马踏棋盘(骑士周游)
题目渊源: 马踏棋盘问题(又称骑士周游或骑士漫游问题)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。 编写代码,实现马踏棋盘的操作,要求用1~64来标注“马”移动的路径。
#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 )
{
*x = *x + 2;
*y = *y - 1;
return 1;
}
break;
case 1:
if( *x+2<=X-1 && *y+1<=Y-1 && chess[*x+2][*y+1]==0 )
{
*x = *x + 2;
*y = *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,每走一步,tag+1
int TravelChessBoard(int x, int y, int tag)
{
int x1=x, y1=y, flag=0, count=0;
chess[x][y] = tag;
// 如果tag==X*Y,则完成整个棋盘的遍历
if( tag == X*Y )
{
print();
return 1;
}
flag = nextxy(&x1, &y1, count);
while( 0==flag && count < 7 )
{
count++;
flag = nextxy(&x1, &y1, count);
}
while( flag )
{
if( TravelChessBoard(x1, y1, tag+1) )
{
return 1;
}
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,马踏棋盘失败鸟~\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;
}