[Leetcode] the shortest conversion distance of word ladder string (Dijkstra)

The shortest distance is required. Use Dijkstra to find the shortest path between nodes.

Note: If the two elements in the enumerated dictionary can be converted, the time-out will occur.

Improvement: For each string, to enumerate the values of each character, it is necessary to enumerate N x 26 times for a string with a length of N.

If it is just a simple enumeration, the duplicate edge will appear:

For example, ABC, BBC, and CBC. After a graph is created, each two nodes have two bidirectional edges. Therefore, there are many redundant edges in the graph stored in the adjacent table.

Solution: each character on each node can only be enumerated after the original character, that is

Enumerate the first character of each string

ABC: BBC, CBC, DBC ,...

BBC: CBC, DBC ,...

CBC: DBC ,....

struct Edge{    int v,next;};class Solution {public:Edge*e;    int *head,len,n;    void addedge(int u,int v){        e[len].v=v;        e[len].next=head[u];        head[u]=len++;        e[len].v=u;        e[len].next=head[v];        head[v]=len++;    }    bool canTrans(const string& p,const string&q){        int i,cnt=0;        for(i=0;i<p.size();++i){            if(p[i]!=q[i]){                cnt++;                if(cnt>1)return 0;            }        }        return 1;    }    int dijk(int st,int ed){        int* dist=new int[n],i,v,j,k;        memset(dist,127,sizeof(int)*n);int unre=dist[0];        for(i=head[st];i!=-1;i=e[i].next){            v=e[i].v;            dist[v]=1;        }        dist[st]=-1;        for(j=1;j<n;++j){            for(i=0,k=-1;i<n;++i)                if(dist[i]>0&&dist[i]!=unre&&(k<0||dist[i]<dist[k]))                    k=i;            if(k<0||k==ed)break;            for(i=head[k];i!=-1;i=e[i].next){                v=e[i].v;                if(dist[v]>=0&&dist[v]>dist[k]+1)                    dist[v]=dist[k]+1;            }            dist[k]=-1;        }        if(k==ed)            k=dist[k];        else k=-1;        delete[]dist;        return k;    }    int ladderLength(string start, string end, unordered_set<string> &dict) {        if(start==end)return 2;        map<string,int>mp;        int cnt=0,i;        mp[start]=cnt++;        mp[end]=cnt++;        unordered_set<string>::iterator bg=dict.begin(),ed=dict.end(),bg2;        for(;bg!=ed;bg++){            if(mp.find(*bg)==mp.end())                mp[*bg]=cnt++;        }        dict.insert(start);        dict.insert(end);n=dict.size();        e=new Edge[n*n];        head=new int[n];len=0;memset(head,-1,sizeof(int)*n);int ch,j;        for(bg=dict.begin(),ed=dict.end();bg!=ed;bg++){            string s=*bg;            for(i=0;i<s.length();++i){                ch=s[i]-'a';                for(j=ch+1;j<26;++j){                    s[i]='a'+j;                    if(dict.find(s)!=dict.end())                        addedge(mp[s],mp[*bg]);                }                s[i]='a'+ch;            }     /*       for(bg2=bg,bg2++;bg2!=ed;bg2++){                if(canTrans(*bg,*bg2)){                    addedge(mp[*bg],mp[*bg2]);                }            }*/        }        i=dijk(mp[start],mp[end]);delete[] e;delete[]head;        return i>=0?i+1:0;    }};

[Leetcode] the shortest conversion distance of word ladder string (Dijkstra)