FIG Applications: Data structures and algorithms (XIV)

Horse riding board (Knights travel around)

Topic origin: horse riding chessboard (also known as the Knight Rider roaming or travel around the problem) is one of the classic problems of algorithm design.

Topics requirements: chess board is 8 * 8 checkerboard, now "horse" at any specified grid in accordance with the rules for playing chess, "horse" will "horse" move. It requires each box can only be entered once, eventually making the "horse" around 64 chessboard squares. Develop code that implements chessboard horse riding operation, requiring 1 to 64 with the path of movement are denoted by "Ma."

#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;
}