Leetcode--word Ladder II

Changed for 3 nights finally no bug ... The result timed out ... I can't figure out how to optimize .... Read the answer, is a layer of traversal, and then push the parent of each word into a vector, I am also a layer, and if the next layer encountered a word, the record word is visited, will not be visited ... Just wondering why it takes a lot more time ... Is it because my code is not concise?!!

classSolution {//with breadth search, a search tree starting from Start is created, and the child node is the parent node that changes only one letter to all the words found in the dictionary .//each child node can record its parent node and its minimum number of layers, guaranteeing the shortest path//after the tree is established, the tree is traversed, and if the node can be converted to end, the path to start is recorded and added to the sequencesPrivate: Vector<vector<string> > sequences;//Record Results Public: Vector<vector<string>> Findladders (stringStartstringEnd, unordered_set<string> &dict) {Unordered_map<string, pair<queue<string,int> > Father;//Record parent node and number of layersunordered_set<string> visited;//record has been visitedvector<string>sequence; Queue<string>que; intShortest (Int_max);//The shortest sequence to end        if(Start.size ()! = end.size () | | start.size () <1)returnsequences; intSize =start.size (); if(Size = =1) {sequence.push_back (start);            Sequence.push_back (end);            Sequences.push_back (sequence); returnsequences; }        //Access root node queue, record parent node extremely layerQue.push (start); Queue<string> FQ; Fq.push (""); Father[start]= Make_pair (FQ,1);  while(!Que.empty ()) {            //out Team            stringWord =Que.front ();            Que.pop (); //Change the alphabet             for(inti =0; i < size; i++){                 for(Charc ='a'; C <='Z'; C++){                    if(c = = Word[i])Continue; stringNewword =Word; Newword= Newword.replace (I,1,1, c); //If it is end, it is the shortest path if it appears for the first time//because it is a layer-by-layer search, the first search to end is the shortest//if the shortest path, record the path, add sequences                    if(Newword = =end) {                        if(Shortest = = Int_max) Shortest = Father[word].second +1; intCurlayer = Father[word].second +1; if(Curlayer =shortest) addsequence (father, start, end, word); }                    //if it is in the dictionary and has not been visited                    if(Dict.find (newword)! = Dict.end () && visited.find (newword) = =Visited.end ()) {                        //queue, record its layer and its parent node                        intLayer =Father[word].second; //If the Newword does not have a parent node, create                        if(Father.find (newword) = =Father.end ())                            {Que.push (Newword); Queue<string>FQ;                            Fq.push (word); Father[newword]= Make_pair (FQ, Layer +1); }//if Newword already has a parent node and the number of layers in Newword is 1 more than the number of layers in word now, increase its parent node                        Else    if(Layer <= (Father[newword].second-1) {Que.push (Newword); Auto Fq=Father[newword].first;                            Fq.push (word); Father[newword]= Make_pair (Fq,layer +1); }//If the Newword has a parent node and the current number of layers is greater than its minimum number of layers, it is marked as accessed                        Else{Visited.insert (Newword); }                                            }                }            }        }        returnsequences; } InlinevoidAddsequence (unordered_map<string, pair<queue<string,int> > &father,stringStartstringEndstringword) {Vector<string>sequence; if(Word! =end) Sequence.push_back (word);        Sequence.push_back (end); Auto Wordfather=Father[word].first; Auto Fatherword=Wordfather.front (); Auto N=Father[word].second; if(Wordfather.size () >1) {wordfather.pop (); Father[word]=Make_pair (Wordfather, N); } Word=Fatherword;  while(Word! = Start && Word! ="") {Sequence.insert (Sequence.begin (), word); Wordfather=Father[word].first; Auto Fatherword=Wordfather.front (); Auto N=Father[word].second; if(Wordfather.size () >1) {wordfather.pop (); Father[word]=Make_pair (Wordfather, N); } Word=Fatherword; }        if(sequence[0] !=start) Sequence.insert (Sequence.begin (), start);    Sequences.push_back (sequence); }};

Leetcode--word Ladder II