leetcode 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:
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.

// ConsoleApplication6.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include<string>#include<vector>#include<map>#include<set>#include<algorithm>#include<iostream>using namespace std;class Solution {public:vector<string> findItinerary(vector<pair<string, string>> tickets) {struct compare{vector<pair<string, string>>vv;compare(vector<pair<string, string>>uu){ vv = uu; }bool operator()(pair<string, string > ds1, pair<string, string>ds2){for (int i = 0; i < 3; i++){if (ds1.first[i] - ds2.first[i] < 0)return true;else if (ds1.first[i] - ds2.first[i] > 0)return false;}for (int i = 0; i < 3; i++){if (ds1.second[i] - ds2.second[i] < 0)return true;else if (ds1.second[i] - ds2.second[i] > 0)return false;}return false;}};std::sort(tickets.begin(), tickets.end(), compare(tickets));map<string, set<string>>pp, ppp;map<pair<string, string>, int>count, count2;for (int i = 0; i < tickets.size(); i++){pp[tickets[i].first].insert(tickets[i].second);count[tickets[i]]++;}vector<string>re;set<vector<string>>aa;for (int i = 0; i < tickets.size(); i++){if (i>0 && tickets[i - 1] == tickets[i])continue;aa.clear();re.clear();count2 = count;ppp = pp;re.push_back(tickets[i].first);re.push_back(tickets[i].second);count2[tickets[i]]--;if (count2[tickets[i]] == 0)ppp[tickets[i].first].erase(ppp[tickets[i].first].find(tickets[i].second));while (true){while (ppp[re.back()].size() > 0){string str = re.back();re.push_back(*ppp[str].begin());count2[pair<string, string>(str, re.back())]--;if (count2[pair<string, string>(str, re.back())] == 0)ppp[str].erase(ppp[str].begin());if (re.size() == tickets.size() + 1)return re;}int kk = re.size() - 2;bool f = false;while (!f&&kk >= 1){while (kk >= 1 && ppp[re[kk]].size() == 0){kk--;}if (kk == 0)break;for (int j = kk; j < re.size() - 1; j++){ppp[re[j]].insert(re[j + 1]);count2[pair<string, string>(re[j], re[j + 1])]++;}aa.insert(vector<string>(re.begin(), re.begin() + kk + 2));re.erase(re.begin() + kk + 1, re.end());for (set<string>::iterator it = ppp[re.back()].begin(); it != ppp[re.back()].end(); it++){vector<string>temp = re;temp.push_back(*it);if (aa.find(temp) == aa.end()){f = true;aa.insert(temp);count2[pair<string, string>(re.back(), *it)]--;if (count2[pair<string, string>(re.back(), *it)] == 0)ppp[re.back()].erase(it);re = temp;break;}}if (!f)kk--;}if (kk == 0)break;}}}};int _tmain(int argc, _TCHAR* argv[]){Solution sl;vector<pair<string, string>>tickets;//tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]/*[["EZE","TIA"],["EZE","HBA"],["AXA","TIA"],["JFK","AXA"],["ANU","JFK"],["ADL","ANU"],["TIA","AUA"],["ANU","AUA"],["ADL","EZE"],["ADL","EZE"],["EZE","ADL"],["AXA","EZE"],["AUA","AXA"],["JFK","AXA"],["AXA","AUA"],["AUA","ADL"],["ANU","EZE"],["TIA","ADL"],["EZE","ANU"],["AUA","ANU"]]*/tickets.push_back(pair<string, string>("EZE", "TIA"));tickets.push_back(pair<string, string>("EZE", "HBA"));tickets.push_back(pair<string, string>("AXA", "TIA"));tickets.push_back(pair<string, string>("JFK", "AXA"));tickets.push_back(pair<string, string>("ANU", "JFK"));tickets.push_back(pair<string, string>("ADL", "ANU"));tickets.push_back(pair<string, string>("TIA", "AUA"));tickets.push_back(pair<string, string>("ANU", "AUA"));tickets.push_back(pair<string, string>("ADL", "EZE"));tickets.push_back(pair<string, string>("ADL", "EZE"));tickets.push_back(pair<string, string>("EZE", "ADL"));tickets.push_back(pair<string, string>("AXA", "EZE"));tickets.push_back(pair<string, string>("AUA", "AXA"));tickets.push_back(pair<string, string>("JFK", "AXA"));tickets.push_back(pair<string, string>("AXA", "AUA"));tickets.push_back(pair<string, string>("AUA", "ADL"));tickets.push_back(pair<string, string>("ANU", "EZE"));tickets.push_back(pair<string, string>("TIA", "ADL"));tickets.push_back(pair<string, string>("EZE", "ANU"));tickets.push_back(pair<string, string>("AUA", "ANU"));vector<string>re = sl.findItinerary(tickets);system("pause");return 0;}

