Shivaya

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*



*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

template<typename T>
struct fenwick {
    int siz;
    vector<T> tree;

    fenwick(int n) {
        siz = n;
        tree = vector<T>(n + 1);
    }

    int lsb(int x) {
        return x & -x;
    }

    void build(vector<T> &a, int n) {
        for (int i = 1; i <= n; ++i) {
            int par = i + lsb(i);
            tree[i] += a[i];

            if (par <= siz) {
                tree[par] += tree[i];
            }
        }
    }

    void pupd(int i, T v) {
        i++;

        while (i <= siz) {
            tree[i] += v;
            i += lsb(i);
        }
    }

    T sum(int i) {
        i++;

        T res = 0;

        while (i) {
            res += tree[i];
            i -= lsb(i);
        }

        return res;
    }

    T query(int l, int r) {
        if (l > r) return 0;
        T res = sum(r) - sum(l - 1);
        return res;
    }
};

template<typename T>
struct segtree {
    // https://codeforces.com/blog/entry/18051

    /*=======================================================*/

    struct data {
        vector<pll> a;
    };

    data neutral = data();

    data merge(data &left, data &right) {
        data curr;

        auto &v1 = left.a, &v2 = right.a, &v3 = curr.a;
        ll n = sz(v1), m = sz(v2);
        ll ptr1 = 0, ptr2 = 0;

        while (ptr1 < n and ptr2 < m) {
            if (v1[ptr1] <= v2[ptr2]) {
                v3.pb(v1[ptr1++]);
            }
            else {
                v3.pb(v2[ptr2++]);
            }
        }

        for (; ptr1 < n; ++ptr1) {
            v3.pb(v1[ptr1]);
        }

        for (; ptr2 < m; ++ptr2) {
            v3.pb(v2[ptr2]);
        }

        return curr;
    }

    void create(int i, T v) {
        tr[i].a = v;
    }

    void modify(int i, T v) {

    }

    /*=======================================================*/

    int n;
    vector<data> tr;

    segtree() {

    }

    segtree(int siz) {
        init(siz);
    }

    void init(int siz) {
        n = siz;
        tr.assign(2 * n, neutral);
    }

    void build(vector<T> &a, int siz) {
        rep(i, siz) create(i + n, a[i]);
        rev(i, n - 1, 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
    }

    void pupd(int i, T v) {
        modify(i + n, v);
        for (i = (i + n) >> 1; i; i >>= 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
    }

    vector<pll> query(int l, int r, ll mn, ll mx) {
        pll px = {mn, -inf2};
        vector<pll> res;

        for (l += n, r += n; l <= r; l >>= 1, r >>= 1) {
            if (l & 1) {
                auto &v = tr[l++].a;
                auto it = upper_bound(all(v), px);
                for (; it != v.end() and it->first <= mx; ++it) {
                    res.pb(*it);
                }
            }

            if (!(r & 1)) {
                auto &v = tr[r--].a;
                auto it = upper_bound(all(v), px);
                for (; it != v.end() and it->first <= mx; ++it) {
                    res.pb(*it);
                }
            }
        }

        return res;
    }
};

void solve(int test_case)
{
    ll n, k; cin >> n >> k;
    vector<pll> a(n);
    rep(i, n) cin >> a[i].ff >> a[i].ss;

    rep(i, n) {
        auto [x, y] = a[i];
        a[i] = {x + y, x - y};
    }

    vector<ll> b;
    rep(i, n) b.pb(a[i].ss);

    sort(all(b));
    b.resize(unique(all(b)) - b.begin());
    ll siz = sz(b);

    vector<array<ll, 3>> inds(n);

    auto get_cnt = [&](ll m) {
        rep(i, n) {
            auto [x, y] = a[i];
            inds[i][0] = lower_bound(all(b), y - m) - b.begin();
            inds[i][1] = lower_bound(all(b), y) - b.begin();
            inds[i][2] = upper_bound(all(b), y + m) - b.begin() - 1;
        }

        vector<array<ll, 3>> events;

        rep(i, n) {
            auto [x, y] = a[i];
            events.pb({x, 0, i});
        }

        rep(i, n) {
            auto [x, y] = a[i];
            events.pb({x - m - 1, 1, i});
            events.pb({x + m, 2, i});
        }

        sort(all(events));
        fenwick<ll> fenw(siz + 5);
        ll res = 0;

        for (auto [x, t, i] : events) {
            if (t == 0) {
                fenw.pupd(inds[i][1], 1);
            }
            else {
                ll c = 1;
                if (t == 1) c = -1;
                ll lx = inds[i][0], rx = inds[i][2];
                res += fenw.query(lx, rx) * c;
            }
        }

        res -= n;
        res /= 2;

        return res;
    };

    ll l = 0, r = 5e9;
    ll mx_cost = inf2;

    while (l <= r) {
        ll mid = (l + r) >> 1;
        if (get_cnt(mid) <= k) {
            mx_cost = mid;
            l = mid + 1;
        }
        else {
            r = mid - 1;
        }
    }

    b.clear();
    rep(i, n) {
        auto [x, y] = a[i];
        b.pb(x - mx_cost), b.pb(x), b.pb(x + mx_cost);
    }

    sort(all(b));
    b.resize(unique(all(b)) - b.begin());
    siz = sz(b);

    vector<vector<pll>> here(siz);
    rep(i, n) {
        auto [x, y] = a[i];
        ll ind = lower_bound(all(b), x) - b.begin();
        here[ind].pb({y, x});
    }

    rep(i, siz) sort(all(here[i]));

    segtree<vector<pll>> st(siz + 5);
    st.build(here, siz);

    vector<ll> ans;

    rep(i, n) {
        auto [x, y] = a[i];
        ll lx = lower_bound(all(b), x - mx_cost) - b.begin();
        ll rx = lower_bound(all(b), x + mx_cost) - b.begin();
        ll mny = y - mx_cost;
        ll mxy = y + mx_cost;

        auto v = st.query(lx, rx, mny, mxy);

        for (auto [y2, x2] : v) {
            if (x == x2 and y == y2) conts;
            ll cost = max(abs(x - x2), abs(y - y2));
            ans.pb(cost);
        }
    }

    sort(all(ans));

    vector<ll> ans2;
    for (int i = 0; i < sz(ans); i += 2) {
        assert(ans[i] == ans[i + 1]);
        ans2.pb(ans[i]);
    }

    ans = ans2;
    while (sz(ans) < k) ans.pb(mx_cost + 1);

    trav(x, ans) cout << x << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

컴파일 시 표준 에러 (stderr) 메시지

road_construction.cpp: In function 'void solve(int)':
road_construction.cpp:337:23: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<long long int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
  337 |     for (int i = 0; i < sz(ans); i += 2) {
      |                       ^
road_construction.cpp:343:20: warning: comparison of integer expressions of different signedness: 'std::vector<long long int>::size_type' {aka 'long unsigned int'} and 'll' {aka 'long long int'} [-Wsign-compare]
  343 |     while (sz(ans) < k) ans.pb(mx_cost + 1);
      |                    ^

• Subtask 1: 5 / 5
• Subtask 2: 6 / 6
• Subtask 3: 7 / 7
• Subtask 4: 20 / 20
• Subtask 5: 27 / 27
• Subtask 6: 35 / 35