1010-Tempter of the Bone

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

参考了其他人的代码；

#include

#include

#include

int N,M,T,s1,s2,d1,d2,sum;

char maze[7][7];

int ans=0,temp=0;

//ans是已经走过的步数，temp是用来判定是否在规定T步到达的开关变量

int step(int x,int y)

//step计算当前坐标到终点坐标的最少步数

{

return abs(x-d1)+abs(y-d2);

}

void dfs(int i,int j)

{

int t,distan=T-step(i,j)-ans;//distan是剩余要“绕弯”走的步数

if(distan%2||distan<0||((T-ans)T||sum

//奇偶剪枝，distan必须是偶数而且大于0（小于0代表剩余步数）不够走到；T不能大于sum（路径中所有.的数量）；ans不能大于T；如果temp=1了，就没必要继续递归下去了（一直在这里TLE）

return ;

if(maze[i][j]=='D')

if(ans==T)

{

temp=1;

return ;

}

if(maze[i][j]=='.'||maze[i][j]=='S')

{

maze[i][j]='X';

ans++;

for(t=-1;t<=1;t+=2)

{

if(0<=(i+t)&&(i+t)

&&maze[i+t][j]!='X')

{

dfs(i+t,j);

}

}

for(t=-1;t<=1;t+=2)

{

if(0<=(j+t)&&(j+t)

&&maze[i][j+t]!='X')

{

dfs(i,j+t);

}

}

maze[i][j]='.';

ans--;

}

}

int main(void)

{

    while(scanf("%d%d%d",&N,&M,&T),N,M,T)

    {

        int i,j;

        sum=1;

        for(i=0; i

        {

        scanf("%s",maze[i]);//用getchar（）的请注意这个地方。

        }

        for(i=0; i

            for(j=0; j

            {

                if(maze[i][j]=='.')

                    sum++;

                if(maze[i][j]=='S')

                {

                    s1=i;

                    s2=j;

                }

                if(maze[i][j]=='D')

                {

                    d1=i;

                    d2=j;

                }

            }

        ans=0;

        temp=0;

        dfs(s1,s2);

        if(temp)

            printf("YES ");

        else

            printf("NO ");

    }

    return 0;

}