Codeforces-solutions/Educational Codeforces Round 3/D. Gadgets for dollars and pounds.cpp

/* Problem URL: https://codeforces.com/contest/609/problem/D */

#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 <class T, class comp = less<>>
using ordered_set = tree<T, null_type , comp , rb_tree_tag , tree_order_statistics_node_update>;

#define V vector

#define rmin(a, b) a = min(a, b)
#define rmax(a, b) a = max(a, b)

#define rep(i, lim) for (int i = 0; i < (lim); i++)
#define nrep(i, s, lim) for (int i = s; i < (lim); i++)

#define repv(i, v) for (auto &i : (v))
#define fillv(v) for (auto &itr_ : (v)) { cin >> itr_; }
#define sortv(v) sort(v.begin(), v.end())
#define all(v) (v).begin(), (v).end()

using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vvvvi = vector<vvvi>;

using ll = long long;

using vl = vector<ll>;
using vvl = vector<vl>;
using vvvl = vector<vvl>;
using vvvvl = vector<vvvl>;

template<class v>
auto operator<<(ostream &os, const vector<v> &vec)->ostream& {
    os << vec[0];
    for (size_t i = 1; i < vec.size(); i++) {
        os << ' ' << vec[i];
    }
    os << '\n';
    return os;
}

template<class v>
auto operator>>(istream &is, vector<v> &vec)->istream& {
    for (auto &i : vec) {
        is >> i;
    }
    return is;
}

template<class v>
auto operator<<(ostream &os, const vector<vector<v>> &vec)->ostream& {
    for (auto &i : vec) {
        os << i[0];
        for (size_t j = 1; j < i.size(); j++) {
            os << ' ' << i[j];
        }
        os << '\n';
    }
    return os;
}

template<class v>
auto operator>>(istream  &is, vector<vector<v>> &vec)->istream& {
    for (auto &i : vec) {
        for (auto &j : i) {
            is >> j;
        }
    }
    return is;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, m, k;
    ll s;
    cin >> n >> m >> k >> s;

    vl cost[2] = {vl(n), vl(n)};
    cin >> cost[0];
    cin >> cost[1];

    nrep(i, 1, n) {
        rmin(cost[0][i], cost[0][i - 1]);
        rmin(cost[1][i], cost[1][i - 1]);
    }

    V<pair<ll, int>> gad[2];

    rep(i, m) {
        int t;
        ll c;
        cin >> t >> c;

        gad[t - 1].emplace_back(c, i);
    }

    sortv(gad[0]);
    sortv(gad[1]);

    ll sum1 = 0;
    ll sum2 = 0;

    int i = 0;
    int j = 0;

    for (; i < min((int)gad[0].size(), k); i++) {
        sum1 += gad[0][i].first;
    }

    for (; j < k - i; j++) {
        sum2 += gad[1][j].first;
    }

    int oo = INT32_MAX >> 1;

    int ansi = -1;
    int ansj = -1;
    int ans = oo;

    while (i >= 0 && j <= gad[1].size()) {
        int low = 0;
        int high = n - 1;
        int act = -1;

        while (low <= high) {
            int mid = (low + high) >> 1;

            ll now = sum1 * cost[0][mid] + sum2 * cost[1][mid];

            if (now <= s) {
                act = mid;
                high = mid - 1;
                continue;
            }

            low = mid + 1;
        }

        if (act == -1) {
            if (j == gad[1].size()) {
                break;
            }
            i--;
            sum1 -= gad[0][i].first;
            sum2 += gad[1][j].first;
            j++;
            continue;
        }

        if (act < ans) {
            ans = act;
            ansi = i;
            ansj = j;
        }

        if (j == gad[1].size()) {
            break;
        }
        i--;
        sum1 -= gad[0][i].first;
        sum2 += gad[1][j].first;
        j++;
    }

    if (ans == oo) {
        cout << "-1\n";
        return 0;
    }

    int choice1 = 0;
    int choice2 = 0;

    nrep(i, 1, ans + 1) {
        if (cost[0][i] < cost[0][i - 1]) {
            choice1 = i;
        }

        if (cost[1][i] < cost[1][i - 1]) {
            choice2 = i;
        }
    }

    cout << ans + 1 << '\n';

    rep(k, ansi) {
        cout << gad[0][k].second + 1 << ' ' << choice1 + 1 << '\n';
    }
    rep(k, ansj) {
        cout << gad[1][k].second + 1 << ' ' << choice2 + 1 << '\n';
    }
}