Submission #731946

#	Time	Username	Problem	Language	Result	Execution time	Memory
731946		GrindMachine	Pairs (IOI07_pairs)	C++17	40 / 100	4056 ms	16904 KiB

pairs

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

    T sum(int 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;
    }
};

void solve(int test_case)
{
    ll b, n, d, m; cin >> b >> n >> d >> m;

    if (b == 1) {
        vector<ll> a(n);
        rep(i, n) cin >> a[i];

        sort(all(a));

        ll ans = 0;
        ll ptr = 0;

        rep(i, n) {
            while (a[i] - a[ptr] > d) {
                ptr++;
            }

            ans += i - ptr;
        }

        cout << ans << endl;
    }

    else if (b == 2) {
        vector<pll> a(n);
        rep(i, n) cin >> a[i].ff >> a[i].ss;

        rep(i, n) {
            ll x = a[i].ff + a[i].ss;
            ll y = a[i].ff - a[i].ss;
            a[i] = {x, y + m};
        }

        vector<ll> here[2 * m + 5];
        vector<pll> queries[2 * m + 5];
        fenwick<ll> fenw(2 * m + 5);

        rep(i, n) {
            auto [x, y] = a[i];
            here[x].pb(y);
            queries[x].pb({y, 1});
            ll x2 = x - d - 1;
            if (x2 >= 0) {
                queries[x2].pb({y, -1});
            }
        }

        ll ans = 0;

        rep(x, 2 * m + 5) {
            trav(y, here[x]) {
                fenw.pupd(y, 1);
            }

            for (auto [y, c] : queries[x]) {
                ll lx = max(y - d, 1ll), rx = min(y + d, 2 * m);
                ans += fenw.query(lx, rx) * c;
            }
        }

        ans -= n;
        cout << ans << endl;
    }

    else {
        vector<vector<ll>> a(n, vector<ll>(b));
        rep(i, n) rep(j, b) cin >> a[i][j];

        ll ans = 0;

        rep(i, n) {
            for (int j = i + 1; j < n; ++j) {
                ll dis = 0;
                rep(k, b) {
                    dis += abs(a[i][k] - a[j][k]);
                }

                ans += (dis <= d);
            }
        }

        cout << ans << endl;
    }
}

int main()
{
    fastio;

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

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

    return 0;
}

• Subtask 1: 6 / 6
• Subtask 2: 6 / 6
• Subtask 3: 6 / 6
• Subtask 4: 6 / 6
• Subtask 5: 6 / 6
• Subtask 6: 0 / 7
• Subtask 7: 0 / 7
• Subtask 8: 0 / 8
• Subtask 9: 0 / 8
• Subtask 10: 10 / 10
• Subtask 11: 0 / 10
• Subtask 12: 0 / 10
• Subtask 13: 0 / 10