graph - While loop, cout only the shortest path C++ -


Shows me the output shortcut of the graph from one peak to another. The production of that inspiration is in all ways, how to move from one over to another. Maybe I'm tired, but I can not solve it, thanks for help zero search path (string source, string trgt) {vector & gt; Path; Top * source = name_to_vertex [src]; Top * target = name_to_vertex [trgt]; 

  path.push_back (pair & lieutenant; header *, interp (source, 0)); From queue & lt; The vector & lt; Pair & lt; Top *, integer & gt; & Gt; & Gt; Q; q.push (path); Int i = 1; While (! Q.empty ()) {path = q.front (); q.pop (); Top * last_odef_path = path [path.size () - 1]. Fast; If (last_nodeof_path == target) {if (path.back (). Second & lt; 1) // second door pozisia etteraa {cout & lt; & Lt; "Path:"; print_path (path); Cout & lt; & Lt; "\ N"; }} Set & lieutenant; Edge * & gt; :: Iterator; (This is = last_odox_path-> edges.bgi); this! = Final_odox_path-> Edges.and (); ++) {if (isadjacency_node_not_present_in_current_path ((this *) -> getOposite (last_nodeof_path), path)) {vector & lieutenant; Pair and lieutenant; Header *, Intimate & gt; & Gt; new_path (path.begin (), path.end ()); Pinnacle * sused = (* it) - & gt; getOposite (last_nodeof_path); Pair and lieutenant; Header *, Int & gt; Equal (sused, (* this) -> gt = dist + new_path [new_path.size () - 1] .second); new_path.push_back (equals); q.push (new_path); }}} Cout & lt; & Lt; "No Path"; }    
 
 
 Try adding a  brake;  If your  if (last_nodeof_path == target)  block, thus you print the answer first and end the execution of the main loop.

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