D. Bananas in a Microwave
dfs and similar; dp; graphs; implementation
You have a malfunctioning microwave in which you want to put some bananas. You have \(n\) time-steps before the microwave stops working completely. At each time-step, it displays a new operation.Let \(k\) be the number of bananas in the microwave currently. Initially, \(k = 0\). In the \(i\)-th operation, you are given three parameters \(t_i\), \(x_i\), \(y_i\) in the input. Based on the value of \(t_i\), you must do one of the following:Type 1: (\(t_i=1\), \(x_i\), \(y_i\)) — pick an \(a_i\), such that \(0 \le a_i \le y_i\), and perform the following update \(a_i\) times: \(k:=\lceil (k + x_i) \rceil\).Type 2: (\(t_i=2\), \(x_i\), \(y_i\)) — pick an \(a_i\), such that \(0 \le a_i \le y_i\), and perform the following update \(a_i\) times: \(k:=\lceil (k \cdot x_i) \rceil\).Note that \(x_i\) can be a fractional value. See input format for more details. Also, \(\lceil x \rceil\) is the smallest integer \(\ge x\).At the \(i\)-th time-step, you must apply the \(i\)-th operation exactly once.For each \(j\) such that \(1 \le j \le m\), output the earliest time-step at which you can create exactly \(j\) bananas. If you cannot create exactly \(j\) bananas, output \(-1\).
The first line contains two space-separated integers \(n\) \((1 \le n \le 200)\) and \(m\) \((2 \le m \le 10^5)\).Then, \(n\) lines follow, where the \(i\)-th line denotes the operation for the \(i\)-th timestep. Each such line contains three space-separated integers \(t_i\), \(x'_i\) and \(y_i\) (\(1 \le t_i \le 2\), \(1\le y_i\le m\)).Note that you are given \(x'_i\), which is \(10^5 \cdot x_i\). Thus, to obtain \(x_i\), use the formula \(x_i= \dfrac{x'_i} {10^5}\).For type 1 operations, \(1 \le x'_i \le 10^5 \cdot m\), and for type 2 operations, \(10^5 < x'_i \le 10^5 \cdot m\).
Print \(m\) integers, where the \(i\)-th integer is the earliest time-step when you can obtain exactly \(i\) bananas (or \(-1\) if it is impossible).
In the first sample input, let us see how to create \(16\) number of bananas in three timesteps. Initially, \(k=0\). In timestep 1, we choose \(a_1=2\), so we apply the type 1 update — \(k := \lceil(k+3)\rceil\) — two times. Hence, \(k\) is now 6. In timestep 2, we choose \(a_2=0\), hence value of \(k\) remains unchanged. In timestep 3, we choose \(a_3=1\), so we are applying the type 1 update \(k:= \lceil(k+10)\rceil\) once. Hence, \(k\) is now 16. It can be shown that \(k=16\) cannot be reached in fewer than three timesteps with the given operations.In the second sample input, let us see how to create \(17\) number of bananas in two timesteps. Initially, \(k=0\). In timestep 1, we choose \(a_1=1\), so we apply the type 1 update — \(k := \lceil(k+3.99999)\rceil\) — once. Hence, \(k\) is now 4. In timestep 2, we choose \(a_2=1\), so we apply the type 2 update — \(k := \lceil(k\cdot 4.12345)\rceil\) — once. Hence, \(k\) is now 17. It can be shown that \(k=17\) cannot be reached in fewer than two timesteps with the given operations.