[CF864E] Fire - 背包dp

[CF864E] Fire - 背包dp

Description

某人的房子着火了,他想从大火中带走价值总和尽量多的物品,每次他只能带走一个,分别给出挽救某物品需要的时间t,该物品开始燃烧的时间d(在d时间开始燃烧就不能再挽救该物品了),该物品的价值p。约束：n(1<=n<=100) ti(1<=ti<=20),di(1<=di<=2000),pi(1<=pi<=20)

Solution

以d为关键字排序，做背包即可

这里学到一种记录方案的偷懒办法，虽然复杂度有点高

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

#define int long long

const int N = 2005;

int f[N];
vector<int> v[N];

int n;

struct item
{
    int t;
    int d;
    int p;
    int id;
    bool operator<(const item &rhs) const
    {
        return d < rhs.d;
    }
} a[N];

signed main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i].t >> a[i].d >> a[i].p;
        a[i].id = i;
    }
    sort(a + 1, a + n + 1);

    for (int i = 1; i <= n; i++)
    {
        int t = a[i].t;
        int d = a[i].d;
        --d;
        int p = a[i].p;
        int id = a[i].id;
        for (int j = d; j >= t; j--)
        {
            if (f[j - t] + p > f[j])
            {
                f[j] = f[j - t] + p;
                v[j] = v[j - t];
                v[j].push_back(id);
            }
        }
    }

    int ans = 0;
    vector<int> ansv;

    for (int i = 0; i < N; i++)
    {
        if (f[i] > ans)
        {
            ans = f[i];
            ansv = v[i];
        }
    }

    cout << ans << endl;
    cout << ansv.size() << endl;
    for (auto i : ansv)
        cout << i << " ";
    cout << endl;
}

标签：背包,Fire,CF864E,int,ans,物品,id,dp
来源： https://www.cnblogs.com/mollnn/p/14615596.html