妈的，超时

过了大概一个月重新看这道题，发现题意没审清楚，JFK要打头，这就简单很多了嘛。然后修改了下

class Solution {public:vector<string> findItinerary(vector<pair<string, string>> tickets) {clock_t start, finish;double Total_time;start = clock();struct compare{vector<pair<string, string>>vv;compare(vector<pair<string, string>>uu){ vv = uu; }bool operator()(pair<string, string > ds1, pair<string, string>ds2){for (int i = 0; i < 3; i++){if (ds1.first[i] - ds2.first[i] < 0)return true;else if (ds1.first[i] - ds2.first[i] > 0)return false;}for (int i = 0; i < 3; i++){if (ds1.second[i] - ds2.second[i] < 0)return true;else if (ds1.second[i] - ds2.second[i] > 0)return false;}return false;}};std::sort(tickets.begin(), tickets.end(), compare(tickets));map<string, set<string>> ppp;map<pair<string, string>, int>count;for (int i = 0; i < tickets.size(); i++){ppp[tickets[i].first].insert(tickets[i].second);count[tickets[i]]++;}vector<string>re;set<vector<string>>aa;re.push_back("JFK");while (true){while (ppp[re.back()].size() > 0){string str = re.back();re.push_back(*ppp[str].begin());count[pair<string, string>(str, re.back())]--;if (count[pair<string, string>(str, re.back())] == 0)ppp[str].erase(ppp[str].begin());if (re.size() == tickets.size() + 1)return re;}int kk = re.size() - 2;if (kk == 0){string s = re[1];re.pop_back();re.push_back(*ppp[re[0]].begin());count[pair<string, string>(re[0], re.back())]--;if (count[pair<string, string>(re[0], re.back())] == 0)ppp[re[0]].erase(ppp[re[0]].begin());ppp[re[0]].insert(s);count[pair<string, string>(re[0], s)]++;}bool f = false;while (!f&&kk >= 1){while (kk >= 1 && ppp[re[kk]].size() == 0){kk--;}if (kk == 0){for (int i = re.size() - 1; i > 1; i--){string s = re.back();re.pop_back();ppp[re.back()].insert(s);count[pair<string, string>(re.back(), s)]++;}string s = re[1];re.pop_back();re.push_back(*ppp[re[0]].begin());count[pair<string, string>(re[0], re.back())]--;if (count[pair<string, string>(re[0], re.back())] == 0)ppp[re[0]].erase(ppp[re[0]].begin());ppp[re[0]].insert(s);count[pair<string, string>(re[0], s)]++;continue;}for (int j = kk; j < re.size() - 1; j++){ppp[re[j]].insert(re[j + 1]);count[pair<string, string>(re[j], re[j + 1])]++;}aa.insert(vector<string>(re.begin(), re.begin() + kk + 2));re.erase(re.begin() + kk + 1, re.end());for (set<string>::iterator it = ppp[re.back()].begin(); it != ppp[re.back()].end(); it++){vector<string>temp = re;temp.push_back(*it);if (aa.find(temp) == aa.end()){f = true;aa.insert(temp);count[pair<string, string>(re.back(), *it)]--;if (count[pair<string, string>(re.back(), *it)] == 0)ppp[re.back()].erase(it);re = temp;break;}}if (!f)kk--;}}finish = clock();Total_time = (double)(finish - start) / CLOCKS_PER_SEC;printf("%f seconds\n", Total_time);}};

accept

不过话说现在看一个月前的代码感觉很low，不过也懒得改了