HDU 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 5S.X...X...XD....3 4 5S.X...X....D0 0 0
剪枝的题目不容易
百度的…………
#include <iostream>  using namespace std;  char map[9][9];        //地图最多不超过7行7列,又从(1,1)计算,所以开辟9,9  int n,m,t,di,dj;       //给定的三个量,以及终点位置  bool escape;           //标识逃生成功  int dir[4][2]={{0,-1},{0,1},{1,0},{-1,0}};  //分别表示左、右、下、上四个方向  void DFS(int si,int sj,int cnt)  {      int i,temp;      if (si>n || sj>m || si<=0 || sj<=0)   //越出边界便不搜索          return;      if (si==di && sj==dj && cnt==t)      {          escape=1;                   //时间正好的时候才能逃生          return;      }      temp=abs(t-cnt)-(abs(di-si)+abs(dj-sj));     //计算当前到终点的最短路与还需要的时间差,若小于0则路径剪枝      if (temp<0 || temp&1)           //temp如果是奇数的话也要剪枝          return;      for (i=0;i<4;i++)      {          if (map[si+dir[i][0]][sj+dir[i][1]]!='X')          {              map[si+dir[i][0]][sj+dir[i][1]]='X';  //把当前点刷为X              DFS(si+dir[i][0],sj+dir[i][1],cnt+1); //搜索该点                          if (escape)                  return;              map[si+dir[i][0]][sj+dir[i][1]]='.';  //如果搜索不到退出来了,则重新把该点刷为'.'          }      }      return ;  }  int main()  {      int i,j,si,sj,wall;      while (cin>>n>>m>>t)      {          if (n==0 && m==0 && t==0)          {              break;          }          wall=0;          for (i=1;i<=n;i++)          {              for (j=1;j<=m;j++)              {                  cin>>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++;                  }              }          }          if (n*m-wall<=t)      //路径剪枝,当走完不含墙的迷宫都还不到T时间将不能逃生          {              cout<<"NO"<<endl;              continue;          }          escape=0;          map[si][sj]='X';      //记得刷为'X'          DFS(si,sj,0);          if (escape)          {              cout<<"YES"<<endl;          }          else          {              cout<<"NO"<<endl;          }      }      return 3;  }