感谢同学用户 - 2315 - 南阳理工学院OJ (fuquan.moe)

using namespace std;
const int N = 1e5 + 10;
long long arr[N], brr[N],l[N],r[N],x[N],n,m;
int axx(int c)//差分
{
	brr[1] = m;
	for (int i = 2; i <= n; i++)
		brr[i] = 0;
	for (int i = 1; i <= c; i++)
	{
		brr[l[i]] -= x[i];
		brr[r[i] + 1] += x[i];
	}
	for (int i = 1; i <= n; i++)
	{
		arr[i] = arr[i - 1] + brr[i];
		if (arr[i] <= 0)return 1;
	}
	return 0;
}
int main()
{
	cin >> n >> m;
	brr[1] = m;
	for (int i = 1; i <= n; i++)
		cin >> l[i] >> r[i] >> x[i];
	int i = 1, j = n;
	while (i < j)//二分时间，总共n秒
	{
		int mad = (i + j) / 2;
		if (axx(mad))j = mad;
		else
			i = mad + 1;
	}
	if (j == n)//特判j=n的情况
	{
		if (axx(n))cout << n;//如果炸断了就输出n
		else
			cout << "impossible";//如果n秒后炸不断，就输出impossible
	}
	else
		cout << j;
}