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我正在尝试打印地铁系统中所有可能的路径,该地铁系统具有从 A 到 L 的车站。换句话说,目标是找出一个人可以采取多少条可能的路线来通过地铁系统而无需越过轨道不止一次。我知道有 640 条可能的路径,因为我之前使用 C 中的邻接矩阵编写了这个程序。现在我正在尝试编写相同的程序,除了使用 C++ 中的类而不使用邻接矩阵。

我遇到的问题是我似乎无法正确实现我的递归函数 SearchRoute,因为我需要打印路径,标记路径,然后再次取消标记路径以允许回溯。当我打印出最终结果时,我只得到了从 A 到 B 的轨迹,这显然意味着某些地方显然是错误的。

这就是我认为的问题所在:我知道在我的 SubwaySystem::SearchRoute 函数中,我使用了 if 和 else 语句,这显然不允许调用我的递归函数,但我尝试使用另一个 if 语句而不是 else 但我'不太确定条件会是什么。

void SubwaySystem::SearchRoute(int Current_Station_ID)
{
    while(Current_Station_ID < 33)
    {
        cout << "In while loop\n";
        // \\ Checking progress
        if(Current_Station_ID == 0)  //Find a successful route to Station L
        {
            count_routes++; //Add 1 into the variable “count_routes”
            cout << "In if statement\n";
            // \\Checking progress
            cout << count_routes << " " << my_track[Current_Station_ID] << endl; //Print out this route
            return;
        }
        else //Get into recursive Function Body
        {
            for(int i = my_station[Current_Station_ID].track_starting_ID; i < my_station[Current_Station_ID].track_starting_ID + my_station[Current_Station_ID].track_size; i++)
            {
                if(my_track[Current_Station_ID].visited == 0)  //if this track is not visited before
                {
                    cout << "In recursive part of function\n";
                    // \\ Checking progress
                    my_track[Current_Station_ID].visited = 1; //mark this track as visited
                    my_track[Current_Station_ID].node_2 = 1; //mark its corresponding track as visited
                    cout << my_track[Current_Station_ID] << endl; //save this track
                    SearchRoute(Current_Station_ID + 1); //Recursive
                    i--; //Backtrack this track
                    my_track[Current_Station_ID].visited = 0;//mark this track as unvisited
                    my_track[Current_Station_ID].node_2 = 0;//mark its corresponding track as unvisited
                }
            }
        }
    }
}

我还尝试在整个程序中添加打印语句以跟踪整个程序的进度。由于我上面指定的原因,我的递归函数永远不会被调用(至少这就是我认为它不起作用的原因)。由于某种原因,我无法弄清楚为什么我的默认和重载轨道构造函数被多次调用。

如果您能帮助我找出代码中的问题区域/告诉我正确的方法,我将不胜感激。我厌倦了思考这个问题。提前致谢。

这是单个 TU 中的其余程序:

//Function Declarations
#include <iostream>
#include <string>

using namespace std;

#ifndef SUBWAY_H
#define SUBWAY_H

class Track
{
public:
    //Default Constructor
    Track();

    //Overload Constructor
    Track(char, char);

    //Destructor
    ~Track();

    //Member variables
    char node_1;
    char node_2;
    bool visited;
};

class Station
{
public:
    //Default Constructor
    Station();

    //Destructor
    ~Station();

    //Overload Constructor
    Station(char, int, int);

    //Member variables
    char station_name;
    int track_starting_ID;
    int track_size;
};

class SubwaySystem
{
public:
    //Default Constructor
    SubwaySystem();

    //Destructor
    ~SubwaySystem();

    //Recursive function
    void SearchRoute(int);

    //Other member functions
    friend ostream& operator<<(ostream& os, const Track& my_track);
    friend ostream& operator<<(ostream& os, const Station& my_station);


    //Member variables
    Track my_track[34];
    Station my_station[12];

    int count_routes;
    int Current_Station_ID;

    //String to save found route
};

#endif

// **cpp**

//Function Definitions
#include <iostream>
#include <string>

//#include "subway.h"

using namespace std;

Track::Track()
{
    visited = 0;
    //cout << "Default Track has been called\n";
    //\\ Checking progress
}


Track::~Track()
{
}


Track::Track(char pass_track1, char pass_track2)
{
    node_1 = pass_track1;
    node_2 = pass_track2;
    visited = false;
    //cout << "Overload Track constructor has been called\n";
    // \\ Checking progress
}


Station::Station()
{
}


Station::~Station()
{
}


Station::Station(char pass_station_name, int pass_start, int pass_size)
{
    station_name = pass_station_name;
    track_starting_ID = pass_start;
    track_size = pass_size;
    //cout << "Overload station has been called\n";
    // \\ Checking progress
}


