Tempter of the Bone
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 129140 Accepted Submission(s):
34882
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：走迷宫 看你是不是可以走出去
直接dfs bfs 都可以过的
我使用的是dfs
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 using namespace std; 7 int h[4][2]={1,0,0,1,-1,0,0,-1}; 8 char mp[10][10]; 9 int vis[10][10]; 10 int lag,t; 11 int n,m; 12 void dfs(int i,int j,int time) 13 { 14 15 if(lag||i<0||i>n||j<0||j>m||time>t) 16 return ; 17 for(int k=0;k<4;k++) 18 { 19 int di=i+h[k][0]; 20 int dj=j+h[k][1]; 21 if(vis[di][dj]==0&&mp[di][dj]=='.') 22 { 23 vis[di][dj]=1; 24 dfs(di,dj,time+1); 25 vis[di][dj]=0; 26 } 27 if(vis[di][dj]==0&&mp[di][dj]=='D'&&time==t) 28 { 29 lag=1; 30 return ; 31 } 32 } 33 } 34 int main() 35 { 36 37 while(~scanf("%d%d%d",&n,&m,&t)&&n&&m&&t) 38 { 39 memset(vis,0,sizeof(vis)); 40 lag=0; 41 //getchar(); 42 for(int i=0; i<n; i++) 43 { 44 getchar(); 45 scanf("%s",mp[i]); 46 } 47 int x,y,dx,dy; 48 for(int i=0; i<n; i++) 49 { 50 for(int j=0; j<m; j++) 51 { 52 if(mp[i][j]=='S') 53 { 54 x=i; 55 y=j; 56 } 57 if(mp[i][j]=='D') 58 { 59 dx=i; 60 dy=j; 61 } 62 } 63 } 64 if(abs(dx-x)+abs(dy-y)>t||(abs(dx-x)+abs(dy-y)+t)%2); 65 else{ 66 vis[x][y]=1; 67 dfs(x,y,1); 68 } 69 70 if(!lag) 71 printf("NO "); 72 else 73 printf("YES "); 74 } 75 return 0; 76 }