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:

1. 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"].
2. All airports are represented by three capital letters (IATA code).
3. 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.

重建行程单，在图中找一条路径，能经过所有的边。

参考：http://www.cnblogs.com/grandyang/p/5183210.html

class Solution{
public:
    vector<string> findItinerary(vector<pair<string,string>> tickets){
        unordered_map<string,multiset<string>> m;
        for(auto t : tickets){
            m[t.first].insert(t.second);
        }
        vector<string> res;
        dfs(m,"JFK",res);
        return vector<string> (res.rbegin(),res.rend());
    }

    void dfs(unordered_map<string,multiset<string>>& m,string s,vector<string>& res){
        while(m[s].size()){
            string t = *m[s].begin();
            m[s].erase(m[s].begin());
            dfs(m,t,res);
        }
        res.push_back(s);
    }
};