## 2 Methods
### Optimization Algorithm Design
Suppose the number of demand points is N, the number of demand points to be removed is M, and the random element is L. Then, the destruction operator step is shown in Figure 4. Based on obtaining the removed demand point set Q and destroying solution P_destroy, the operator repair is carried out, as shown in Figure 5.

Insertion cost: Under the condition of not violating constraints, randomly select a demand point in Q and insert it back to a certain insertion position in P_destroy, forming the repaired solution P_repair. The insertion cost calculation formula for inserting the demand point into this position is shown in Equation (36). \[Cost\_insert=F_{P\_repair}-F_{P\_destroy} \tag{36}\]

where, \(Cost\_insert\) represents the insertion cost, \(F_{P\_repair}\) represents the objective function value of the repaired solution, and \(F_{P\_destroy}\) represents the objective function value of the damaged solution. Regret value: Under the condition of not violating the constraint, if the total number of feasible insertion positions for demand point \(i\) in Q that can be inserted back into \(P\_destroy\) is \(\updownarrow\), then \(\updownarrow\) insertion positions correspond to \(\updownarrow\) insertion costs. The insertion costs are sorted from small to large, and the sorted insertion costs are defined as \(Cost\_new\_insert\). The

Figure 4: Destroy operator description graph. Figure 5: Repair operator description graph. formula for calculating the regret value of inserting demand point \(i\) back into \(P\_destroy\) is shown in Equation (37). \[c^{\prime}_{ij}=\frac{c_{ij}}{\max c_{ij}} \tag{34}\]0

where, \(Value\_regret\) represents the regret value, \(Cost\_new\_insert(2)\) represents the insertion cost of ranking second in the back row, and \(Cost\_new\_insert(1)\) represents the insertion cost of ranking first in the back row. ## 3 Results

### Case Analysis

#### 3.1.1 Relevant Data

The basic attribute of emergency materials is emergency, so it is required that emergency materials be delivered to the disaster area in the shortest time with the fastest speed to play the role of emergency materials. During the new coronavirus epidemic, home isolation is adopted for patients with solid bodies and mild symptoms. Traditional manual door-to-door delivery is highly risky and has a significant risk of infection, and vehicle delivery cannot complete accurate door-to-door delivery service. At this time, the use of drones for \"last kilometer\" emergency supplies distribution is essential. Based on this, this paper takes Ehang Falcon B (Logistics version) as an experimental UAV to analyze the distribution of materials during the COVID-19 outbreak in Changchun, China. The relevant parameters are shown in Table 2, the specifications of emergency materials are shown in Table 3, and the data of demand points are shown in \[c^{\prime}_{ij}=\frac{c_{ij}}{\max c_{ij}} \tag{34}\]1 \begin{table}
\begin{tabular}{c c c c c c c c c c} \hline \hline \multirow{2}{*}{**Number**} & \multirow{2}{*}{**X**} & \multirow{2}{*}{**Y**} & \multicolumn{3}{c}{**Demand**} & \multirow{2}{*}{**Recycling**} & \multirow{2}{*}{**Recycling**} & \multirow{2}{*}{**Left Time**} & \multirow{2}{*}{**Right Time**} & \multirow{2}{*}{**Service**} \\ \cline{3-4} \cline{6-9}  & & & **A** & & & & & **Quantity/kg** & & & **Volume/m3** & **Window** & **Time/min** \\ \hline
0 & 57 & 61 & 0 & 0 & 0 & 0 & 0 & 9:00 & 11:00 & 10 \\ \hline
1 & 52 & 62 & 4 & 14 & 3 & 0.25 & 0.00641 & 9:00 & 9:30 & 10 \\ \hline
2 & 55 & 55 & 17 & 13 & 11 & 0.85 & 0.00426 & 9:10 & 9:40 & 12 \\ \hline
3 & 54 & 68 & 3 & 16 & 13 & 1 & 0.00789 & 9:10 & 9:30 & 14 \\ \hline
4 & 58 & 62 & 17 & 9 & 20 & 0.5 & 0.00430 & 9:20 & 9:50 & 15 \\ \hline \hline \end{tabular}
\end{table} 
Table 4: Data related to demand points.