Leetcode---word ladder

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

1. Only one letter can be changed at a time
2. Each intermediate word must exist in the dictionary

For example,

Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.

Have you been asked this question in an interview?

直观反应是图，所有词（包含start和end）组成顶点，“相似”的两个词之间有边，边没有权重。所以这是一个无权无向图。
这道题是求最短路径的，所以使用BFS，code如下：

int ladderLength(string start, string end, unordered_set<string> &dict) {
        if(0==lookslike(start,end)){
                return 0;
        }
        if(1==lookslike(start,end)){
                return 2;
        }
        dict.erase(start);
        dict.erase(end);
        queue< pair<string,int> > q;
        q.push(make_pair(start, 1));
        unordered_map<string,int>::iterator iter;
        while(!q.empty()){
            pair<string,int> p = q.front();
            q.pop();
            string first=get<0>(p);
            int length=get<1>(p);
            if(1==lookslike(first,end)){
                return length+1;
            }
            else{
                string tmp=first;
                for(int i=0;i<tmp.length();i++){
                    for(char c='a';c<='z';c++){
                        //tmp[i]=c;
                        tmp.replace(i,1,1,c);
                        if(dict.find(tmp)!=dict.end()){
                            q.push(make_pair(tmp, length+1));
                            dict.erase(tmp);
                        }
                    }
                    tmp=first;
                }
            }
        }
        return 0;
    }
    int lookslike(string s1, string s2){
        int diff = 0;
        for (int i = 0; i < s1.length(); i++) {
            if (s1[i] != s2[i]) {
                diff++;
                if (diff >= 2)
                    return 2;
            }
        }
        return diff;
    }