LeetCode332. Reconstruct Itinerary

题目：

https://leetcode.com/problems/reconstruct-itinerary/

Given a list of airline tickets represented by pairs of departure and arrival airports [from, to], reconstruct the itinerary in order. All of the tickets belong to a man who departs from JFK. Thus, the itinerary must begin with JFK.

Note:
If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string. For example, the itinerary [“JFK”, “LGA”] has a smaller lexical order than [“JFK”, “LGB”].
All airports are represented by three capital letters (IATA code).
You may assume all tickets form at least one valid itinerary.

Example 1:
tickets = [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Return ["JFK", "MUC", "LHR", "SFO", "SJC"].

Example 2:
tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Return ["JFK","ATL","JFK","SFO","ATL","SFO"].
Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"]. But it is larger in lexical order.

思路：

题意：给一些机票，出发机场和到达机场表示方式[from,to]，用完这些机票，并且行程必须从JFK开始。
注意：如果存在多条有效路径，选择字典序最小的那个。

感谢：http://blog.youkuaiyun.com/murmured/article/details/50641289

诶，STL还是不咋会用啊，题还是简单。
建图的时候，顺便按字典序排好。
然后开始不断的dfs选地点加入到栈里面，进入到下一层里面，根据之前选的地点，找与它相连的点，判断是否能加入到结果栈中，如果能又继续递归往下找，如果最后不符合，回溯还原处理。

代码：

bool dfs(vector<string>& res,string curPos,int n,unordered_map<string,map<string,int>>& m){
    if (res.size() == n){
        return true;
    }
    for (auto ticket = m[curPos].begin(); ticket != m[curPos].end();ticket++){
        if (ticket->second != 0){
            res.push_back(ticket->first);
            ticket->second--;
            if (dfs(res, ticket->first, n, m)){
                return true;
            }
            res.pop_back();
            ticket->second++;
        }
    }
    return false;
}
vector<string> findItinerary(vector<pair<string, string>> tickets) {
    unordered_map<string, map<string, int>> m;
    vector<string> res;
    int len = tickets.size();
    if (len == 0){
        return res;
    }
    for (int i = 0; i <len; i++){
        m[tickets[i].first][tickets[i].second]++;
    }
    string start = "JFK";
    res.push_back(start);
    dfs(res,start,len+1,m);
    return res;
}