题目渊源:
马踏棋盘问题(又称骑士周游问题或骑士漫游问题)是算法设计的经典问题之一。
题目要求:
国际象棋的棋盘为8*8的方格棋盘,现将"马"放在任意指定的方格中,按照"马"走棋的规则将"马"进行移动。要求每个方格只能进入一次,最终使得"马"走遍棋盘64个方格。
cpp
#include <stdio.h>
#include <time.h>
#define X 8
#define Y 8
int chess[X][Y];
//找到基于(x,y)位置的下一个可走的位置
int nextxy(int *x,int *y,int count)
{
switch(count)
{
case 0:
if(*x+2<=X-1 && *y-1>=0 && chess[*x+2][*y-1]==0)
{
*y+=2;
*y-=1;
return 1;
}
break;
case 1:
if(*x+2<=X-1 && *y+1<=Y-1 && chess[*x+2][*y+1]==0 )
{
*x+=2;
*y+=1;
return 1;
}
break;
case 2:
if(*x+1<=X-1 && *y-2>=0 && chess[*x+1][*y-2]==0 )
{
*x=*x+1;
*y=*y-2;
return 1;
}
break;
case 3:
if(*x+1<=X-1 && *y+2<=Y-1 && chess[*x+1][*y+2]==0)
{
*x = *x+1;
*y= *y+2;
return 1;
}
break;
case 4:
if(*x-2>=0 && *y-1>=0 && chess[*x-2][*y-1]==0)
{
*x= *x-2;
*y= *y+1;
return 1;
}
break;
case 5:
if(*x-2>=0 && *y+1<=Y-1 && chess[*x-2][*y+1]==0 )
{
*x= *x-2;
*y = *y+1;
return 1;
}
break;
case 6:
if(*x-1>=0 && *y-2>=0 && chess[*x-1][*y-2]==0)
{
*x = *x - 1;
*y = *y - 2;
return 1;
}
break;
case 7:
if(*x-1>=0 && *y+2<=Y-1 && chess[*x-1][*y+2]==0)
{
*x = *x -1;
*y = *y +2;
return 1;
}
break;
default:
break;
}
return 0;
}
void print()
{
int i,j;
for(i=0;i<X;i++)
{
for(j=0;j<Y;j++)
{
printf("%2d\t",chess[i][j]);
}
printf("\n");
}
printf("\n");
}
//深度优先遍历棋盘
//(x,y)为位置坐标
//tag是标记变量
int TravelChessBoard(int x,int y,int tag)
{
int x1= x,y1=y,count =0,flag =0;
chess[x][y] = tag;
if(x*Y == tag)
{
//打印棋盘
print();
return 1;
}
//找到马的下一个可走的坐标(x1,y1)
flag = nextxy(&x1,&y1,count);
while(0==flag && count<7)
{
count++;
}
while(flag)
{
if(TravelChessBoard(x1,y1,tag+1))
{
return 1;
}
//出现意外,找到马的下一步可走坐标(x1,y1)
x1=x;
y1=y;
count++;
flag = nextxy(&x1,&y1,count);
while(0==flag && count < 7)
{
count++;
flag = nextxy(&x1,&y1,count);
}
}
if(0 == flag)
{
chess[x][y] =0;
}
return 0;
}
int main()
{
int i,j;
clock_t start,finish;
start = clock();
for(i=0;i<X;i++)
{
for(j=0;j<Y;j++)
{
chess[i][j]=0;
}
}
if(TravelChessBoard(2,0,1))
{
printf("抱歉,马踏棋盘失败!\n");
}
finish = clock();
printf("\n本次计算一共耗时:%f秒\n\n",(double)(finish - start)/CLOCKS_PER_SEC);
return 0;
}