leecode 解题总结：332. Reconstruct Itinerary

#include <iostream>
#include <stdio.h>
#include <vector>
#include <string>
#include <unordered_map>
#include <set>
#include <algorithm>
#include <sstream>
#include <queue>
using namespace std;
/*
问题:
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.分析:题目给定了多个[from,to]的二元组，现在希望能够重新构建原来的航线，
起点都是"JFK"，并且希望构建的航线在字典序上是最小的
本质上就是当前的二元组中的to是下一个而元组的from。
然后不断查找。查找可以用哈希建立映射
<from ,to>，但是一个from可能对应多个to，因此建立的映射是
<from , [to1 , to2, ...]>
另外一个问题是，可能存在环，from指向to之后，to可能指向from
而且应该是递归来做，否则不能处理，
递归来做后，获得每一条最终路线
每生成一段路线，需要将一个二元组删除，在对应建立的映射中删除该映射即可。输入:
4(票数)
MUC LHR
JFK MUC
SFO SJC
LHR SFO
5
JFK SFO
JFK ATL
SFO ATL
ATL JFK
ATL SFO
3
JFK KUL
JFK NRT
NRT JFK
输出:
JFK MUC LHR SFO SJC
JFK ATL JFK SFO ATL SFO
JFK NRT JFK KUL报错:
Input:[["JFK","KUL"],["JFK","NRT"],["NRT","JFK"]]
Output:["JFK","KUL"]
Expected:["JFK","NRT","JFK","KUL"]
也就是说优先拼出较长的路线，然后在较长的路线中选择字典序较小的
这样就需要初始的时候递归处理，
然后对于多个结果比较报错:
Input:
[["EZE","AXA"],["TIA","ANU"],["ANU","JFK"],["JFK","ANU"],["ANU","EZE"],["TIA","ANU"],["AXA","TIA"],["TIA","JFK"],["ANU","TIA"],["JFK","TIA"]]
Output:
["JFK","ANU","EZE","AXA","TIA","ANU","JFK","TIA","JFK"]
Expected:
["JFK","ANU","EZE","AXA","TIA","ANU","JFK","TIA","ANU","TIA","JFK"]
存在重复的路线，我只算了一次。这是set本身自带的去重属性导致的，
如果不想去重，采用优先级队列或者multiset关键:
1
参考别人的解法：http://www.cnblogs.com/zmyvszk/p/5657056.html
题目意思理解错了，是要遍历完所有的边，不允许剩余。
java中优先级队列的poll是移除队头并返回。本题需要设定
unordered_map<string , priority<string>>的数据结构
注意递归存储的结果需要逆序2注意C++优先级队列默认大顶堆，输出最大元素，改为小顶堆用greater<int> (例如整型)
或者用仿函数自定义比较函数
priority_queue<Type , Container , Functional>
*/struct Compare
{bool operator()(string& a , string& b ){return a > b;}
};class Solution {
public:void dfs(unordered_map<string , priority_queue<string , vector<string> , Compare> >& airlineMap , string& from , vector<string>& result){//不断获取剩余遍历的地方，然后删除优先级队列中最小的元素，递归处理，保存结果while( airlineMap.find(from) != airlineMap.end() && (!airlineMap[from].empty()) ){string to = airlineMap[from].top();airlineMap[from].pop();dfs(airlineMap , to , result);}result.push_back(from);}vector<string> findItinerary(vector<pair<string, string>> tickets) {vector<string> result;if(tickets.empty()){return result;}unordered_map<string , priority_queue<string , vector<string>  ,Compare > > airlineMap;int size = tickets.size();//生成 起点到终点的映射for(int i = 0 ; i < size ; i++){pair<string , string> ticket = tickets.at(i);if(airlineMap.find(ticket.first) != airlineMap.end()){airlineMap[ticket.first].push(ticket.second);}else{priority_queue<string , vector<string>  , Compare> toPlaces;toPlaces.push(ticket.second);airlineMap[ticket.first] = toPlaces;}}//递归处理string sFrom("JFK");dfs(airlineMap , sFrom , result );//保存的结果需要逆置reverse(result.begin() , result.end());return result;}
};void print(vector<string>& result)
{if(result.empty()){cout << "no result" << endl;return;}int size = result.size();for(int i = 0 ; i < size ; i++){cout << result.at(i) << " " ;}cout << endl;
}void process()
{vector<pair<string , string> > tickets;int num;Solution solution;vector<string> result;while(cin >> num ){tickets.clear();for(int i = 0 ; i < num ; i++){pair<string , string> ticket;cin >> ticket.first >> ticket.second;tickets.push_back(ticket);}result = solution.findItinerary(tickets);print(result);}
}int main(int argc , char* argv[])
{process();getchar();return 0;
}/*
class Result
{
public:Result(){}Result(string& result , vector<string>& results){_result = result;_results = vector<string>(results.begin() ,results.end());}bool operator < (const Result& result) const {int value = strcmp(_result.c_str() , result._result.c_str());// value < 0 ,说明 str1 < str2if(value < 0 ){return true;}else{return false;}}
public:string _result;vector<string> _results;
};class Solution {
public:void dfs(unordered_map<string , set<string> >& airlineMap , string& from , vector<string>& result, vector< vector<string> >& results){//如果当前位置找不到下一个结束位置，返回结果if(airlineMap.find(from) == airlineMap.end()){results.push_back(result);return;}set<string> toPlaces= airlineMap[from];for(set<string>::iterator it = toPlaces.begin() ; it != toPlaces.end() ; it++){string to = *it;//将当前结果加入到结果集result.push_back(to);//删除当前选中的<from, to>if(1 == toPlaces.size()){airlineMap.erase(from);}else{airlineMap[from].erase(to);}dfs(airlineMap , to , result , results);//插入回溯result.pop_back();//删除回溯if(airlineMap.find(from) != airlineMap.end()){airlineMap[from].insert(to);}else{set<string> places;places.insert(to);airlineMap[from] = places; }}}vector<string> findItinerary(vector<pair<string, string>> tickets) {vector<string> result;if(tickets.empty()){return result;}unordered_map<string , set<string> > airlineMap;int size = tickets.size();//生成 起点到终点的映射for(int i = 0 ; i < size ; i++){pair<string , string> ticket = tickets.at(i);if(airlineMap.find(ticket.first) != airlineMap.end()){airlineMap[ticket.first].insert(ticket.second);}else{set<string> toPlaces;toPlaces.insert(ticket.second);airlineMap[ticket.first] = toPlaces;}}//递归处理string sFrom("JFK");vector< vector<string> > results;result.push_back(sFrom);dfs(airlineMap , sFrom , result , results);size = results.size();int maxLen = INT_MIN;vector< Result > candidateResults;for(int i = 0 ; i < size ; i++){maxLen = max(maxLen , (int)results.at(i).size());}//如果最大长度等于结果长度，说明for(int i = 0 ; i < size ; i++){if(maxLen == results.at(i).size()){stringstream stream;for(int j = 0 ; j < maxLen ; j++){stream << results.at(i).at(j);}string temp = stream.str();Result result(temp , results.at(i));candidateResults.push_back(result);}}//排序sort(candidateResults.begin() , candidateResults.end());result = candidateResults.at(0)._results;if(result.size() > 1){return result;}else{return vector<string>();}}
};
*/

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