直接按公式排就可以了。

AC code：

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

struct Node {
    string name;
    double k;
};

bool cmp(const Node& a, const Node& b) {
    return a.k > b.k;
}

int main() {
    freopen("football.in", "r", stdin);
    freopen("football.out", "w", stdout);
    int T;
    cin >> T;
    while (T--) {
        int a;
        cin >> a;
        vector<Node> Nodes(a);
        for (int i = 0; i < a; i++) {
            string name;
            long long m, n;
            int t;
            cin >> name >> m >> n >> t;
            double k = (0.6 * m + 0.3 * n) * 0.1 * t;
            Nodes[i].name = name, Nodes[i].k = k;
        }
        sort(Nodes.begin(), Nodes.end(), cmp);
        for (int i = 0; i < a; i++) {
            if (i > 0) {
                cout << " ";
            }
            cout << Nodes[i].name;
        }
        cout << endl;
    }
    return 0;
}

看清总评价的计算公式后利用 struct 和 sort 排序即可。

为了避免精度误差，建议按 $k=(2m+n)t$（将原来的公式除以 $0.03$）计算总评价，此时需要使用 long long。（当然了，直接使用 double 或 long double 也是一个不错的选择。）

#include <bits/stdc++.h>
using namespace std;
const int A = 20 + 2;
struct Team
{
	string b;
	int m;
	int n;
	int t;
	long long k;
};
Team team[A];
bool cmp(Team x, Team y)
{
	return x.k > y.k;
}
int main()
{
	freopen("football.in", "r", stdin);
	freopen("football.out", "w", stdout);
	int T;
	cin >> T;
	while (T--)
	{
		int a;
		cin >> a;
		for (int i = 1; i <= a; i++)
		{
			cin >> team[i].b >> team[i].m >> team[i].n >> team[i].t;
			team[i].k = 1ll * team[i].t * (2ll * team[i].m + 1ll * team[i].n);
		}
		sort(team + 1, team + a + 1, cmp);
		for (int i = 1; i <= a; i++)
		{
			cout << team[i].b;
			if (i != a) cout << ' ';
		}
		cout << endl;
	}
	return 0;
}