/*对马踏棋盘的一点研究*/
/* 制作人  ：波上清风*/
/* QQ：164094487          */
/* Email: fygood@163.com  */
/*欢迎与我联系，讨论问题  */


/*本人先后编了两次，第二次进行了改进。改进的思想主要是注意到棋盘上每一点的下一可到达点的个数
（下称为权值）不同，对于可到达点较少（权值小）的点应该先跳上去，这样后来跳的点可跳的方向就比
较多，回溯的现象就比较少，这样就可以大幅度提高速度*/

/*第一次*/
/*原始的马踏棋盘，未加权值，有些点速度很慢*/

#include "stdio.h"
#define N 8
int w=0;
int way1[8]={-2,-1,1,2, 2, 1,-1,-2};
int way2[8]={ 1,2, 2,1,-1,-2,-2,-1};
int ch[N*N]={0};
int a[N*N+1][3]={0};
int st=1;
char c='y';

void print()
{
    int x,y;
    
    printf("\n------%d answer----\n",++w);
    
    for(x=1;x<N+1;x++)
    {
        printf("\n");
        for(y=1;y<N+1;y++)
            printf("%2d ",ch[(x-1)*N+y-1]);
        printf("\n");
    }
    printf("\nPress n to quit ,press any other key to continue.\n");
    c=getchar();        /*询问是否继续输出结果*/
}

main()   
{
    int x,y,way;
    printf("Please enter the row and column of the starting point.\n");
    scanf("%d,%d",&a[1][0],&a[1][1]);/*输入行数和列数*/
    getchar();                       /*接收回车符*/
    x=a[1][0],y=a[1][1];             
    ch[(x-1)*N+y-1]=1;               /*在ch数组中对相应点赋值*/
    
    
    while(1)
    {
        if(a[1][2]>=8)                   /*出发点的八个方向都已走过，表示所有的方法均已找出*/
            break;
      if(a[st][2]>=8)            /*此点的八个方向都已走过，应该退回到上一次走的点*/
        {
            x=a[st][0];
            y=a[st][1];
        ch[(x-1)*N+y-1]=0;          /*将这一点被走过的痕迹抹去*/
            a[st][0]=a[st][1]=a[st][2]=0;
        a[st-1][2]++;               /*使上一次走的点走的方向发生变化*/
            st--;                       /*步数减一*/
        }
        
    else                             /*此点的八个方向未全走过，应走此方向*/
        {
            way=a[st][2];
            a[st][2]++;                  /*确定下次应走的方向*/
            x=a[st][0]+way1[way];        
    y=a[st][1]+way2[way];         /*确定按这次的方向走应走到的x,y坐标*/
            
            
        if(x<1||y<1||x>N||y>N||ch[(x-1)*N+y-1]!=0)/*此点不满足要求*/
                continue;
            ch[(x-1)*N+y-1]=++st;        /*走到这一点 */
            a[st][0]=x;
            a[st][1]=y;
            a[st][2]=0;                   /*标记这一步*/
            
            if(st==N*N)                   /*步数已满*/
            {
                print();                  /*输出结果*/
                if(c=='n')
                    break;
                ch[(x-1)*N+y-1]=0;
                a[st][0]=a[st][1]=a[st][2]=0;
                a[st-1][2]++;
                st--;                      /*退回前一步*/
            }
        }
    }
}

/*
第二次：

改进后的马踏棋盘，加入了每点的权值因素，速度极快*/
#include "stdio.h"
#define N 8
int w=0;
int way1[8]={-2,-1,1,2, 2, 1,-1,-2};
int way2[8]={ 1,2, 2,1,-1,-2,-2,-1};
int ch[N*N]={0};
int a[N*N+1][3]={0};
int dir[N][N][8];
int st=1;
char c='y';
int weight[N][N];    


void caculate();
void dirctions();
void print();
int  check(int i,int j);

void caculate()      /*计算各点的权值*/
{
int i,j,k;
for(i=1;i<=N;i++)
   for(j=1;j<=N;j++)
       for(k=0;k<N;k++)
       {
           int x,y;
           x=i+way1[k];
           y=j+way2[k];
           if(x>=1&&x<=N&&y>=1&&y<=N)
               weight[i-1][j-1]++;
       }

