Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58940 Accepted Submission(s): 16038

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
题目链接点击打开链接

这道题的意思是小狗从S点出发走到D点，每次走一步而且只能是相邻的，X代表不可走，.代表可走，
问在T步内可不可以到达终点D，这道题可以用深搜解决，不过深搜很容易超时，要加上一定的剪枝
才可以不说了，看代码解释
#include
   
    
#include
    
      #include
     
       #include
      
        using namespace std; char map[10][10]; int dir[4][2]={{-1,0},{1,0},{0,1},{0,-1}}; int s1,s2,e1,e2,N,M,T,flag; int dfs(int a,int b,int step1) { if(step1==T&&a==e1&&b==e2)//判断是否到达终点，是否步数为T，是就返回递归 return 1; if(T-step1
       
        >map[i][j];//如果用scanf会发现超时，应该是输入的问题 if(map[i][j]=='S')//找到开始的坐标跟结束的坐标 { s1=i; s2=j; } if(map[i][j]=='D') { e1=i; e2=j; } } } map[s1][s2] = 'X'; if(dfs(s1,s2,0)) printf("YES\n"); else printf("NO\n"); } return 0; }

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Tempter of the Bone