迷宫求解（栈实现）

自己写的，外面别人的看了一下，感觉不少都是递归解决这类问题的，然后我尝试用非递归解决的。混合了类的运用，多说不如直接上代码了。

#include<iostream>#include<stdio.h>#include<stack>#define M 8using namespace std;struct Pos {int i;int j;};class Maze{    public:    Maze()=default;    Maze(int a1,int b1,int a2,int b2)    {        start_.i=a1;        start_.j=b1;        end_.i=a2;        end_.j=b2;    }    ~Maze()=default;    void CreateMaze();    void PrintMaze();    bool MakeIt();    private:        int Ma_ze[M][M];        Pos start_;        Pos end_;};void Maze::CreateMaze( ){    int i=0,j=0;    int a=0;    for (i=0;i<M;i++)    {        for (j=0;j<M;j++)        {         cin>>a;         Ma_ze[i][j]=a;        }    }    Ma_ze[start_.i][start_.j]=1;}void Maze::PrintMaze( ){    int i=0,j=0;    for (i=0;i<M;i++)    {        for (j=0;j<M;j++)        {         if (Ma_ze[i][j]==2)         {             cout<<"●";         }         else if(Ma_ze[i][j]==0)         {             cout<<"  ";         }         else if (Ma_ze[i][j]==1)         {             cout<<"**";         }        }        cout<<endl;    }}bool Maze::MakeIt()//完成路径寻找{    bool issuccess=false;    stack <Pos> make;    Pos alit,blit;    make.push(start_);    while (1)    {        if (make.empty()==true)        {            break;        }        alit=make.top();        //cout<<alit.i<<"  "<<alit.j<<endl;        //make.pop();        if (Ma_ze[alit.i+1][alit.j]==0)        {            blit.i=alit.i+1;            blit.j=alit.j;            Ma_ze[alit.i+1][alit.j]=1;            if (blit.i==end_.i&&blit.j==end_.j)            {                issuccess=true;                break;            }            make.push(blit);        }        else  if (Ma_ze[alit.i][alit.j+1]==0)        {            blit.i=alit.i;            blit.j=alit.j+1;            Ma_ze[alit.i][alit.j+1]=1;            if (blit.i==end_.i&&blit.j==end_.j)            {                issuccess=true;                break;            }            make.push(blit);        }        else  if (Ma_ze[alit.i-1][alit.j]==0)        {            blit.i=alit.i-1;            blit.j=alit.j;            Ma_ze[alit.i-1][alit.j]=1;               if (blit.i==end_.i&&blit.j==end_.j)            {                issuccess=true;                break;            }            make.push(blit);        }        else  if (Ma_ze[alit.i][alit.j-1]==0)        {            blit.i=alit.i;            blit.j=alit.j-1;            Ma_ze[alit.i][alit.j-1]=1;            if (blit.i==end_.i&&blit.j==end_.j)            {                issuccess=true;                break;            }            make.push(blit);        }        else        {            make.pop();        }    }    return issuccess;}int main(){    Maze m(1,1,6,6);    freopen("maze.txt","r",stdin);    m.CreateMaze();    m.PrintMaze();    if (m.MakeIt()==true)    {        m.PrintMaze();    }    else    {        cout<<"error"<<endl;    }}

这是测试用的数据，“2”表示墙壁，“0”表示可以走。

2 2 2 2 2 2 2 2
2 0 0 2 0 2 0 2
2 2 0 0 0 0 0 2
2 0 0 2 0 2 0 2
2 2 2 2 0 2 0 2
2 2 0 0 0 2 0 2
2 0 0 2 0 0 0 2
2 2 2 2 2 2 2 2

反思就是代码还有很多可以写的更简洁的地方，比较懒想等到以后了。