}

int  check(int i,int j) /*检查(i,j)是否在棋盘内*/
{
if(i<1||i>8||j<1||j>8)
    return 0;
return 1;
}


void directions()        /*求出各点的最佳方向序列，即优先向权值小的方向*/
{
int i,j,k,m,n1,n2,x1,y1,x2,y2,way_1,way_2;
for(i=0;i<N;i++)
   for(j=0;j<N;j++)
       {
         for(k=0;k<8;k++)
             dir[i][j][k]=k;
         for(k=0;k<8;k++)
         {
             
            for(m=k+1;m<8;m++) /*对每个方向考察看有没有更好的*/
             {
              way_1=dir[i][j][k];
              x1=i+way1[way_1];
              y1=j+way2[way_1];
              way_2=dir[i][j][m];
              x2=i+way1[way_2];
              y2=j+way2[way_2];
              n1=check(x1+1,y1+1);
              n2=check(x2+1,y2+1);
              if(   
                  ( n1==0 && n2 )   ||   /*k方向不可达到，而m方向可达到*/
                  ( n1 && n2&&weight[x1][y1]>weight[x2][y2] )/*都可达到但m方向权值小*/
               )
              {
                
                dir[i][j][k]=way_2;
                dir[i][j][m]=way_1;      /*交换两个方向值*/
              }
             }
         }




        
        
        }

}


void print()
{
    int x,y;
    
    printf("\n------%d answer----\n",++w);
    
    for(x=1;x<N+1;x++)
    {
        printf("\n");
        for(y=1;y<N+1;y++)
            printf("%2d ",ch[(x-1)*N+y-1]);
        printf("\n");
    }
    printf("\nPress n to quit ,press any other key to continue.\n");

    c=getchar();        /*询问是否继续输出结果*/
}

main()   
{
    int x,y,way,way0;
    caculate();
    directions();
    printf("Please enter the row and column of the starting point.\n");
    scanf("%d,%d",&a[1][0],&a[1][1]);/*输入行数和列数*/
    getchar();                       /*接收回车符*/
    x=a[1][0],y=a[1][1];             
    ch[(x-1)*N+y-1]=1;               /*在ch数组中对相应点赋值*/
    
    
    while(1)
    {
        if(a[1][2]>=8)                   /*出发点的八个方向都已走过，表示所有的方法均已找出*/
            break;
        
        if(a[st][2]>=8)                    /*此点的八个方向都已走过，应该退回到上一次走的点*/
        {
            x=a[st][0];
            y=a[st][1];
            ch[(x-1)*N+y-1]=0;          /*将这一点被走过的痕迹抹去*/
            a[st][0]=a[st][1]=a[st][2]=0;
            a[st-1][2]++;               /*使上一次走的点走的方向发生变化*/
            st--;                       /*步数减一*/
        }
        
        else                             /*此点的八个方向未全走过，应走此方向*/
        {
            way0=a[st][2];
            a[st][2]++;                    /*确定下次应走的方向*/
            x=a[st][0];
            y=a[st][1];
            way=dir[x-1][y-1][way0];
            x=a[st][0]+way1[way];        
            y=a[st][1]+way2[way];         /*确定按这次的方向走应走到的x,y坐标*/
            
            
            if(x<1||y<1||x>N||y>N||ch[(x-1)*N+y-1]!=0)/*此点不满足要求*/
                continue;
            ch[(x-1)*N+y-1]=++st;        /*走到这一点 */
            a[st][0]=x;
            a[st][1]=y;
            a[st][2]=0;                   /*标记这一步*/
            
            if(st==N*N)                   /*步数已满*/
            {
                print();                  /*输出结果*/
                if(c=='n')
                    break;
                ch[(x-1)*N+y-1]=0;
                a[st][0]=a[st][1]=a[st][2]=0;
                a[st-1][2]++;
                st--;                      /*退回前一步*/
            }
        }
    }
}