leetcode 332 重新安排行程

本是古典何须时尚 2022-08-28 10:45 142阅读 0赞

## 前言 ##

题目：[332. 重新安排行程][332.]

参考题解：[重新安排行程-代码随想录][-]

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## 提交代码 ##

题目要求：给定一系列pair，一笔将其全部串联起来。

思路一：一个点，有不同的到达点。要选出其中最小可达。我先统计出所有可达路径，再选择最小可达。

代码实现如下，通过为11/80。我将超时的测试用例作为输入，打印生成的可达路径。思路和代码实现上可能没有问题，可能是测试用例比较复杂。

#include <iostream>
    #include <vector>
    #include <map>
    
    using namespace std;
    
    class Solution {
    public:
        void backTraing(vector<vector<string>>& tickets, map<string,vector<pair<string,bool>>>& tickets_map, string start ,vector<string>& path, vector<vector<string>>& all_path){
            // 每张票只能用一次，当所有票串联使用一次时候，为一条航线
            // 节点 = 边 + 1
            if(path.size() == tickets.size()+1){
                // for(auto str : path)
                //     cout<<str<<" ";
                cout<<endl;
                all_path.push_back(path);
                return;
            }
    
            vector<pair<string,bool>>& ends = tickets_map[start];
            int size = ends.size();
            for(int i=0; i<size; i++){
                if(ends[i].second) // 已经使用过的票，不可再用
                    continue;
                ends[i].second = true;
                path.push_back(ends[i].first); // 新的票加入路径
                backTraing(tickets,tickets_map,ends[i].first,path,all_path); // 使用当前票的目标点，作为新的起点
                path.pop_back();
                ends[i].second = false;
            }
        }
    
        vector<string> findItinerary(vector<vector<string>>& tickets) {
            // 数据预处理，将数据处理为map<start,vector<pair<end,uesd>>> -- 起点，重点，是否使用
            map<string,vector<pair<string,bool>>> tickets_map;
            for(auto tickets : tickets){
                string start = tickets[0];
                string end = tickets[1];
                tickets_map[start].push_back(make_pair(end,false));
            }
    
            // 回溯遍历，将所有票可一次串联成功的路径，放入all_path中
            string start("JFK");
            vector<string> path; 
            vector<vector<string>> all_path;
            path.push_back(start);
            backTraing(tickets,tickets_map,start,path,all_path);
    
            // cout<<"<<<<<<<<<<<<<<<<<<<<<<"<<endl;
            // for(auto path : all_path){
            //     for(auto str : path)
            //         cout<<str<<" ";
            //     cout<<endl;
            // }
            // cout<<">>>>>>>>>>>>>>>>>>>>>>"<<endl;
    
            // 返回字典序排序最小的组合
            // 注意：以字典序比较两个字符串在C++20中被移除
            vector<string>& min_result = all_path[0]; // 题目已经限制，至少存在一条路径
            int size = all_path.size();
            for(int i=1; i<size; i++){
                vector<string>& now_path = all_path[i];
                for(int j=0; j<tickets.size(); j++){
                    if(now_path[j] < min_result[j]){
                        min_result = now_path;
                        break;
                    }else if(now_path[j] > min_result[j])
                        break;
                }
            }
    
            // 返回结果
            vector<string> result = min_result;
            return result;
        }
    };
    
