Algorithm 1: SP Algorithm

Begin Initialize the value of all the decision variables to be zero; Construct the nodes and links in the 2-D planar graph; Let the node position of the evacuating person be the start node; While $\left(t\le \left|T\right|\right)$ //looping at each time slot t Begin  Collect the temperature data at each node and the link cost data at each link at time slot $t$;  Calculate the best path from the start node to every exit node by using Dijkstra’s algorithm and let ${\lambda }_e=$ the shortest path cost to every exit $e\in E$; //Step 2  Let $\tilde{\lambda }=\mathrm{Min} Ar{g}_{e\in E}{\lambda }_e$; //the exit number that has the smallest shortest path cost, evacuation path is the best path to exit $\tilde{\lambda }$  If (the hop distance from the start node to the exit number $\tilde{\lambda }$ on the shortest path $\le$ $\mathsf{\Lambda }$) //reach the exit  Begin   Let $\mathsf{\Omega }$ = $\mathsf{\Omega }$ + (The link cost from the start node to the exit $\tilde{\lambda }$ on this evacuation path);   Report $\mathsf{\Omega }$ and the evacuation path;   $\rho$= 1;   Break from the While loop;  End // If (the hop  Else // have not reached the exit, then move the start node $\mathsf{\Lambda }$ hop distance closer to exit $\tilde{\lambda }$ on the emergency path  Begin   Let $\mathsf{\Omega }$ = $\mathsf{\Omega }$ + (All the link cost from the start node to the node with hop distance $\mathsf{\Lambda }$on the evacuation path);   Move the start node to the new position with $\mathsf{\Lambda }$ hop distance on the evacuation path;  End // Else End; //While loop If ($\rho$== 0)   Report no$\mathrm{feasible} \mathrm{solutions}$; End