leetcode 332 Reconstruct Itinerary

题目连接

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

Reconstruct Itinerary

Description

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.

题目大意: 从"JFK"出发找出一个序列使得所有的tickets用完,注意所求的结果要求字典序最小.
思路: 深搜,注意要标记已经访问过的节点(回溯时要记得节点还原)。
PS: 用完$n$张票当然需要用$n+1$个点

class Solution {
public:
    vector<string> findItinerary(vector<pair<string, string>> tickets) {
        vector<string> res;
        if(tickets.empty()) return res;
        for(auto &r: tickets) mp[r.first].insert({ r.second, false });
        res.push_back("JFK");
        dfs("JFK", res, tickets.size());
        return res;
    }
    bool dfs(string from, vector<string> &res, int n) {
        if((int)res.size() == n + 1) return true;
        for(auto r = mp[from].begin(); r != mp[from].end(); ++r) {
            string temp = (*r).first;
            auto c = mp[from].find({ temp, false });
            if(c == mp[from].end()) continue;
            res.push_back(temp);
            mp[from].erase(c);
            mp[from].insert({ temp, true });
            if(dfs(temp, res, n)) return true;
            res.pop_back();
            c = mp[from].find({ temp, true });
            if(c == mp[from].end()) continue;
            mp[from].erase(c);
            mp[from].insert({ temp, false });
        }
        return false;
    }
private:
    unordered_map<string, multiset<pair<string, bool>>> mp;
};

转载于:https://www.cnblogs.com/GadyPu/p/5612534.html