布线问题其实可以看成是二维矩阵的赋值问题

#include"math.h"

#include<iostream>
#include<math.h>
using namespace std;

class Position
{
 friend bool FindPath(Position start,Position finish,int &Pathlen, Position * &path);
 public:
  Position(int x=0,int y=0)
  {
   this->x=x;
   this->y=y;  
  }
 private:
 int x;
 int y;
};
bool FindPath(Position start,Position finish,int &Pathlen, Position * &path)
{
 if((start.x==finish.x)&&(start.y=finish.y))
 {
  Pathlen=0;
  return 1;
 }
 int i,j,grid[9][9];
 for(i=0;i<9;i++)
    for(j=0;j<9;j++)
    grid[i][j]=0;
 for(i=0;i<=8;i++)
 grid[0][i]=grid[8][i]=1;
 for(i=0;i<=8;i++)
 grid[i][0]=grid[i][8]=1;
 grid[1][3]=1;grid[2][3]=1;grid[2][4]=1;grid[3][5]=1;grid[4][4]=1;grid[4][5]=1;grid[7][3]=1;
 grid[5][1]=1;grid[5][5]=1;grid[6][1]=1;grid[6][2]=1;grid[6][3]=1;grid[7][1]=1;grid[7][2]=1;
 Position offset[4]; //offset是四个移动方向的相对位移矩阵
    offset[0].x = 0; offset[0].y = 1; // 右
    offset[1].x = 1; offset[2].y = 0;//下
 offset[2].x = 0; offset[2].y = -1; // 左
    offset[3].x = -1; offset[3].y = 0; // 上
    int NumOfNbrs=4;
    Position here, nbr;
    here.x=start.x;
    here.y=start.y;
    grid[start.x][start.y] = 2; //起始点距离为2
    LinkedQueue < Position > Q;
    do {  // 标记可达相邻方格
    for (int i = 0; i < NumOfNbrs; i++)
      { // NumOfNbrs为相邻方格数
         nbr.x = here.x + offset[i].x;  // here是正在走线的方格，
         nbr.y = here.y + offset[i].y;    // nbr是可考虑走线方格
         if (grid[nbr.x][nbr.y] == 0)
  { // 该方格未标记
            grid[nbr.x][nbr.y] = grid[here.x][here.y] + 1;
                              // grid[][]中记录该方格距离起始方格距离
         if ((nbr.x == finish.x) && (nbr.y == finish.y))   
            break; // 完成布线
         Q.Add(nbr);} //队列中加入新的扩展结点
       }

if ((nbr.x == finish.x) && (nbr.y == finish.y))    break;
     if(Q.IsEmpty())
  return false;
     Q.Delete(here);
 }while(1);
 
 Pathlen=grid[finish.x][finish.y]-2;
 path =new Position[Pathlen];
 here=finish;
 for(int j=Pathlen-1;j>=0;j--)
 {
 path[j]=here;
 for(i=0;i<NumOfNbrs;i++)
 {
  nbr.x=here.x+offset[i].x;
  nbr.y=here.y+offset[i].y;
  if(grid[nbr.x][nbr.y]==j+2) break;
 }
 here=nbr;
 cout<<"position:"<<"("<<here.x<<","<<here.y<<")"<<endl;
    }
    return 1;
}

int main()
{
Position path[20];
int a,b,c,d;
cout<<"the start:  ";
cin>>a>>b;
Position start(a,b);
cout<<endl<<"the finish:   ";
cin>>c>>d;
Position finish(c,d);
int Pathlen;
if(FindPath(start,finish,Pathlen, path)==1)
cout<<"finished!";
return 1; 
}