    int main(void){
        // vector<vector<string>> tickets = {
       {"MUC","LHR"},{"JFK","MUC"},{"SFO","SJC"},{"LHR","SFO"}};
        // vector<vector<string>> tickets = {
       {"JFK","SFO"},{"JFK","ATL"},{"SFO","ATL"},{"ATL","JFK"},{"ATL","SFO"}};
        vector<vector<string>> tickets = {
       {"AXA","EZE"},{"EZE","AUA"},{"ADL","JFK"},{"ADL","TIA"},{"AUA","AXA"},{"EZE","TIA"},{"EZE","TIA"},{"AXA","EZE"},{"EZE","ADL"},{"ANU","EZE"},{"TIA","EZE"},{"JFK","ADL"},{"AUA","JFK"},{"JFK","EZE"},{"EZE","ANU"},{"ADL","AUA"},{"ANU","AXA"},{"AXA","ADL"},{"AUA","JFK"},{"EZE","ADL"},{"ANU","TIA"},{"AUA","JFK"},{"TIA","JFK"},{"EZE","AUA"},{"AXA","EZE"},{"AUA","ANU"},{"ADL","AXA"},{"EZE","ADL"},{"AUA","ANU"},{"AXA","EZE"},{"TIA","AUA"},{"AXA","EZE"},{"AUA","SYD"},{"ADL","JFK"},{"EZE","AUA"},{"ADL","ANU"},{"AUA","TIA"},{"ADL","EZE"},{"TIA","JFK"},{"AXA","ANU"},{"JFK","AXA"},{"JFK","ADL"},{"ADL","EZE"},{"AXA","TIA"},{"JFK","AUA"},{"ADL","EZE"},{"JFK","ADL"},{"ADL","AXA"},{"TIA","AUA"},{"AXA","JFK"},{"ADL","AUA"},{"TIA","JFK"},{"JFK","ADL"},{"JFK","ADL"},{"ANU","AXA"},{"TIA","AXA"},{"EZE","JFK"},{"EZE","AXA"},{"ADL","TIA"},{"JFK","AUA"},{"TIA","EZE"},{"EZE","ADL"},{"JFK","ANU"},{"TIA","AUA"},{"EZE","ADL"},{"ADL","JFK"},{"ANU","AXA"},{"AUA","AXA"},{"ANU","EZE"},{"ADL","AXA"},{"ANU","AXA"},{"TIA","ADL"},{"JFK","ADL"},{"JFK","TIA"},{"AUA","ADL"},{"AUA","TIA"},{"TIA","JFK"},{"EZE","JFK"},{"AUA","ADL"},{"ADL","AUA"},{"EZE","ANU"},{"ADL","ANU"},{"AUA","AXA"},{"AXA","TIA"},{"AXA","TIA"},{"ADL","AXA"},{"EZE","AXA"},{"AXA","JFK"},{"JFK","AUA"},{"ANU","ADL"},{"AXA","TIA"},{"ANU","AUA"},{"JFK","EZE"},{"AXA","ADL"},{"TIA","EZE"},{"JFK","AXA"},{"AXA","ADL"},{"EZE","AUA"},{"AXA","ANU"},{"ADL","EZE"},{"AUA","EZE"}};
        Solution s;
        vector<string> result = s.findItinerary(tickets);
        for(auto str : result)
            cout<<str<<" ";
    }

思路二：在思路一的基础上，能否优先选择哪些可能是最小可达的路径呢？

参考题解中，选择的数据结构，比我上面使用的数据结构要好。它使用`unordered_map`,建立一对多，默认有序的映射。这样找到的第一条路径，即为最小可达路径。

我没有重写一遍，下面代码来自上面参考题解。

class Solution { 
    private:
    // unordered_map<出发机场, map<到达机场, 航班次数>> targets
    unordered_map<string, map<string, int>> targets;
    bool backtracking(int ticketNum, vector<string>& result) { 
        if (result.size() == ticketNum + 1) { 
            return true;
        }
        for (pair<const string, int>& target : targets[result[result.size() - 1]]) { 
            if (target.second > 0 ) {  // 记录到达机场是否飞过了
                result.push_back(target.first);
                target.second--;
                if (backtracking(ticketNum, result)) return true;
                result.pop_back();
                target.second++;
            }
        }
        return false;
    }
    public:
        vector<string> findItinerary(vector<vector<string>>& tickets) { 
            targets.clear();
            vector<string> result;
            for (const vector<string>& vec : tickets) { 
                targets[vec[0]][vec[1]]++; // 记录映射关系
            }
            result.push_back("JFK"); // 起始机场
            backtracking(tickets.size(), result);
            return result;
        }
    };

[332.]: https://leetcode-cn.com/problems/reconstruct-itinerary/
[-]: https://github.com/youngyangyang04/leetcode-master/blob/master/problems/0332.%E9%87%8D%E6%96%B0%E5%AE%89%E6%8E%92%E8%A1%8C%E7%A8%8B.md