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;
}