实现用于查找路径的递归函数的问题

时间:2013-10-20 03:11:30

标签: c++ function recursion

我正在尝试在地铁系统中打印出所有可能的路径,这些路径有从A到L的站点。换句话说,目的是找出一个人可以通过地铁系统可以走多少路线而不会过去不止一次的赛道。我知道自从我之前在C中使用邻接矩阵编写此程序以来,有640条可能的路径。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,然后再次标记路径以允许回溯。当我打印出最终结果时,我只得到从A到B的曲目,这显然意味着某些事情显然是错误的。

这就是我认为问题所在:我知道在我的SubwaySystem :: SearchRoute函数中我使用了if和else语句,它显然不允许调用我的递归函数但是我尝试使用另一个if语句而不是否则我不太确定条件是什么。

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
                }
            }
        }
    }
}

此外,我还尝试在整个程序中使用打印语句来跟踪进度。我的递归函数永远不会被我上面指定的原因调用(至少这就是为什么我认为它不起作用)。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。

如果您能帮助我弄清楚我的代码中的问题区域/向我展示正确的方法,我将不胜感激。我厌倦了这个想法。提前谢谢。

以下是单个TU中程序的其余部分:

//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”打印语句使得阅读代码变得更加困难,我很抱歉。

2 个答案:

答案 0 :(得分:1)

你永远不会输入递归:

// 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
        }

答案 1 :(得分:0)

基本上没有为你编写代码我建议:

你的递归函数根本不应该有while语句(递归函数的整个点就是它本身就是一种while语句)。

你的递归函数中的第一件事应该是检查你的站点id是否紧邻'L'站,或者是否已经采取了唯一的路径。如果它是相邻的,你可以返回从该站到L的轨道。

如果没有,则为所有相邻的站点调用递归函数,这些站点还有尚未到达的路径,并将其结果用作当时所在站点的计算的一部分

使用基于您的数据的示例澄清:

  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'是否有一个'l'的路径(它没有,因此该函数不返回)。
  3. 然后,该函数自动调用递归,其中所有站点都有来自当前站点'a'的路径。 (仅'b')
  4. 从'b'开始,仍然没有立即跟踪'l',所以这次函数继续用'a','c','d','e','f'调用自己。
  5. 从新调用带有'a'的函数,第一个检查发现唯一可用的路径'b'已被采用,因此它应该返回否定响应而不打印任何内容。
  6. 当一个对某个电台的功能的调用发现它有一条“l”的直接路径时,它应该打印(或存储)该路径段并返回true。
  7. 递归函数从其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. 确切的实现取决于程序所需的输出。如果需要打印所有路径,则可以保留有效路径列表,每当出现新分支(多个连接站)时添加路径,并在从递归函数接收到负返回值时删除路径。当最后一个呼叫返回时,您可以打印出您收集的路径。
  9. 这是否澄清了事情?