Tempter of the Bone
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 54670 Accepted Submission(s): 14740
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.
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.
'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
题意；能否在规定的步数恰好到D点。奇偶剪枝：到D点最少步数与剩余步数奇偶性相同。
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 int R,C,T,door_x,door_y,dog_x,dog_y; 7 char maze[10][10]; 8 int vis[10][10]; 9 bool Go; 10 void dfs(int x,int y,int step) 11 { 12 if(Go == true)return ; 13 if(maze[x][y] == 'D' && step == T){Go = true;return;} 14 int flag = T-step -fabs(door_x-x) - fabs(door_y-y); 15 if(step > T || flag&1 || flag < 0)return ; 16 if(x>=0 && x<R && y>=0 && y<C && maze[x][y] != 'X' && !vis[x][y]){ 17 vis[x][y] = 1; 18 dfs(x+1,y,step+1); 19 dfs(x-1,y,step+1); 20 dfs(x,y-1,step+1); 21 dfs(x,y+1,step+1); 22 vis[x][y] = 0; 23 } 24 return; 25 } 26 int main() 27 { 28 //freopen("in.txt","r",stdin); 29 int i,j,sum; 30 while(~scanf("%d%d%d",&R,&C,&T),(R||C||T)) 31 { 32 sum = 0; 33 for(i=0; i<R; i++){ 34 for(j=0; j<C; j++){ 35 cin>>maze[i][j]; 36 if(maze[i][j]=='S'){dog_x = i,dog_y = j;} 37 if(maze[i][j]=='D'){door_x = i;door_y = j;} 38 if(maze[i][j]=='.'){sum++;} 39 } 40 } 41 Go = false; 42 memset(vis,0,sizeof(vis)); 43 int odd = fabs(door_x-dog_x)+fabs(door_y-dog_y); 44 if( T >= odd && sum+1 >= T && !((T - odd)%2) ) //到D点最短时间大于T 45 dfs(dog_x,dog_y,0); 46 if(Go) 47 puts("YES"); 48 else 49 puts("NO"); 50 } 51 return 0; 52 }