Leetcode -- 332. Reconstruct Itinerary --优先级队列 / 图的后序遍历

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本文介绍了一种基于深度优先搜索(DFS)的后序遍历算法，用于解决从JFK机场出发的机票行程重构问题。通过使用优先级队列维护字典序，确保了输出行程的最小字典序。算法中特别注意了每条边的遍历，避免了重复路径，最终输出倒序的行程列表。

332. 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:
Input: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]
Example 2:
Input: [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Output: ["JFK","ATL","JFK","SFO","ATL","SFO"]
Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"].
             But it is larger in lexical order.

使用后序遍历，最后倒序输出，每条边都要遍历一次，使用优先级队列来维持字典序，注意priority_queue默认为less-->大顶堆

所以我们需要定义成greater

不能用set，应为有可能有重复的两条路径！

class Solution {
public:
	unordered_map<string,priority_queue<string,vector<string>,greater<string>>> m;
	void dfs(string cur, vector<string>&res)
	{
		auto&s = m[cur];
		while(!s.empty()){
			string des = s.top();
			s.pop();
			
			dfs(des,res);
		}
		res.push_back(cur);
	}
	vector<string> findItinerary(vector<vector<string>>& tickets) {
		
		unordered_map<string, bool> visited;
		for (int i = 0; i < tickets.size(); i++)
		{
			m[tickets[i][0]].push(tickets[i][1]);
		}
		string start = "JFK";
		vector<string>res;
		dfs(start,res);
		return { res.rbegin(),res.rend() };
	}
};