This costs minimization problem, is it solvable in polynomial time?
It's a bad practice to use the same variable for two purposes, so I'll say that you have $C$ conflicts $(c_1, c_2, \dots, c_C)$ and similarly $M$ conflict resolution methods. To simplify, let $r_{ji} = 0$ if $c_i$ and $m_j$ are disconnected.
Suppose $t_j$ is a variable indicating how many times method $m_j$ was used we can phrase your problem as
\begin{align}
\min \quad&amp;\sum_{j}p_jt_j\\
\text{s.t.}\quad&amp;\forall i: \sum_j r_{ji}t_j \geq s_i\,,\\
&amp;\forall j:0\leq t_j \leq u_j\,,\\
&amp; \forall j:t_j \in \mathbb{N}_0\,.
\end{align}
What if instead we assume we always start by using every $m_j$ as much as possible, and $n_j$ represents the number of times we do not use $m_j$ (compared to the maximum allowed). Note that $t_j = u_j - n_j$. Substituting gives us
\begin{align}
\min \quad&amp;\sum_{j}p_j(u_j-n_j)\\
\text{s.t.}\quad&amp;\forall i: \sum_j r_{ji}(u_j-n_j) \geq s_i\,,\\
&amp;\forall j:0\leq(u_j-n_j) \leq u_j\,,\\
&amp; \forall j:n_j \in \mathbb{N}_0\,,
\end{align}
which is equivalent to
\begin{align}
\max \quad&amp; \sum_{j}p_jn_j\\
\text{s.t.}\quad&amp;\forall i:  \sum_j r_{ji}n_j \leq -s_i + \sum_j r_{ji}u_j\,,\\
&amp;\forall j:0 \leq n_j \leq u_j\,,\\
&amp; \forall j:n_j \in \mathbb{N}_0\,.
\end{align}
But this is exactly the $M$-dimensional bounded knapsack problem. Thus your problem is NP-hard, but you also get to benefit from the existing literature that give very good practical solvers.