HDOJ--1010

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 95038    Accepted Submission(s): 25787

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

 

Sample Output

NO YES

 

走迷宫的一类题型，思路是先扫描迷宫的入口和出口，再进行dfs，对迷宫进行搜索，记得搜索完后一定要把迷宫还原。

注意思维的完整性：

特殊情况：

迷宫的入口就是出口！

以下是我的AC代码：

#include
#include
#define max 8
using namespace std;
 char map[max][max];
 int axisx[]={0,-1,0,1};
 int axisy[]={-1,0,1,0};
 int dx,dy;/*记录迷宫出去的门*/
 int t;
 int n,m;
 bool escape;
 void dfs(int sx,int sy,int cnt)
 {
     if(sx==dx && sy==dy && cnt==t)/*讨论特殊情况:门就是入口*/
     {
         escape=true;
         return;
     }
     int temp=(t-cnt)-abs(sx-dx)-abs(sy-dy);
     if(temp<0 || temp&1)/*奇偶剪枝*/
     {
         escape=false;
         return;
     }
     for(int i=0;i<4;++i)
     {
         if(sx+axisx[i]<0 || sx+axisx[i]>n-1 || sy+axisy[i]<0 || sy+axisy[i]>m-1)continue;
         if(map[sx+axisx[i]][sy+axisy[i]]!='X')
         {
             map[sx+axisx[i]][sy+axisy[i]]='X';
             dfs(sx+axisx[i],sy+axisy[i],cnt+1);
             map[sx+axisx[i]][sy+axisy[i]]='.';/*此处要还原迷宫*/
         }
         if(escape)
         break;
     }
 }
 int main()
 {
     while(cin>>n>>m>>t)
     {
         if(n==0&&m==0&&t==0)break;
         escape=false;
         int sx,sy;                          /*记录迷宫的起始位置：sx为行，sy为列*/
         int wallnum=0;
         for(int i1=0;i1>map[i1][i2];
                 if(map[i1][i2]=='S')
                 {
                     sx=i1;sy=i2;map[sx][sy]='X';
                 }
                 else if(map[i1][i2]=='D')
                 {
                     dx=i1;dy=i2;
                  } 
                  else if(map[i1][i2]=='X')
                  wallnum++;
             }
         }
         if(!(n*m-wallnum