Tempter of the Bone（dfs+奇偶剪枝）

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 94669    Accepted Submission(s):
25669

Problem Description

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

 

 #include <iostream>
 #include <cstdio>
 #include <queue>
 #include <cstring>
 #include <algorithm>
 #define LL long long
 using namespace std;
 char map[][];
 bool vis[][];
 int dx[]={,,-,},dy[]={,-,,};
 int n,m,t;
 int sx,sy,ex,ey;
 bool flag;
 bool dfs(int x,int y,int tim)
 {
     if(flag)    return ;
     if(x<||y<||x>=n||y>=m||vis[x][y]==||tim>t||map[x][y]=='X')
         return ;
     if(tim<t&&map[x][y]=='D')    return ;
     int w=abs(x-ex)+abs(y-ey);
     if(t-tim<w)    return ;
     int tem=w-abs(t-tim);
     if(tem>||tem%)    return ;
     if(tim==t&&map[x][y]=='D')    {flag=;return ;}
     for(int i=;i<;i++)
     {
         vis[x][y]=;
         int nx=x+dx[i];
         int ny=y+dy[i];
         dfs(nx,ny,tim+);
         vis[x][y]=;
     }
     return ;
 }
 int main()
 {
     freopen("in.txt","r",stdin);
     while(scanf("%d%d%d",&n,&m,&t))
     {
         getchar();
         int tim=,z=;
         flag=;
         if(n==&&m==&&t==)    break;
         for(int i=;i<n;i++)
         {
             for(int j=;j<m;j++)
             {
                 scanf("%c",&map[i][j]);
                 if(map[i][j]=='S')
                 {
                     sx=i,sy=j;
                 }
                 if(map[i][j]=='D')
                 {
                     ex=i,ey=j;
                 }
             //    if(map[i][j]=='X')    z++;
             }
             getchar();
         }
         //if(n*m-z<t){printf("NO\n");continue;}
         memset(vis,,sizeof(vis));
         dfs(sx,sy,);
         if(flag)    printf("YES\n");
         else    printf("NO\n");
     }
 }