c++ - 使用回溯的骑士​​之旅

Question

我正在使用回溯在 C++ 中开发一个骑士之旅程序这是我的代码:

#include <iostream>
#include <iomanip>
using namespace std;

template <class T>
class Stack
{
public:
    Stack(int len = 100)
    {
        length = len;
        sp = -1;
        a = new T [length];
    }

    int getLength()
    {
        return length;
    }

    void push(T v)
    {
        if(isFull())
        {
            throw std::exception("Stack Overflow!");
            return;
        }
        a[++sp] = v;
    }
    bool isFull()
    {
        if(sp < getLength()-1)
            return false;
        return true;
    }
    T pop()
    {
        if(isEmpty())
        {
            throw std::exception("Stack Underflow!");
        }
        return a[sp--];
    }
    bool isEmpty()
    {
        if(sp == -1)
            return true;
        return false;
    }
    T getTop()
    {
        if(isEmpty())
        {
            throw std::exception("Stack Underflow!");
        }
        return a[sp];
    }

private:
    T * a;
    int sp;
    int length;

};

class MoveStk
{
public:
    MoveStk()
    {
        set(0,0,0);
    }

    MoveStk(int r , int c , int d)
    {
        set(r,c,d);
    }

    void set(int r , int c , int d)
    {
        row = r;
        col = c;
        dir = d;
    }

    void get(int& r , int & c , int & d)
    {
        r = row;
        c = col;
        d = dir;
    }

private:
    int row , col , dir;
};

class Position
{
public:
    int x , y;
    Position(int i = 0 , int j = 0)
    {
        x = i;
        y = j;
    }
};


bool move(Position pos[],int posLen,int ** chess ,int n, int col, int row , int&newRow , int & newCol , int moveNo )
{
    if (row+pos[moveNo].x>=0 && row+pos[moveNo].x<n && col+pos[moveNo].y>=0 && col+pos[moveNo].y<n && chess[row+pos[moveNo].x][col+pos[moveNo].y] == 0 )
    {
        newRow = row+pos[moveNo].x;
        newCol = col+pos[moveNo].y;
        return true;
    }
    return false;
}

int main()
{

    int n = 8;

    //create stack
    Stack<MoveStk> myStack(100);



    //create dynamic array
    int ** chess = new int* [n];
    for (int i = 0; i < n; i++)
        chess[i] = new int[n];

    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            chess[i][j] = 0;

    //setup moves
    Position pos[8];

    pos[0] = Position(-2,1);
    pos[1] = Position(-2,-1);
    pos[2] = Position(2,1);
    pos[3] = Position(2,-1);
    pos[4] = Position(-1,-2);
    pos[5] = Position(-1,2);
    pos[6] = Position(1,2);
    pos[7] = Position(1,-2);


    //input for start (must be optional) :
    int row = 4;
    int col = 3;


    int count = 1 , moveChoose = 0;
    int newCol , newRow;

    MoveStk temp(row,col,0);//adding first move to stack
    myStack.push(temp);

    chess[row][col] = count;

    while(count<=(n*n) && !myStack.isEmpty())
    {
        temp = myStack.getTop();
        temp.set(row,col,moveChoose);

        while(moveChoose<8 && !move(pos,8,chess,n,col,row,newRow,newCol,moveChoose))//avalin khoone azad baray karkat
        {
            moveChoose++;
        }

        if(moveChoose != 8)//there is somewhere we can go
        {
            myStack.pop();
            temp.set(col,row,moveChoose+1);//adding next possible move for returning back
            myStack.push(temp);
            chess[newRow][newCol] = ++count;
            temp.set(newRow,newCol,0);
            myStack.push(temp);
        }
        else//we must return
        {
            myStack.pop();
            chess[newRow][newCol] = 0;
            count--;
        }

    }

    printArray(chess);

    cin.ignore();
    cin.get();
}

问题是，这个应用程序只适用于第一步，我的堆栈在游戏结束前就变空了。我找不到哪里错了？!

可能是什么问题？

Answer 1

首先，我认为您需要更改行:

while(count<=(n*n) && !myStack.isEmpty())
{
    temp = myStack.getTop();
    temp.set(row,col,moveChoose);

收件人:

while(count<=(n*n) && !myStack.isEmpty())
{
    temp = myStack.getTop();
    temp.get(row,col,moveChoose);

按照目前的情况，每次循环都覆盖了最后一步和位置。你应该做的是从你离开的地方开始，尝试从以前的位置开始下一步。

此外，我注意到您稍后对 set 调用的参数有误:

        temp.set(col,row,moveChoose+1);//adding next possible move for returning back

应该是row然后是col。