C_i + f_1F_1$ ($k \ge 2$), and $L_{\rm red}$ is comb-shaped, i.e., $C_i$’s are disjoint cross-sections.

Case (iii). $L = \sum_{i=1}^k c_iC_i$ ($k \ge 2$).

Recall that $K_X \sim -2 C_1 -(d+2) F$. The nefness of $K_X + L$ implies: $$0 \le (K_X+L ). F = -2 + \sum c_i (C_i . F),$$ $$0 \le (K_X + L) . C_1 = d- 2 - d c_1 + \sum_{i \ge 2} c_i (C_i . C_1) + \sum f_j,$$ $$\sum c_i (C_i . F) \ge 2,$$ $$\sum f_j \ge 2 + (c_1-1) d - \sum_{i \ge 2} c_i (C_i . C_1).$$

In Case (i), the above inequalities imply $c_1 \ge 2$ and $\sum f_j \ge 2 + (c_1 - 1) d \ge d+2$, whence $\lceil L \rceil = \lceil c_1\rceil C_1 + \sum \lceil f_j \rceil F_j
\sim \lceil c_1 \rceil C_1 + (\sum \lceil f_j \rceil) F
\ge 2 C_1 + (\sum f_j) F \ge 2 C_1 + (d+2) F_1\sim -K_X$. Hence $H^0(X, K_X + \lceil L \rceil) \ne 0$.

Consider Case (ii). Then one sees easily that $k = 2$ and $C_2 \sim C_1 + d F_1$ (see \[Ha, Chapter V, §2\]). By the displayed inequalities, we have $c_1 \ge 