AcWing 4549. 做作业

这道题直接状压就行，\(f_{S}\) 表示当前状态是 \(S\) 最少扣分数。

// #define FILE_INPUT
#include <iostream>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define rep(i, a, b) for (int i = a, END##i = b; i <= END##i; i++)
#define per(i, a, b) for (int i = a, END##i = b; i >= END##i; i--)

void Init();
void Solve();

signed main() {
    cin.sync_with_stdio(0);
    cin.tie(0), cout.tie(0);

    #ifdef FILE_INPUT
        freopen("input.in", "r", stdin);
    #endif

    int T = 1;
    cin >> T;
    while (T--) {
        Init();
        Solve();
    }
    return 0;
}

using LL = long long;
using ULL = unsigned long long;

const int Mod = 1e9 + 7;
const int Inf = 0x3f3f3f3f;
const LL InfLL = 0x3f3f3f3f3f3f3f3f;

const int N = 20, M = 1 << N;
int n, f[M], ti[M], pre[M];
struct Node {
    string sbt; // subject
    int d, c;
}a[N];

void Init() {
}

void Solve() {
    cin >> n;
    rep(i, 0, n - 1) cin >> a[i].sbt >> a[i].d >> a[i].c;
    int m = (1 << n) - 1;
    memset(ti, 0, sizeof(ti));
    rep(i, 0, m) {
        rep(j, 0, n - 1) if (i >> j & 1)
            ti[i] += a[j].c;
    }
    memset(f, 0x3f, sizeof(f));
    f[0] = 0;
    rep(now, 0, m) {
        rep(j, 0, n - 1) {
            if (!(now >> j & 1)) {
                int state = now | (1 << j);
                int val = max(0, ti[now] + a[j].c - a[j].d);
                if (f[state] > f[now] + val) {
                    f[state] = f[now] + val;
                    pre[state] = now;
                }
            }
        }
    }
    cout << f[m] << "\n";
    int cur = m;
    vector<string> ans;
    while (cur) {
        // int nxt;
        // rep(i, 0, n - 1) {
        //     if ((cur >> i & 1) && !(pre[cur] >> i & 1))
        //         nxt = i;
        // }
        int nxt = __lg(pre[cur] ^ cur);
        ans.push_back(a[nxt].sbt);
        cur = pre[cur];
    }
    per(i, ans.size() - 1, 0) cout << ans[i] << "\n";
}