SubwaySystem::SubwaySystem()
{
    //Initialize tracks
    //node_1, node_2
    my_track[0] = Track('a', 'b');
    my_track[1] = Track('b', 'a');
    my_track[2] = Track('b', 'c');
    my_track[3] = Track('b', 'd');
    my_track[4] = Track('b', 'e');
    my_track[5] = Track('b', 'f');
    my_track[6] = Track('c', 'b');
    my_track[7] = Track('c', 'e');
    my_track[8] = Track('d', 'b');
    my_track[9] = Track('d', 'e');
    my_track[10] = Track('e', 'b');
    my_track[11] = Track('e', 'c');
    my_track[12] = Track('e', 'd');
    my_track[13] = Track('e', 'g');
    my_track[14] = Track('e', 'h');
    my_track[15] = Track('f', 'b');
    my_track[16] = Track('f', 'h');
    my_track[17] = Track('g', 'e');
    my_track[18] = Track('g', 'k');
    my_track[19] = Track('h', 'e');
    my_track[20] = Track('h', 'f');
    my_track[21] = Track('h', 'i');
    my_track[22] = Track('h', 'j');
    my_track[23] = Track('h', 'k');
    my_track[24] = Track('i', 'h');
    my_track[25] = Track('i', 'k');
    my_track[26] = Track('j', 'h');
    my_track[27] = Track('j', 'k');
    my_track[28] = Track('k', 'g');
    my_track[29] = Track('k', 'h');
    my_track[30] = Track('k', 'i');
    my_track[31] = Track('k', 'j');
    my_track[32] = Track('k', 'l');
    my_track[33] = Track('l', 'k');
    //Initialize stations
    //station_name, track_starting_ID, track_size
    my_station[0] = Station('a', 0, 1);
    my_station[1] = Station('b', 1, 5);
    my_station[2] = Station('c', 6, 2);
    my_station[3] = Station('d', 8, 2);
    my_station[4] = Station('e', 10, 5);
    my_station[5] = Station('f', 15, 2);
    my_station[6] = Station('g', 17, 2);
    my_station[7] = Station('h', 19, 5);
    my_station[8] = Station('i', 24, 2);
    my_station[9] = Station('j', 26, 2);
    my_station[10] = Station('k', 28, 5);
    my_station[11] = Station('l', 33, 1);
    //Initiaize other members
    count_routes = 0;
    Current_Station_ID = 0;
    //cout << "SubwaySystem constructor called\n";
    // \\ Checking progress
}


SubwaySystem::~SubwaySystem()
{
}

ostream& operator<<(ostream& os, const Track& my_track)
{
    os << my_track.node_1 << '.' << my_track.node_2;
    return os;
}

ostream& operator<<(ostream& os, const Station& my_station)
{
    os << my_station.station_name << '.' << my_station.track_starting_ID << '.' << my_station.track_size;
    return os;
}


//This is where the above recursive function SearchRoute goes. I posted it separately so it's easier to read.



// **main**

#include <iostream>
#include <string>

//#include "subway.h"

using namespace std;

int main(int argc, char **argv)
{
    SubwaySystem Test;
    Test.SearchRoute(0);
}

如果“检查进度”打印语句使代码更难阅读,我很抱歉。

4

2 回答 2

1

You never enter recursion:

// this method is called with value 0
void SubwaySystem::SearchRoute(int Current_Station_ID)
{
    while(Current_Station_ID < 33)
    {
        cout << "In while loop\n";
        // \\ Checking progress
        if(Current_Station_ID == 0)  //Find a successful route to Station L
        {
            count_routes++; //Add 1 into the variable “count_routes”
            cout << "In if statement\n";
            // \\Checking progress
            cout << count_routes << " " << my_track[Current_Station_ID] << endl; //Print out this route
            return; // ERROR: here you return from the very 1st method call, breaking the search
        }
于 2013-10-20T03:32:02.190 回答
0

基本上无需为您编写代码,我建议这样做:

您的递归函数根本不应该有 while 语句(递归函数的全部意义在于它本身就是一种 while 语句)。

递归函数中的第一件事应该是检查您的车站 ID 是否紧邻“L”车站,或者是否已经采取了唯一的路径。如果它是相邻的,您可以返回从该站到 L 的轨道。

如果没有,则为所有相邻站点调用递归函数,这些站点具有您尚未经过的路径,并将它们的结果用作您当时所在站点的计算的一部分。

根据您的数据举例说明:

  1. 您首先使用“a”作为参数调用该函数。
  2. 该函数首先检查 'a' 是否有到 'l' 的路径(它没有,所以函数不会返回)。
  3. 然后,该函数递归地调用所有从我们当前站“a”开始的路径的站。(只有“b”)
  4. 从 'b' 到 'l' 仍然没有直接的轨迹,所以这次函数继续用 'a'、'c'、'd'、'e'、'f' 调用自身。
  5. 从使用 'a' 对函数的新调用开始,第一次检查发现到 'b' 的唯一可用路由已经被采用,因此它应该返回否定响应并且不打印任何内容。
  6. 当一个站对函数的调用发现它有一条到“l”的直接路径时,它应该打印(或存储)那个路径段并返回真。
  7. 递归函数从它自己的递归调用中收集所有结果,并将路径添加到这些站的结果中。
  8. 确切的实现取决于您需要的程序输出。如果您需要打印出所有路径,您可以保存一个有效路径列表,在出现新分支(多个连接的站点)时添加路径,并在从递归函数收到负返回值时删除路径。当最终调用返回时,您可以打印出您收集的路径。

这能澄清事情吗?

于 2013-10-21T08:11:01.103 回答