杭电OJ——Tempter of the Bone

Tempter of the Bone

                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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 5S.X...X...XD....3 4 5S.X...X....D0 0 0

 

Sample Output

NOYES

 

Author

ZHANG, Zheng

 

Source

ZJCPC2004

 

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//超级逆天的迷宫算法，我真心服了！还是多看看数据结构这本书为好！/********************************************di所指的方位：0 上；1 右；2 下；3 左；********************************************/#include<iostream>using namespace std;const int MAX=100;//max要开大点，不然出现ACCESS_VIOLATION非法访问，即指针越界！！const int SIZE=8;typedef struct{    int i;//当前方块的行号    int j;//当前方块的列号    int di;//di记录下一可走方位的方位号}Stack;//栈定义Stack st[MAX];int mg[SIZE][SIZE];int length[MAX];int mgpath(int x1,int y1,int x2,int y2)//起点终点的坐标值{  int i,j,di,find,k=-1;  int top=-1;//初始化栈指针  top++;  st[top].i=x1; st[top].j=y1; st[top].di=-1; mg[x1][y1]=-1;//初始化，起点入栈  while(top>-1)//栈不为空时循环  {      i=st[top].i; j=st[top].j;       di=st[top].di;//绝妙一笔，记录下该方块的下一个可走方位，并且di只可以增，且小于4，可以保证不走曾经走过的路      if(i==x2 && j==y2)//找到了终点      {          k++;          length[k]=top;//记录下这条路径的长度          mg[x2][y2]=2;          top--;//出栈，重新寻找另一条路径          i=st[top].i; j=st[top].j; di=st[top].di;      }      find=0;      while(di<4 && find==0)//算法最妙的地方      {          di++;//先加加，指向下一个方向!          switch(di)          {              case 0: i=st[top].i-1; j=st[top].j; break;              case 1: i=st[top].i; j=st[top].j+1; break;              case 2: i=st[top].i+1; j=st[top].j; break;              case 3: i=st[top].i; j=st[top].j-1; break;          }          if(mg[i][j]==2) find=1;      }      if(find==1)//找到了下一个可走方块      {          st[top].di=di;//修改原栈顶元素的di值          top++;//下一个可走方块进栈          st[top].i=i; st[top].j=j; st[top].di=-1;          mg[i][j]=-1;//走过的方块标记为-1，避免重复走到该方块      }      else//没有路径可走，就退栈      {          mg[st[top].i][st[top].j]=2;//让该位置变为其他路径可走方块          top--;      }  }  return k+1;}int main(){    int m,n,time;    int i,j,k;    char str[SIZE];//字符串数组    int a,b,c,d;//记录起点终点的值    while(cin>>m>>n>>time && (m!=0 && n!=0 && time!=0))    {    //    cout<<"m="<<m<<"  "<<"n="<<n<<"   "<<"time="<<time<<endl;         for(i=0;i<m;i++)//m行         {            cin>>str;//输入字符串            for(k=0;k<n;k++)//n列            {                if(str[k]=='S')                     {                     a=i; b=k;                 }                if(str[k]=='D')                     {                     c=i; d=k;                 }                switch(str[k])                {                       case 'S' :                     case 'D':                    case  '.':  mg[i][k]=2;   break;                    case 'X' :   mg[i][k]=-1;  break;                    }            }         }         /*         for(i=0;i<m;i++)         {             for(j=0;j<n;j++)             cout<<mg[i][j]<<"   ";             cout<<endl;         }         */         /*         cout<<"a="<<a<<endl;         cout<<"b="<<b<<endl;         cout<<"c="<<c<<endl;         cout<<"d="<<d<<endl;         */        j=mgpath(a,b,c,d);        //cout<<"j="<<j<<endl;        for(i=0;i<j;i++)        {            //cout<<length[i]<<endl;            if(length[i]==time)             {             cout<<"YES"<<endl;             break;            }        }        if(i==j)         cout<<"NO"<<endl;    }    return 0;}