杭电1010



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
  

 

Author

ZHANG, Zheng

 

Source

ZJCPC2004

 

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JGShining   |   We have carefully selected several similar problems for you:   1016  1241  1242  1072  1312 

题意：给定一个矩阵大小，有一个小狗从S进入，从D走出，途中只能走上下左右，遇到“."才可以走，遇到X会被挡住不能再往下走，让你判断能否在恰好在t时刻走出D。

下面是我AC了的代码：

#include <stdio.h>
#include <string.h>
#include <math.h>
char a [ 8 ][ 8 ];
int xx [ 4 ]= { 1 ,- 1 , 0 , 0 } ;//定义四个方向
int yy [ 4 ]= { 0 , 0 ,- 1 , 1 } ;
int n ,m ,t ;
int tag = 0 ;
int tt ;
int sx ,sy ,ex ,ey ;
int ok ( int x , int y )
{
    if (x < 0 ||x >=n ||y < 0 ||y >=m )
    {
        return 0 ;
    }
    if (a [x ][y ]== 'X' )
    {
        return 0 ;
    }
    return 1 ;
}
//对节点的判断
void dfs ( int x , int y )
{
    int i ;
    if (tag == 1 )
    {
        return ;
    }
    if (x ==ex &&y ==ey &&tt ==t )
    {
      tag = 1 ;
      return ;
    }
    if (tt >t )
    {
        return ;
    }//对时间的判断
    for (i = 0 ;i < 4 ;i ++)
    {
        if ( ok (x +xx [i ],y +yy [i ]))
        {
            a [x ][y ]= 'X' ; tt ++;
            dfs (x +xx [i ],y +yy [i ]);
            a [x ][y ]= '.' ;
            tt --;//返回的时候需要对其进行还原
        }
    }
}
int main ()
{
    int i ,W ,j ;
    while ( scanf ( "%d%d%d" ,&n ,&m ,&t )!=EOF )
    {
        tag = 0 ;W = 0 ;
        if (n ==m &&m ==t &&t == 0 ) { break ; }
         tt = 0 ;
        for (i = 0 ;i <n ;i ++)
        {
            scanf ( "%s" ,a [i ]);
            getchar ();
            for (j = 0 ;j <m ;j ++)
            {
                if (a [i ][j ]== 'S' )
                {
                    sx =i ;
                    sy =j ;
                }
                if (a [i ][j ]== 'D' )
                {
                    ex =i ;
                    ey =j ;
                }
                if (a [i ][j ]== 'X' )
                {
                    W ++;
                }
            }
        }
        if (t >(n *m -W ))
        {
        printf ( "NO \n " );
        continue ;
        }//当所给时间大于所有的”.“的和就直接结束
        if (( abs (sx -ex )+ abs (sy -ey )+t )% 2 == 1 )
        {
            printf ( "NO \n " );
            continue ;
        }//对两者的距离和所给的时间进行判断，为奇数直接结束，所有的时间都是在最短距离上增加，且必须是2的倍数，这个很容易发现
        dfs (sx ,sy );
        if (tag == 1 )
        {
            printf ( "YES \n " );
        }
        else
        {
            printf ( "NO \n " );
        }
    }
    return 0 ;
}

哎，这道题WA了很多次，主要时间超时，就是没有想到那个对奇偶的剪枝，这道题剪枝还是很强，自己很菜，做了很长时间。