[CF3B] Lorry

[CF3B] Lorry

Description

有一辆载重量为 v 的货车，准备运送两种物品。物品 A 的重量为 1，物体 B 的重量为 2，每个物品都有一个价值。求货车可以运送的物品的最大价值。

Solution

枚举运多少个 B 物品，剩下的排序即可

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

#define int long long
const int N = 1000005;

signed main()
{
    ios::sync_with_stdio(false);

    int n, v;
    cin >> n >> v;

    vector<pair<int, int>> a, b;
    for (int i = 1; i <= n; i++)
    {
        int t, p;
        cin >> t >> p;
        if (t == 1)
            a.push_back({p, i});
        else
            b.push_back({p, i});
    }

    sort(a.begin(), a.end());
    reverse(a.begin(), a.end());
    sort(b.begin(), b.end());
    reverse(b.begin(), b.end());

    vector<int> sa(n + 2), sb(n + 2);
    for (int i = 1; i <= a.size(); i++)
        sa[i] = sa[i - 1] + a[i - 1].first;
    for (int i = 1; i <= b.size(); i++)
        sb[i] = sb[i - 1] + b[i - 1].first;
    for (int i = a.size() + 1; i <= n; i++)
        sa[i] = sa[i - 1];
    for (int i = b.size() + 1; i <= n; i++)
        sb[i] = sb[i - 1];

    int ans = 0, ans_a = 0, ans_b = 0;
    for (int i = 0; i <= n; i++)
    {
        int w_b = 2 * i;
        int v_b = sb[i];
        int w_a = v - w_b;
        if (i > b.size())
            continue;
        if (w_a < 0)
            break;
        if (w_a > a.size())
            w_a = a.size();
        int v_a = sa[w_a];
        ans = max(ans, v_a + v_b);
        if (ans == v_a + v_b)
            ans_a = w_a, ans_b = i;
    }

    cout << ans << endl;
    for (int i = 1; i <= ans_a; i++)
        cout << a[i - 1].second << " ";
    for (int i = 1; i <= ans_b; i++)
        cout << b[i - 1].second << " ";
}