最小操作数

给定一个单词集合Dict，其中每个单词的长度都相同。现从此单词集合Dict中抽取两个单词A、B，我们希望通过若干次操作把单词A变成单词B，每次操作可以改变单词的一个字母，同时，新产生的单词必须是在给定的单词集合Dict中。求所有行得通步数最少的修改方法。

举个例子如下：

Given:
A = "hit"
B = "cog"
Dict = ["hot","dot","dog","lot","log"]
Return
[
["hit","hot","dot","dog","cog"],
["hit","hot","lot","log","cog"]
]

即把字符串A = "hit"转变成字符串B = "cog"，有以下两种可能：

"hit" -> "hot" -> "dot" -> "dog" -> "cog"；

"hit" -> "hot" -> "lot" -> "log" ->"cog"。

class Solution {
public:
	vector > findLadders(string start, string end,
			set& dict) {
		vector > result, result_temp;
		if (dict.erase(start) == 1 && dict.erase(end) == 1) {
//			为当前方向上所有点与其下一结点的配对
			map > kids_from_start;
			map > kids_from_end;

			//开始方向经过的点集
			set reach_start;
			reach_start.insert(start);
			//终点方向经过的点集
			set reach_end;
			reach_end.insert(end);

			set meet;
			//BFS直到相交
			while (meet.empty() && !reach_start.empty() && !reach_end.empty()) {
				if (reach_start.size() < reach_end.size()) {
					//从开始方向搜索子结点和其分支
					search_next_reach(reach_start, reach_end, meet,
							kids_from_start, dict);
					/*cout << "start ";
					 printMap(kids_from_start);*/
				} else {
					search_next_reach(reach_end, reach_start, meet,
							kids_from_end, dict);
					/*	cout << "end ";
					 printMap(kids_from_end);*/
				}
//				printset(meet);
			}

			if (!meet.empty()) {
				//由相遇点拓展可走路径数
				for (set::iterator it = meet.begin(); it != meet.end();
						++it) {
					//参数1是空间,参数2是值
					vector words(1, *it);
					result.push_back(words);
				}

				//由相遇点向起点拓展分支
				walk(result, kids_from_start);
				/*printvv(result);
				 printMap(kids_from_start);*/
				for (size_t i = 0; i < result.size(); ++i) {
					reverse(result[i].begin(), result[i].end());
				}
				walk(result, kids_from_end);
				/*printvv(result);
				 printMap(kids_from_end);*/
			}
		}

		return result;
	}

private:
	/*reach为该方向已经经过的点集,
	 * other_reach为反方向,
	 * meet为两个方向点集相交点集,
	 * path为当前方向上所有点与其下一结点的配对,(下一元素:{当前元素})
	 * dict为字典集合*/
	void search_next_reach(set& reach, const set& other_reach,
			set& meet, map >& path,
			set& dict) {
		//reach必须从端点开始遍历,所以必须重新添加,不然会重复
		set temp;
		reach.swap(temp);

		/*对该方向已出现的元素匹配下一元素
		 * 匹配方案：
		 * 尝试依次改变指定位置上的元素来匹配*/
		for (set::iterator it = temp.begin(); it != temp.end(); ++it) {
			string s = *it;
			for (size_t i = 0; i < s.length(); ++i) {
				char back = s[i];
				for (s[i] = 'a'; s[i] <= 'z'; ++s[i]) {
					if (s[i] != back) {
						//已出现在当前点集
						if (reach.count(s) == 1) {
							path[s].push_back(*it);
						} else if (dict.erase(s) == 1) {	//存在于字典中,删除后加入点集
							path[s].push_back(*it);
							reach.insert(s);
						} else if (other_reach.count(s) == 1) {	//存在于相反方向集合,说明点集相交
							path[s].push_back(*it);
							reach.insert(s);
							meet.insert(s);
						}
					}
				}
				s[i] = back;
			}
		}
	}

	void walk(vector >& all_path,
			map > kids) {
		vector > temp;
		//到达起止点则为空
		while (!kids[all_path.back().back()].empty()) {
			//每个路径的结尾元素可以对应多个子元素,则拼接路径有多条,所以必须先清空再添加
			all_path.swap(temp);
			all_path.clear();
			for (vector >::iterator it = temp.begin();
					it != temp.end(); ++it) {
				vector& one_path = *it;
				//取得当前路径尾元素的下一元素集合
				vector& p = kids[one_path.back()];
				//将下一元素集合的每一元素与尾元素分别拼接，存为新路径
				for (size_t i = 0; i < p.size(); ++i) {
					all_path.push_back(one_path);
					all_path.back().push_back(p[i]);
				}
			}
		}
	}
};