AcWing 4889. 空调II

考虑到空调的数量M<= 10，可以使用暴力枚举的方法，暴力枚举每个空调可以使用二进制枚举的方式来简化，用1/0来表示这个空调开/不开 ，然后枚举每一个牛的牛栏是否达到降温需求，若达到，就和答案取min

int n,m;
struct Cow
{
	LL s,t,c;
}cows[N];

struct Ac
{
	LL a,b,p,m;
}acs[N];

int w[N];
void solve()
{
	cin >> n >> m;
	for (int i = 0;i < n;i ++) 
	{
		int s,t,c; cin >> s >> t >> c;
		cows[i] = {s,t,c};
	}
	
	for (int i = 0;i < m;i ++)
	{
		int a,b,p,m; cin >> a >> b >> p >> m;
		acs[i] = {a,b,p,m};
	}

	LL ans = inf;
	for (int i = 0;i < 1 << m;i ++)
	{
		memset(w,0,sizeof w);//每次将牛栏的降温情况清空
		LL cost = 0;
		for (int j = 0;j < m;j ++)//枚举每个空调开不开
			if (i >> j & 1) 
			{
				int a = acs[j].a,b = acs[j].b,p = acs[j].p,m = acs[j].m;
				cost += m;
				for (int k = a;k <= b;k ++)
					w[k] += p;//降温
			}
		
		bool f = true;
		for (int i = 0;i < n;i ++)
		{
			int s = cows[i].s,t = cows[i].t,c = cows[i].c;
			for (int j = s;j <= t;j ++)
				if (w[j] < c)
				{
					f = false;
					break;
				}
		}
		
		if (f) ans = min(ans,cost);
	}	
	
	cout << ans << endl;
}