zoj2110

Tempter of the Bone

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of this maze.

The maze was a rectangle with sizes N by M. There was a door in the maze. At the beginning, the door was closed and it would open at the T-th second for a short period of time (less than 1 second). Therefore the doggie had to arrive at the door on exactly the T-th second. In every second, he could move one block to one of the upper, lower, left and right neighboring blocks. Once he entered a block, the ground of this block would start to sink and disappear in the next second. He could not stay at one block for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him.

Input

The input consists of multiple test cases. The first line of each test case contains three integers N, M, and T (1 < N, M < 7; 0 < T < 50), which denote the sizes of the maze and the time at which the door will open, respectively. The next N lines give the maze layout, with each line containing M characters. A character is one of the following:

'X': a block of wall, which the doggie cannot enter; 
'S': the start point of the doggie; 
'D': the Door; or
'.': an empty block.

The input is terminated with three 0's. This test case is not to be processed.

Output

For each test case, print in one line "YES" if the doggie can survive, or "NO" otherwise.

Sample Input

4 4 5
S.X.
..X.
..XD
....
3 4 5
S.X.
..X.
...D
0 0 0

Sample Output

NO
YES

#include <cstdio>
#include <cmath>	//用到了求绝对值的函数fabs

char map[9][9];	//迷宫地图
int n, m, t;	//迷宫的大小，及迷宫的门会在第t秒开启
int di, dj;		//(di,dj)：门的位置
bool escape;	//是否成功逃脱的标志，escape为1表示能成功逃脱
int dir[4][2] = { {0,-1}, {0,1}, {1,0}, {-1,0} }; //分别表示下、上、左、右四个方向

//已经到达(si,sj)位置，且已经花费cnt秒
void dfs( int si, int sj, int cnt )
{
	int i, temp;

	if( si>n || sj>m || si<=0 || sj<=0 )  return;	//边界

	if( si==di && sj==dj && cnt==t )	//成功逃脱
	{
		escape = 1;  return;
	}
	
	//abs(x-ex) + abs(y - ey)表示现在所在的格子到目标格子的距离(不能走对角线)
	//t-cnt是实际还需要的步数，将他们做差
	//如果temp < 0或者temp为奇数，那就不可能到达！
	temp = (t-cnt) - fabs(si-di) - fabs(sj-dj);
	if( temp<0 || temp%2 )  return;	//搜索过程当中的剪枝

	for( i=0; i<4; i++ )
	{
		if( map[ si+dir[i][0] ][ sj+dir[i][1] ] != 'X')
		{
			//前进方向！将当前方格设置为墙壁‘X’
			map[ si+dir[i][0] ][ sj+dir[i][1] ] = 'X';
			dfs(si+dir[i][0], sj+dir[i][1], cnt+1);	//从下一个位置继续搜索
			if(escape)  return;
			map[ si+dir[i][0] ][ sj+dir[i][1] ] = '.'; //后退方向！恢复现场！
		}
	}
	return;
}

int main( )
{
	int i, j;	//循环变量
	int si, sj;	//小狗的起始位置
	while( scanf("%d%d%d", &n, &m, &t) )
	{
		if( n==0 && m==0 && t==0 )  break;	//测试数据结束
		int wall = 0;
		char temp;
		scanf( "%c", &temp );	//见下面的备注
		for( i=1; i<=n; i++ )
		{
			for( j=1; j<=m; j++ )
			{ 
				scanf( "%c", &map[i][j] );
				if( map[i][j]=='S' ){  si=i;  sj=j;  } 
				else if( map[i][j]=='D' ){  di=i;  dj=j;  }
				else if( map[i][j]=='X' )  wall++;
			}
			scanf( "%c", &temp );
		}
		if( n*m-wall <= t )	//搜索前的剪枝
		{
			printf( "NO
" );  continue;
		}
		escape = 0;
		map[si][sj] = 'X';
		dfs( si, sj, 0 );
		if( escape )  printf( "YES
" );	//成功逃脱
		else  printf( "NO
" );
	}
	return 0;
}